Arithmetic Progressions Class 10 Previous Year Questions: 1 Mark (Chapter 5)

1-mark questions on Arithmetic Progressions

This page has 32 1-mark questions on Arithmetic Progressions from CBSE Class 10 board papers of 2023, 2024, 2025 and 2026. Of these, 23 appeared only in Standard papers and 9 appeared only in Basic papers. Question formats: 32 MCQs. 11 of them were asked in two or more paper sets.

0 of 32 practised

All 32 Arithmetic Progressions 1-mark questions

Q1. If the sum of the first \(n\) terms of an A.P be \(3n^2 + n\) and its common difference is 6, then its first term is

Mark practised

(A) 2
(B) 3
(C) 1
(D) 4

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2023
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2023 Mathematics Standard, Set 30/6/2, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 4, because the first term is S₁ = 3(1)² + 1 = 4.
Why? In an AP, the sum of the first one term, S₁, is exactly the first term a, so substituting 1 for n gives a.

The full 1 mark is for the correct option.

Answer:

(D) 4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/2: option (D).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing the coefficient of n² as the first term without using S₁.
Treating 3n² + n as the nth term instead of the sum; the first term is still S₁.
Putting n as zero in the given expression and taking the result as the first term.

Why this works

The first term of an AP is S₁, the sum of the first one term. So a = S₁ = 3(1)² + 1 = 4. Check: with first term 4 and the given common difference, the second term is 4 + 6 = 10, so S₂ = 4 + 10 = 14; the given formula also gives 3(2)² + 2 = 14. Another correct method also earns full marks.

Key idea to remember

Sum of First n Terms of an AP: see Key ideas used on this page.

Explanation

The sum of the first n terms of an AP with first term a and common difference d is S = n/2[2a + (n − 1)d]. Substituting 1 for n makes the sum contain only the first term, so S₁ = a. Therefore the first term is found by substituting 1 for n in the given expression.

Important formulas

S = n/2[2a + (n − 1)d] · Sum of the first n terms of an AP with first term a and common difference d.
S₁ = a · The sum of the first one term is the first term.

Q2. The common difference of an A.P., if \(a_{23} - a_{19} = 32\) , is :

Mark practised

(A) 8
(B) - 8
(C) - 4
(D) 4

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Mathematics Basic, Set 430/4/1, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 8, because a₂₃ − a₁₉ = 4d = 32, so d = 32 ÷ 4 = 8.
Why? By the nth term formula, aₙ = a + (n − 1)d, so a₂₃ − a₁₉ = 4d. Equating 4d to 32 gives d = 8.

The full 1 mark is for the correct option.

Answer:

(A) 8

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/4/1: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking 32 as the common difference itself, without dividing by 4.
Choosing −8 because the 23rd term is later than the 19th term, without noticing that a₂₃ − a₁₉ is positive 4d, so d must be positive.

Why this works

Write the two terms using aₙ = a + (n − 1)d: a₂₃ = a + 22d and a₁₉ = a + 18d. Subtracting gives a₂₃ − a₁₉ = 4d. The question gives this difference as 32, so 4d = 32, and d = 8. Quick check: from the 19th term to the 23rd term there are four common differences, and 4 × 8 = 32, which matches. The other options give −32, −16 and 16, not 32.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

For an AP with first term a and common difference d, the nth term is aₙ = a + (n − 1)d. This question uses the formula for two terms and subtracts them, which cancels a and leaves only a multiple of d, so d can be found directly.

Important formulas

aₙ = a + (n − 1)d · The nth term of an AP with first term a and common difference d.
aₘ − aₙ = (m − n)d · Follows from the nth term formula; subtracting two terms removes the first term.

Q3. Which term of the A.P. -29, -26, -23, ..., 61 is 16 ?

Mark practised

(A) 11\(^{th}\)
(B) 16\(^{th}\)
(C) 10\(^{th}\)
(D) 31\(^{st}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Mathematics Standard, Set 30/2/1, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 16th, because the AP has first term −29 and common difference 3, so aₙ = −29 + (n − 1) × 3 = 16 gives n = 16.
Why? The nth term formula aₙ = a + (n − 1)d links the value of a term to its position, so solving aₙ = 16 gives the required position n.

The full 1 mark is for the correct option.

Answer:

(B) 16th

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/2/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Picking 16 as the answer because the question says 'is 16', without solving aₙ = 16.
Forgetting to add 1 after dividing 45 by 3, which gives 15 instead of 16.
Taking the common difference as negative because the first terms are negative; here each term increases by 3, so the common difference is 3.

Q4. In an A.P., if \(d = -4\) and \(a_7 = 4\), then the first term 'a' is equal to

Mark practised

(A) 6
(B) 7
(C) 20
(D) 28

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Mathematics Basic, Set 430/2/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 28, because a₇ = a + 6d = a + 6(−4) = a − 24, so a = 4 + 24 = 28.
Why? By the nth term formula for an AP, substituting the given term and common difference leaves the first term as the only unknown.

The full 1 mark is for the correct option.

Answer:

(D) 28

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/2/1: option (D).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing the seventh term as a + 7d instead of a + 6d.
Reading the common difference as positive 4 because the sign is missed, so the first term comes out wrong.
Choosing 20 by adding 4 and 4 × 4 without applying the nth term formula.

Q5. The common difference of the A.P.
\(\frac{1}{2x}, \frac{1-4x}{2x}, \frac{1-8x}{2x}, \dots\) is :

Mark practised

(A) \(-2x\)
(B) \(-2\)
(C) \(2\)
(D) \(2x\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2024 Mathematics Standard, Set 30/3/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) -2, because \(\frac{1-4x}{2x} - \frac{1}{2x} = \frac{-4x}{2x} = -2\).
Why? In an AP, the common difference is the fixed number obtained by subtracting a term from its succeeding term; subtracting the first term from the second term gives this common difference.

The full 1 mark is for the correct option.

Answer:

(B) -2

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Subtracting in the wrong order, taking the first term minus the second term, which reverses the sign of the common difference.
Choosing an option that still contains x, without simplifying the ratio to a constant.

Why this works

An AP is a list of numbers in which each term is obtained by adding the same fixed number to the preceding term. That fixed number is the common difference. To find it, subtract any term from its succeeding term. Here the second term and the first term have the same denominator, so subtract the numerators and divide by that denominator; the x cancels and the common difference is a constant.

Key idea to remember

Common difference of an Arithmetic Progression (AP): see Key ideas used on this page.

Explanation

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term. This fixed number is called the common difference. The solution uses d = a_{k+1} - a_k, subtracting the first term from the second term.

Important formulas

d = a_{k+1} - a_k · The common difference is found by subtracting the kth term from the (k+1)th term.
a, a+d, a+2d, a+3d, ... · General form of an AP; d is the fixed number added each time.

Q6. The number of terms in the A.P. 3, 6, 9, 12, ..., 111 is :

Mark practised

(A) 36
(B) 40
(C) 37
(D) 30

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Mathematics Standard, Set 30/5/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 37, because (111 − 3) ÷ 3 + 1 = 37.
Why? The nth term of an AP is aₙ = a + (n − 1)d; solving for n when the last term is known gives the number of terms.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using 111 ÷ 3 + 1 as the number of terms; since 111 ÷ 3 = 37, 111 is the 37th term, so adding 1 counts one extra.
Using the sum formula S = n/2[2a + (n − 1)d] when the question only asks for the number of terms.
Taking the common difference as 6 because the terms are multiples of 3; the common difference is the fixed number added, 6 − 3 = 3.

Q7. The next (4\(^{th}\)) term of the A.P. \(\sqrt{18}\) , \(\sqrt{50}\) , \(\sqrt{98}\) , ... is :

Mark practised

(A) \(\sqrt{128}\)
(B) \(\sqrt{140}\)
(C) \(\sqrt{162}\)
(D) \(\sqrt{200}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2024 Mathematics Standard, Set 30/5/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) √162, because √18 = 3√2, √50 = 5√2, √98 = 7√2, so the 4th term is 7√2 + 2√2 = 9√2 = √162.
Why? In an AP, each term is obtained by adding the same fixed number d to the preceding term; the nth term is aₙ = a + (n − 1)d.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Finding the common difference by subtracting the radicands, 50 − 18 = 32 and 98 − 50 = 48, and concluding the terms are not in AP.
Adding 2√2 to √98 as if the root covers the whole sum, so reaching √100 instead of combining 7√2 and 2√2 to get 9√2.
Forgetting to square the 9 when converting 9√2 to a single surd, so writing √18 instead of √162.

Why this works

Write each surd as a multiple of √2: √18 = √(9 × 2) = 3√2, √50 = √(25 × 2) = 5√2, √98 = √(49 × 2) = 7√2. So the AP is 3√2, 5√2, 7√2, ... . The common difference d is 5√2 − 3√2 = 2√2 (also 7√2 − 5√2 = 2√2). Hence the 4th term is 7√2 + 2√2 = 9√2, or by the formula a₄ = a + 3d, the 4th term is 3√2 + 3(2√2) = 9√2. Since 9√2 = √(9² × 2) = √162, the required term is √162. A quick check: the given terms are about 4.24, 7.07, 9.90, rising by about 2.83 each time; the next term is about 12.73, which is √162.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

The nth term of an AP with first term a and common difference d is given by aₙ = a + (n − 1)d. Here the AP is rewritten as 3√2, 5√2, 7√2, so the first term is 3√2 and the common difference is 2√2; hence the 4th term is 3√2 + 3(2√2), which is 9√2, i.e. √162.

Important formulas

aₙ = a + (n − 1)d · nth term of an AP with first term a and common difference d.
d = aₖ₊₁ − aₖ · The common difference of an AP is the difference between any term and its preceding term.

Q8. If the first term of an A.P. is 'a' and its common difference is 'b', then its 10\(^{th}\) term is :

Mark practised

(A) \(a + 10b\)
(B) \(10a + b\)
(C) \(a + 9b\)
(D) \(9a + b\)

Marks: 1
Level: Basic
Asked in: 2 paper sets
Compartment: 2025
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2025 Compartment Mathematics Basic, Set 430/S/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) a + 9b, because a_n = a + (n − 1)d, so a_10 = a + (10 − 1)b = a + 9b.
Why? The nth term of an AP is obtained by adding the common difference once less than n times to the first term.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 430/S/1: option (C).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Putting n = 10 as a + 10b, which actually gives the 11th term because the formula has (n − 1).
Choosing 10a + b by multiplying the first term by the term number.
Using n = 11 for the 10th term, because the first term is counted as the 1st term.

Q9. The 6\(^{th}\) term of the AP \(\sqrt{27}\) , \(\sqrt{75}\) , \(\sqrt{147}\) , ... is :

Mark practised

(A) \(\sqrt{243}\)
(B) \(\sqrt{363}\)
(C) \(\sqrt{300}\)
(D) \(\sqrt{507}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Compartment: 2025
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2025 Compartment Mathematics Standard, Set 30/S/2, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) √507, because √27 = 3√3, √75 = 5√3, √147 = 7√3, so a = 3√3, d = 2√3 and the 6th term is 3√3 + 5 × 2√3 = 13√3 = √507.
Why? By the definition of an AP, the nth term is the first term plus (n − 1) times the common difference, and each square-root term must be simplified before comparing terms.

The full 1 mark is for the correct option.

Answer:

(D) √507

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 30/S/2: option (D).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Using n = 5 for the 6th term; that gives 11√3, but the 6th term needs n = 6.
Finding d by subtracting the square roots before simplifying, as if √75 − √27 could be combined directly; simplify first to get 5√3 − 3√3 = 2√3.
Stopping at 13√3 without squaring it to see that (13√3)² = 507, which is option D.

Q10. If \(n^{\text{th}}\) term of an A.P. is \(5n - 6\), then its common difference is :

Mark practised

(A) \(-6\)
(B) \(5n\)
(C) 5
(D) 6

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025
Topic: 5.3 nth Term of an AP
Source: (CBSE 2025 Mathematics Basic, Set 430/4/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 5, because a₂ − a₁ = (5 × 2 − 6) − (5 × 1 − 6) = 5.
Why? The common difference of an AP is the fixed number obtained by subtracting any term from its succeeding term, d = aₖ₊₁ − aₖ.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/1: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 6 because the constant term is −6, even though the common difference is the coefficient of n.
Choosing −6 because the minus sign is attached to 6 in 5n − 6.
Choosing 5n because 5n appears in the nth term, instead of picking only its coefficient 5.

Why this works

The expression 5n − 6 gives the nth term, so substituting the first two counting numbers gives two consecutive terms of the AP. Their difference is the common difference. Also, the nth term formula aₙ = a + (n − 1)d can be rewritten as aₙ = dn + (a − d), so the number multiplying n is always d. Here that number is 5. Quick check: a₁ = 5(1) − 6 = −1, a₂ = 5(2) − 6 = 4, a₃ = 5(3) − 6 = 9, and the differences are 4 − (−1) = 5 and 9 − 4 = 5.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

The nth term of an AP with first term a and common difference d is aₙ = a + (n − 1)d. Since aₙ = 5n − 6 has the same form, the coefficient of n gives the common difference. Equivalently, subtracting consecutive terms also gives d.

Important formulas

aₙ = a + (n − 1)d · nth term of an AP, where a is the first term and d is the common difference.
d = aₖ₊₁ − aₖ · Common difference is the fixed difference between consecutive terms.

Q11. The common difference of the AP : \(\sqrt{2}\), \(2\sqrt{2}\), \(3\sqrt{2}\), \(4\sqrt{2}\), ... is :

Mark practised

(A) \(\sqrt{2}\)
(B) 1
(C) \(2\sqrt{2}\)
(D) \(-\sqrt{2}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2026
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2026 Mathematics Standard, Set 30/1/1, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) √2, because the common difference is a₂ − a₁ = 2√2 − √2 = √2.
Why? By the definition of an AP, the common difference is d = a_{k+1} − a_k for any two consecutive terms, so subtracting the first term from the second gives d.

The full 1 mark is for the correct option.

Answer:

(A) √2

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/1/1: option (A).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 2√2, which is the second term of the AP, not the common difference.
Choosing 1, because the coefficient of √2 is 1, but the common difference is √2 itself.
Choosing −√2, from subtracting the second term from the first; the common difference is a₂ − a₁, not a₁ − a₂.

Why this works

The given list is √2, 2√2, 3√2, 4√2, ... . Each term is obtained by adding √2 to the previous term: √2 + √2 = 2√2, 2√2 + √2 = 3√2, 3√2 + √2 = 4√2. So the fixed number added, the common difference, is √2. Quick check: subtract the first term from the second: 2√2 − √2 = √2; subtract the second from the third: 3√2 − 2√2 = √2. The same value appears every time. Option B is 1, which ignores the √2 factor; option C is the second term itself; option D is the reverse subtraction, which is wrong because d is found by subtracting the kth term from the (k+1)th term, not the other way round.

Key idea to remember

Common difference: see Key ideas used on this page.

Explanation

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term. That fixed number is called the common difference, denoted by d. Here d = a₂ − a₁ = √2.

Important formulas

d = a_{k+1} − a_k · The common difference of an AP is found by subtracting the kth term from the (k+1)th term.

Q12. If \(x, 2x + 9, 4x + 3\) are three consecutive terms of an A.P., then the value of \(x\) is :

Mark practised

(A) 3
(B) 10
(C) 13
(D) 15

Marks: 1
Level: Standard
Compartment: 2023
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/1, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 15, because (2x + 9) − x = (4x + 3) − (2x + 9), so x + 9 = 2x − 6, giving x = 15.
Why? In an AP, the difference between any two consecutive terms is the same fixed number, called the common difference.

The full 1 mark is for the correct option.

Answer:

(D) 15

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/1: option (D).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Equating (4x + 3) − x with 2x + 9 instead of using the two consecutive differences; this leads to the wrong value 6.
Expanding (4x + 3) − (2x + 9) as 2x + 12 instead of 2x − 6, because the 9 is not subtracted.
Changing −6 to +6 while solving x + 9 = 2x − 6, which gives the wrong value 3.

Why this works

Three consecutive terms of an AP have the same common difference. Subtract the first term from the second and the second from the third, then equate: (2x + 9) − x = (4x + 3) − (2x + 9). This gives x + 9 = 2x − 6, so x = 15. Check: putting x = 15, the terms are 15, 39 and 63, and 39 − 15 = 24, 63 − 39 = 24. Since the two differences are equal, the terms are in AP. Another correct method also earns full marks: use the arithmetic mean, 2x + 9 = (x + 4x + 3)/2.

Key idea to remember

Common difference: see Key ideas used on this page.

Explanation

An AP is a list of numbers in which each term after the first is obtained by adding a fixed number d to the preceding term. So for any three consecutive terms, the second term minus the first term equals the third term minus the second term. The solution uses this equality to form a linear equation in x and solve it.

Important formulas

a_{k+1} − a_k = d · In an AP, every pair of consecutive terms has the same difference d.
b = (a + c)/2 · If a, b, c are in AP, then b is the arithmetic mean of a and c.

Q13. If 2x, x + 10, 3x + 2 are three consecutive terms of an A.P., then the value of x is :

Mark practised

(A) 4
(B) 5
(C) 6
(D) 8

Marks: 1
Level: Standard
Compartment: 2023
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/2, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 6, because (x + 10) − 2x = (3x + 2) − (x + 10), so 10 − x = 2x − 8, giving x = 6.
Why? In an AP, the common difference is the same for every pair of consecutive terms, so the two differences must be equal.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/2: option (C).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Simplifying (x + 10) − 2x as 10 + x instead of 10 − x, which flips the sign of x.
From 10 − x = 2x − 8, moving −x to the other side without changing its sign and then getting a wrong value for x.

Why this works

An AP is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term. So the difference between the second and first terms must equal the difference between the third and second terms. Here, (x + 10) − 2x = 10 − x and (3x + 2) − (x + 10) = 2x − 8. Equating these gives 10 − x = 2x − 8. Add x and 8 to both sides: 18 = 3x, so x = 6. Check: for x = 6 the terms are 12, 16, 20, and 16 − 12 = 4 and 20 − 16 = 4, so the common difference is 4.

Key idea to remember

Common difference of an AP: see Key ideas used on this page.

Explanation

An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference. For consecutive terms, a_{k+1} − a_k = d. Since 2x, x + 10 and 3x + 2 are three consecutive terms of the same AP, the two differences must be equal. Equating them forms a linear equation in x, and solving it gives x = 6.

Important formulas

a_{k+1} - a_k = d · In an AP, every pair of consecutive terms has the same difference d.
a, a + d, a + 2d, ... · The general form of an AP; the three given terms are consecutive terms of one AP.

Q14. If \(x + 1\), \(3x\) and \(4x + 2\) are three consecutive terms of an A.P., then the value of \(x\) is :

Mark practised

(A) 2
(B) 3
(C) 4
(D) 5

Marks: 1
Level: Standard
Compartment: 2023
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/3, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 3, because in an AP, 3x − (x + 1) = (4x + 2) − 3x, so 2x − 1 = x + 2 and x = 3.
Why? In an AP, the common difference is the same for every pair of consecutive terms, so the difference between the 2nd and 1st terms equals the difference between the 3rd and 2nd terms.

The full 1 mark is for the correct option.

Answer:

(B) 3

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/3: option (B).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Treating the middle term as the sum of the outer terms instead of their arithmetic mean; this gives x = −1.5, which is not one of the options.
Substituting x = 2 and seeing the terms 3, 6, 10 all increase, without checking that 6 − 3 = 3 but 10 − 6 = 4.
Transposing −1 to the right side with the wrong sign, and getting x = 1 instead of 3.

Why this works

For three consecutive terms of an AP, the same fixed number is added each time. So the jump from the first term to the second must equal the jump from the second to the third. Write this equality: 3x − (x + 1) = (4x + 2) − 3x. Simplify: the left side is 3x − x − 1 = 2x − 1, and the right side is 4x + 2 − 3x = x + 2. So 2x − 1 = x + 2, which gives x = 3. Check: with x = 3, the terms are 4, 9, 14, and the differences are 9 − 4 = 5 and 14 − 9 = 5, so the terms are in AP. A quick exam check is to substitute each option into the equality; only x = 3 makes both sides equal.

Key idea to remember

Arithmetic Progression (AP): see Key ideas used on this page.

Explanation

An AP is a list of numbers in which each term is obtained by adding a fixed number to the preceding term. This fixed number is called the common difference. For three consecutive terms, the difference between the second and first terms must equal the difference between the third and second terms.

Important formulas

a2 − a1 = a3 − a2 · The condition used when three consecutive terms of an AP are given.

Q15. The seventh term of an A.P. whose first term is 28 and common difference \(-4\) , is

Mark practised

(A) 0
(B) 4
(C) 52
(D) 56

Marks: 1
Level: Basic
Years: 2023
Topic: 5.3 nth Term of an AP
Source: (CBSE 2023 Mathematics Basic, Set 430/1/1, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 4, because a₇ = 28 + (7 − 1)(−4) = 28 − 24 = 4.
Why? By the formula for the nth term of an AP, aₙ = a + (n − 1)d; substituting the given first term, common difference and n gives this term.

The full 1 mark is for the correct option.

Answer:

(B) 4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using 7 steps instead of 6 and getting 0; the 7th term is reached after (7 − 1) = 6 additions of the common difference.
Taking the common difference as +4 and getting 52, although the given d is −4.

Q16. The seventh term of an A.P. whose first term is 28 and common difference – 4, is

Mark practised

(A) 0
(B) 4
(C) 52
(D) 56

Marks: 1
Level: Basic
Years: 2023
Topic: 5.3 nth Term of an AP
Source: (CBSE 2023 Mathematics Basic, Set 430/1/2, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 4, because a₇ = 28 + (7 − 1)(−4) = 28 − 24 = 4.
Why? By the nth term formula aₙ = a + (n − 1)d, the 7th term is the first term plus six times the common difference.

The full 1 mark is for the correct option.

Answer:

(B) 4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/2: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 0 by subtracting 7 lots of 4 from 28, although the 7th term needs only 6 subtractions from the first term.
Choosing 52 by adding 6 × 4 to 28, ignoring that the common difference is −4.
Choosing 56 by adding 7 × 4 to 28, ignoring both the minus sign and using 7 instead of 6.

Q17. If \(a, b, c\) form an A.P. with common difference \(d\), then the value of \(a - 2b - c\) is equal to

Mark practised

(A) \(2a + 4d\)
(B) 0
(C) \(-2a - 4d\)
(D) \(-2a - 3d\)

Marks: 1
Level: Standard
Years: 2023
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2023 Mathematics Standard, Set 30/1/2, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) −2a − 4d, because in an AP, b = a + d and c = a + 2d, so a − 2(a + d) − (a + 2d) = −2a − 4d.
Why? In an AP, each term after the first is obtained by adding the common difference d to the preceding term, so b = a + d and c = a + 2d; substituting them makes the expression simplify.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/2: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking c = a + d instead of c = a + 2d, which leads to option D, −2a − 3d.
Expanding −(a + 2d) as −a + 2d (leaving the sign of 2d unchanged), so the d terms cancel and the result is −2a instead of −2a − 4d.
Substituting b = a + d but forgetting the minus sign in front of 2b, giving a + 2(a + d) − (a + 2d) = 2a.

Why this works

Since a, b, c are consecutive terms of an AP with common difference d, b = a + d and c = a + 2d. Substituting: a − 2(a + d) − (a + 2d) = a − 2a − 2d − a − 2d = −2a − 4d. Quick check: with a as 1 and d as 2, the AP is 1, 3, 5; a − 2b − c = 1 − 6 − 5 = −10, and −2a − 4d = −2 − 8 = −10. Another correct method also earns full marks.

Key idea to remember

General form of an AP: see Key ideas used on this page.

Explanation

An AP with first term a and common difference d is written as a, a + d, a + 2d, a + 3d, ... . Here b is the second term and c is the third term, so b = a + d and c = a + 2d.

Important formulas

a, a + d, a + 2d, a + 3d, ... · General form of an AP with first term a and common difference d.
a_n = a + (n − 1)d · nth term of the AP; used here for the second and third terms.

Q18. The 8\(^{th}\) term of an A.P., whose first two terms are - 5 and 2 respectively, is :

Mark practised

(A) - 54
(B) 44
(C) - 61
(D) - 33

Marks: 1
Level: Basic
Compartment: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430(B), Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 44, because d = 2 − (−5) = 7 and a₈ = −5 + 7 × 7 = 44.
Why? In an AP, the common difference is found by subtracting a term from its succeeding term, and the nth term is a + (n − 1)d.

The full 1 mark is for the correct option.

Answer:

(B) 44

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430(B): option (B).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking d as −7 (i.e. −5 − 2 instead of 2 − (−5)) would make a₈ = −5 + 7 × (−7) = −54, option A.
Using the first term as 2 and the common difference as −5 instead of −5 and 7 would make a₈ = 2 + 7 × (−5) = −33, option D.
Putting 8 in place of the 7 jumps, that is using −5 + 8 × 7 = 51, gives a number not among the options; the formula uses (n − 1)d, so only 7 jumps are added.

Why this works

The AP has first term a = −5. The second term is obtained by adding the common difference d, so d = 2 − (−5) = 7. The 8th term is the first term plus 7 common differences: a₈ = −5 + 7 × 7 = −5 + 49 = 44. This matches option B. A quick check: adding 7 repeatedly gives −5, 2, 9, 16, 23, 30, 37, 44, so the 8th term is 44.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term. That fixed number is the common difference d. Once the first term a and d are known, the nth term is a + (n − 1)d. Here the first two terms give the first term −5 and the common difference 7, so the 8th term is found by adding 7 seven times to −5.

Important formulas

a_n = a + (n − 1)d · The nth term of an AP with first term a and common difference d.
d = a_{k+1} − a_k · Common difference is a term minus the term before it.

Q19. The 30th term of the A.P. -3, -7, -11, ... is :

Mark practised

(A) 113
(B) -117
(C) -119
(D) -120

Marks: 1
Level: Basic
Years: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Mathematics Basic, Set 430/3/2, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) −119, because a₃₀ = −3 + (30 − 1)(−4) = −119.
Why? The nth term formula of an AP counts the common difference n − 1 times; applying it to the 30th term gives the term asked for.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/3/2: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using 30 differences instead of 29, so the calculation overshoots the correct value by 4.
Subtracting in the wrong order, for example taking the common difference as −3 − (−7), which is 4, instead of −7 − (−3), which is −4.
Multiplying 30 by 4 and ignoring the first term −3, then choosing −120.

Q20. If the sum of first \(n\) terms of an A.P. is \(3n^2 + 4n\) and its common difference is 6, then its first term is :

Mark practised

(A) 7
(B) 4
(C) 6
(D) 3

Marks: 1
Level: Standard
Years: 2024
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2024 Mathematics Standard, Set 30/3/2, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 7, because S₁ = 3(1)² + 4(1) = 7 and S₁ is the first term.
Why? The sum of the first 1 term of an AP is the first term itself, so S₁ gives the first term directly.

The full 1 mark is for the correct option.

Answer:

(A) 7

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/2: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using the given common difference 6 as the first term; d is the fixed difference between consecutive terms, not the first term.
Putting 2 in place of n and treating S₂, the sum of the first two terms, as the first term.
Choosing 3 because it is the coefficient of n² in 3n² + 4n; the first term is S₁, not a coefficient.

Why this works

The symbol Sₙ stands for the sum of the first n terms of the AP. When n is 1, S₁ is just the first term. So put 1 for n in the given Sₙ = 3n² + 4n: S₁ = 3(1)² + 4(1) = 7. Therefore the first term is 7. The common difference 6 is not needed for this step. As a check, an AP with first term 7 and common difference 6 has Sₙ = n/2[2(7) + (n − 1)(6)] = n/2[6n + 8] = 3n² + 4n, which matches the question.

Key idea to remember

Sum of First n Terms of an AP: see Key ideas used on this page.

Explanation

The sum of the first n terms of an AP is written as Sₙ. Since S₁ is the sum of the first 1 term only, S₁ equals the first term. Putting 1 for n in the given sum formula therefore gives the first term directly.

Important formulas

Sₙ = n/2 [2a + (n − 1)d] · Sum of the first n terms of an AP, where a is the first term and d is the common difference.
S₁ = a · The sum of the first 1 term is the first term itself.

Q21. Three numbers in A.P. have the sum 30. What is its middle term ?

Mark practised

(A) 4
(B) 10
(C) 16
(D) 8

Marks: 1
Level: Standard
Years: 2024
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2024 Mathematics Standard, Set 30/4/1, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 10, because the middle term of an AP is the average of the three numbers, so 30 ÷ 3 = 10.
Why? In an AP, the middle term is the arithmetic mean of the other two terms, so the sum of the three terms is three times the middle term.

The full 1 mark is for the correct option.

Answer:

(B) 10

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/4/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 16 because the options 4, 10, 16 add to 30 and look like an AP, but 16 is the third term, not the middle term.
Dividing 30 by 3 to get 10, then subtracting 2 because you assume the common difference is 2 from the options; the common difference is not given, so the middle term stays 10.

Why this works

Think of three numbers in AP: they are equally spaced on the number line. The middle term sits exactly halfway between the first and third terms, so it is their average. Consequently, the sum of the three terms is three times the middle term. The sum is 30, so the middle term is 30 ÷ 3 = 10. The option is (B).

Key idea to remember

Arithmetic mean of three numbers in AP: see Key ideas used on this page.

Explanation

If a, b, c are in AP, then b = (a + c)/2. That is, the middle term is the average of the other two. Substituting this into the sum gives a + b + c = 3b. So the middle term is one-third of the sum of the three terms.

Important formulas

b = (a + c)/2 · For three numbers a, b, c in AP, b is the arithmetic mean of a and c.
a + b + c = 3b · Because a + c = 2b, the sum of the three terms equals three times the middle term.

Q22. The 14th term from the end of the A.P. - 11, - 8, - 5, ..., 49 is :

Mark practised

(A) 7
(B) 10
(C) 13
(D) 28

Marks: 1
Level: Standard
Years: 2024
Topic: 5.3 nth Term of an AP
Source: (CBSE 2024 Mathematics Standard, Set 30/5/3, Q9, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 10, because the AP has 21 terms, so the 14th term from the end is the 8th term, and a₈ = −11 + 7 × 3 = 10.
Why? In a finite AP with n terms, the rth term from the end is the (n − r + 1)th term from the beginning.

The full 1 mark is for the correct option.

Answer:

(B) 10

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/3: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Finding the 14th term from the beginning, −11 + 13 × 3 = 28, instead of from the end.
Counting the 14th term from the end as the 7th term (21 − 14) instead of the 8th term (21 − 14 + 1).
Reversing the AP to 49, 46, 43, ... but then using +3 as the common difference; the reversed AP has common difference −3.

Why this works

The AP has first term −11 and common difference −8 − (−11) = 3. The last term is 49. Using the formula aₙ = a + (n − 1)d with the last term 49 gives n = 21, so there are 21 terms. The 14th term from the end is the (21 − 14 + 1)th = 8th term from the beginning. Then a₈ = −11 + (8 − 1) × 3 = −11 + 21 = 10. Quick check: counting back from the last term, the 14th term from the end is 49 − 13 × 3 = 10. Another correct method also earns full marks.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

The nth term of an AP with first term a and common difference d is aₙ = a + (n − 1)d. In a finite AP with n terms, the rth term from the end is the (n − r + 1)th term from the beginning. Here the AP has 21 terms, so the 14th term from the end is the 8th term from the beginning, and a₈ = −11 + 7 × 3 = 10.

Important formulas

a_n = a + (n - 1)d · nth term of an AP.
rth term from the end = (n − r + 1)th term from the beginning · For a finite AP with n terms.

Q23. The next term of the A.P. \(\sqrt{7}\) , \(\sqrt{28}\) , \(\sqrt{63}\) , ... is :

Mark practised

(A) \(\sqrt{81}\)
(B) \(\sqrt{126}\)
(C) \(\sqrt{112}\)
(D) \(5\sqrt{7}\)

Marks: 1
Level: Basic
Compartment: 2025
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2025 Compartment Mathematics Basic, Set 430/S/3, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) √112, because the terms simplify to √7, 2√7, 3√7, so the next term is 4√7 = √112.
Why? In an AP, every term is obtained by adding the same fixed number, the common difference, to the preceding term. The common difference here is √7, so the fourth term is the third term plus √7.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 430/S/3: option (C).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing the common difference as √28 − √7 and then as √21, which wrongly treats √a − √b as √(a−b); simplify first: √28 = 2√7.
Choosing option D, 5√7, because the terms look like √7, 2√7, 3√7 and the next multiple seems to be 5√7; the next term is actually 4√7.
Adding √63 and √28 directly without simplifying, which gives a wrong surd instead of √112.

Why this works

Write each term as a multiple of √7: √28 = √(4×7) = 2√7 and √63 = √(9×7) = 3√7. So the AP is √7, 2√7, 3√7, ... The common difference is 2√7 − √7 = √7, and also 3√7 − 2√7 = √7. Add the common difference to the third term: 3√7 + √7 = 4√7 = √(16×7) = √112. Option D, 5√7, would be the term after the next one, not this next term.

Key idea to remember

Arithmetic Progression: see Key ideas used on this page.

Explanation

An AP is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term. That fixed number is the common difference. Here the terms simplify to √7, 2√7, 3√7, so the common difference is √7 and the next term is 4√7 = √112.

Important formulas

d = a_{k+1} − a_k · The common difference of an AP is found by subtracting any term from its succeeding term.

Q24. If the sum of the first 'n' terms of an AP is \(2n^2 + 5n\), then its common difference is :

Mark practised

(A) 2
(B) 4
(C) 5
(D) 7

Marks: 1
Level: Standard
Compartment: 2025
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2025 Compartment Mathematics Standard, Set 30/S/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 4, because S₁ = 2(1)² + 5(1) = 7, S₂ = 2(2)² + 5(2) = 18, so d = S₂ − 2S₁ = 18 − 14 = 4.
Why? The first term is a₁ = S₁ and the second term is a₂ = S₂ − S₁, so the common difference is d = a₂ − a₁ = S₂ − 2S₁.

The full 1 mark is for the correct option.

Answer:

(B) 4

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 30/S/1: option (B).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 2 because 2 is the coefficient of n² in 2n² + 5n; the coefficient of n² is d/2, not d.
Choosing 5 because 5 is the coefficient of n in 2n² + 5n; that coefficient is a − d/2, not d.
Thinking the second term (11) is the common difference; the common difference is the difference between consecutive terms.

Why this works

Put the first two values of n into Sₙ = 2n² + 5n. S₁ = 2(1)² + 5(1) = 7 is the first term a₁. S₂ = 2(2)² + 5(2) = 18 is the sum of the first two terms, so the second term is a₂ = S₂ − S₁ = 18 − 7 = 11. Then d = a₂ − a₁ = 11 − 7 = 4. Quick check: the sum formula Sₙ = (n/2)[2a + (n − 1)d] expands to (d/2)n² + (a − d/2)n. Comparing with 2n² + 5n gives d/2 = 2, so d = 4. Another correct method also earns full marks.

Key idea to remember

Nth term of an AP is the difference of the sum to first n terms and the sum to first (n − 1) terms: see Key ideas used on this page.

Explanation

The sum Sₙ is the total of the first n terms. To get a single term, subtract the total before it: aₙ = Sₙ − Sₙ₋₁. Here a₁ = S₁ and a₂ = S₂ − S₁, and then d = a₂ − a₁.

Important formulas

Sₙ = (n/2)[2a + (n − 1)d] · Sum of the first n terms of an AP, where a is the first term and d is the common difference.
aₙ = Sₙ − Sₙ₋₁ · The nth term equals the sum of the first n terms minus the sum of the first (n − 1) terms.
d = aₖ₊₁ − aₖ · Common difference of an AP from any two consecutive terms.

Q25. Which term of the AP 7, 10, 13, ... is 52 ?

Mark practised

(A) 13\(^{th}\)
(B) 15\(^{th}\)
(C) 16\(^{th}\)
(D) 17\(^{th}\)

Marks: 1
Level: Standard
Compartment: 2025
Topic: 5.3 nth Term of an AP
Source: (CBSE 2025 Compartment Mathematics Standard, Set 30/S/2, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 16th, because 7 + (n − 1) × 3 = 52 gives n = 16.
Why? In an AP, the nth term is aₙ = a + (n − 1)d, so substituting a = 7, d = 3 and aₙ = 52 and solving for n gives the position of 52.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 30/S/2: option (C).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting that (52 − 7) ÷ 3 = 15 gives n − 1, not n, and writing the 15th term instead of the 16th.
Using a + nd instead of a + (n − 1)d; the position comes out one less, so the student may mark the 15th term instead of the 16th.

Why this works

An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term; this fixed number is the common difference. Here d = 10 − 7 = 3. Let 52 be the nth term. Using aₙ = a + (n − 1)d with a = 7 and d = 3, the equation is 7 + (n − 1) × 3 = 52. Then 3(n − 1) = 45, so n − 1 = 15 and n = 16. Therefore 52 is the 16th term. Quick check: the 16th term is 7 + (16 − 1) × 3 = 7 + 45 = 52.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

The nth term (general term) of an AP with first term a and common difference d is aₙ = a + (n − 1)d. To find which term equals 52, put 52 in place of aₙ and solve for n. Here n = 16, so 52 is the 16th term.

Important formulas

aₙ = a + (n − 1)d · nth term (general term) of an AP; put the given term value in place of aₙ and solve for n.

Q26. If the 23rd term of an AP exceeds its 16th term by 21, then the common difference is :

Mark practised

(A) 1
(B) 2
(C) 3
(D) 7

Marks: 1
Level: Standard
Compartment: 2025
Topic: 5.3 nth Term of an AP
Source: (CBSE 2025 Compartment Mathematics Standard, Set 30/S/3, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 3, because a₂₃ − a₁₆ = (a + 22d) − (a + 15d) = 7d = 21, so d = 3.
Why? By the formula for the nth term of an AP, aₙ = a + (n − 1)d, the difference between the 23rd and 16th terms is (23 − 16)d = 7d.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 30/S/3: option (C).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 7 because 23 − 16 = 7, without dividing the given excess 21 by 7.
Choosing 21, treating the given excess as the common difference itself.

Why this works

The nth term of an AP is aₙ = a + (n − 1)d. So the 23rd term is a + 22d and the 16th term is a + 15d. Subtracting gives a₂₃ − a₁₆ = 7d. The question states this difference is 21, so 7d = 21, which gives d = 3. Quick check: (23 − 16) × 3 = 21, so the excess is 21 as required. A common difference of 1 gives 7, of 2 gives 14, and of 7 gives 49; none equals the required 21.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

In an AP, the nth term is aₙ = a + (n − 1)d. When two terms of the same AP are subtracted, the first term a cancels, and only the common difference times the difference of the term numbers remains. Here the term numbers differ by 23 − 16 = 7, so 7d equals the given excess 21.

Important formulas

aₙ = a + (n − 1)d · The nth term (general term) of an AP with first term a and common difference d.
a₂₃ − a₁₆ = (23 − 16)d = 7d · Subtracting the two nth-term expressions cancels the first term a.

Q27. If the sum of first m terms of an AP is 2m² + 3m, then its second term is :

Mark practised

(A) 10
(B) 9
(C) 12
(D) 4

Marks: 1
Level: Standard
Years: 2025
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2025 Mathematics Standard, Set 30/1/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 9, because the second term is S₂ − S₁ = (2 × 2² + 3 × 2) − (2 × 1² + 3 × 1) = 14 − 5 = 9.
Why? The nth term of an AP is the difference between the sum of the first n terms and the sum of the first n − 1 terms; for n = 2 this difference is the second term.

The full 1 mark is for the correct option.

Answer:

(B) 9

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 14, the sum of the first two terms, instead of the second term.
Choosing 4, the common difference, instead of the second term.
Treating the expression 2m² + 3m as the mth term itself; the second term must be found from S₂ − S₁.

Why this works

The given Sₘ = 2m² + 3m is the sum of the first m terms. The first term is S₁ = 2 × 1² + 3 × 1 = 5. The sum of the first two terms is S₂ = 2 × 2² + 3 × 2 = 14. The second term is the extra amount added when going from 1 term to 2 terms, so a₂ = S₂ − S₁ = 14 − 5 = 9. Quick check: if the first term is 5 and the second is 9, the common difference is d = 9 − 5 = 4. Then Sₘ = m/2[2 × 5 + (m − 1) × 4] = m/2(4m + 6) = 2m² + 3m, which matches the given sum.

Key idea to remember

Nth term of an AP is the difference of the sum to first n terms and the sum to first (n − 1) terms: see Key ideas used on this page.

Explanation

The sum of the first n terms and the sum of the first n − 1 terms differ by exactly the nth term, because the second sum is the first sum without its last term. So aₙ = Sₙ − Sₙ₋₁; for the second term, a₂ = S₂ − S₁.

Important formulas

aₙ = Sₙ − Sₙ₋₁ · The nth term is the difference of the sum to first n terms and the sum to first n − 1 terms.
Sₙ = n/2[2a + (n − 1)d] · Sum of the first n terms of an AP; used here to verify the given Sₘ.

Q28. The \(11^{\text{th}}\) and \(13^{\text{th}}\) term of an AP are 39 and 45, respectively. What is the common difference of the AP ?

Mark practised

(A) 42
(B) 21
(C) 6
(D) 3

Marks: 1
Level: Standard
Years: 2025
Topic: 5.2 Arithmetic Progressions
Source: (CBSE 2025 Mathematics Standard, Set 30/1/3, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 3, because the 13th and 11th terms are 2 steps apart, so 2d = 45 − 39 = 6 and d = 3.
Why? In an AP, the gap between two terms equals the common difference multiplied by the number of steps between them. Two steps between the 13th and 11th terms give 2d, so d is half of their difference.

The full 1 mark is for the correct option.

Answer:

(D) 3

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/3: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 6 because 45 − 39 = 6, without dividing by the 2 steps between the 13th and 11th terms.
Choosing 42 because it is the average of 39 and 45; that is the 12th term, not the common difference.

Q29. The 9\(^{th}\) term from the end (towards first term) of the AP 7, 11, 15, 19, ..., 147 is :-

Mark practised

(A) 135
(B) 125
(C) 115
(D) 39

Marks: 1
Level: Standard
Years: 2025
Topic: 5.3 nth Term of an AP
Source: (CBSE 2025 Mathematics Standard, Set 30/2/2, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 115, because the 9th term from the end is 147 + (9 − 1)(−4) = 115.
Why? An AP is a list of numbers in which each term after the first is obtained by adding a fixed number, the common difference. Reading an AP backwards makes the last term the first term and changes the sign of the common difference, so the 9th term from the end is the 9th term of the reversed AP.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/2/2: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 39, which is the 9th term from the beginning of the AP, instead of the 9th term from the end.
Treating the last term as the 0th term from the end, so subtracting 9 steps of 4 from 147 instead of 8 steps.
Using the original common difference +4 while counting from the end, which moves away from the first term instead of towards it.

Why this works

The common difference of the AP is d = 11 − 7 = 4. Reading the AP backwards makes 147 the first term, and each step now subtracts 4, so the reversed common difference is −4. The 9th term from the end is the 9th term of the reversed AP: 147 + (9 − 1)(−4) = 147 − 32 = 115. Quick check: 147 − 8 × 4 = 115.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

The nth term of an AP with first term a and common difference d is a_n = a + (n − 1)d. To get a term from the end, read the AP in reverse order: the last term becomes the first term, and the common difference changes from d to −d. So the term k places from the end is l + (k − 1)(−d), where l is the last term.

Important formulas

a_n = a + (n − 1)d · The nth term of an AP with first term a and common difference d.
term k places from the end = l + (k − 1)(−d) · Read the AP backwards; l is the last term and d is the original common difference.

Q30. The 31st term of the A.P. : \(\frac{-5}{6}, \frac{-3}{4}, \frac{-2}{3}, \frac{-7}{12}, \dots\) is

Mark practised

(A) \(\frac{-5}{3}\)
(B) \(\frac{5}{3}\)
(C) \(\frac{12}{20}\)
(D) \(\frac{-12}{20}\)

Marks: 1
Level: Standard
Years: 2026
Topic: 5.3 nth Term of an AP
Source: (CBSE 2026 Mathematics Standard, Set 30(B), Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5/3, because the common difference is 1/12 and a₃₁ = −5/6 + 30 × 1/12 = 5/3.
Why? In an AP, each term is obtained by adding the fixed common difference d to the preceding term, so the nth term is a + (n − 1)d.

The full 1 mark is for the correct option.

Answer:

(B) 5/3

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30(B): option (B).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Subtracting in the wrong order: a₁ − a₂ = −5/6 − (−3/4) gives −1/12 as the common difference; the correct order is a₂ − a₁, which gives 1/12.
Choosing option A, −5/3, just because the first term is negative; the common difference 1/12 is positive, so the terms increase and the 31st term is positive.
Reading the answer 20/12 as 12/20 and selecting option C, though 20/12 simplifies to 5/3.

Why this works

First find the common difference: d = a₂ − a₁ = −3/4 − (−5/6) = 1/12. Check with the next pair: −2/3 − (−3/4) = 1/12, so the AP increases by 1/12 each time. Now use the nth term formula aₙ = a + (n − 1)d with first term −5/6, common difference 1/12, and n = 31: a₃₁ = −5/6 + 30 × 1/12 = −5/6 + 5/2 = 10/6 = 5/3. A quick check: writing the first terms in twelfths gives −10/12, −9/12, −8/12, −7/12, so the 31st term is (−10 + 30)/12 = 20/12 = 5/3. Options C and D are 12/20 and −12/20, which simplify to 3/5 and −3/5, so they are not the answer.

Key idea to remember

Nth Term of an AP: see Key ideas used on this page.

Explanation

The nth term of an AP with first term a and common difference d is aₙ = a + (n − 1)d. Once d is found, this formula gives any term directly; here it finds the 31st term without writing all 31 terms.

Important formulas

aₙ = a + (n − 1)d · Gives the nth term of an AP from the first term a and common difference d.
d = aₖ₊₁ − aₖ · The common difference is a term minus its preceding term.

Q31. The first term of an AP is \(p\) and the common difference is \(q\), then its 10th term is :

Mark practised

(A) \(q - 9p\)
(B) \(p - 9q\)
(C) \(p + 9q\)
(D) \(2p + 9q\)

Marks: 1
Level: Standard
Years: 2026
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2026 Mathematics Standard, Set 30/1/2, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) p + 9q, because the 10th term of an AP with first term p and common difference q is p + (10 − 1)q, which is p + 9q.
Why? In an AP, each term is formed by adding the common difference to the previous term, so the 10th term has the common difference added 9 times.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/1/2: option (C).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option B, p − 9q, treats the common difference as if it is subtracted each time, but an AP is formed by adding q.
Choosing option D, 2p + 9q, uses the first term twice; the first term p appears only once.
Writing the 10th term as p + 10q, forgetting that the common difference is added 9 times, not 10.

Q32. The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P. is :

Mark practised

(A) 6
(B) - 6
(C) 18
(D) - 18

Marks: 1
Level: Standard
Years: 2026
Topic: 5.4 Sum of First n Terms of an AP
Source: (CBSE 2026 Mathematics Standard, Set 30/2/2, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 6, because 2(x + 10) = 2x + (3x + 2), so 18 = 3x and x = 6.
Why? For three consecutive terms of an AP, the sum of the first and third terms equals twice the middle term, because the common difference is the same.

The full 1 mark is for the correct option.

Answer:

(A) 6

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/2/2: option (A).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting that the middle term is counted twice, and using only (x + 10) on the left side instead of 2(x + 10).
Checking only the first gap: for option B, (x + 10) − 2x = 16 but (3x + 2) − (x + 10) = −20, so the gaps are not equal.
Making a sign error while solving 2(x + 10) = 2x + (3x + 2): after subtracting 2x from both sides, writing 5x instead of 3x.

Why this works

The three terms are in AP only when the common difference is the same for both gaps: (x + 10) − 2x = (3x + 2) − (x + 10). This simplifies to 2(x + 10) = 2x + (3x + 2), i.e. 2x + 20 = 5x + 2, so 18 = 3x and x = 6. Quick check: with x = 6, the middle term is 16, twice it is 32; the first and third terms are 12 and 20, whose sum is 32. The gaps are 16 − 12 = 4 and 20 − 16 = 4, so both gaps have the same common difference. The other options give unequal gaps, so they do not form an AP.

Key idea to remember

Arithmetic Progression (AP): see Key ideas used on this page.

Explanation

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term. That fixed number is the common difference. Here the two gaps, (x + 10) − 2x and (3x + 2) − (x + 10), must be equal, which gives the equation used to find x.

Important formulas

2b = a + c, for three consecutive terms a, b, c of an AP · Equivalently, b − a = c − b.
d = next term − previous term · In an AP this difference is the same for every pair of consecutive terms.

1 / 32

Key ideas used on this page

Nth Term of an AP
Used in questions 2 3 4 6 7 8 9 10 15 16 18 19 22 25 26 28 29 30 31
Sum of First n Terms of an AP
Used in questions 1 20
Common difference
Used in questions 11 12
Arithmetic Progression (AP)
Used in questions 14 32
Nth term of an AP is the difference of the sum to first n terms and the sum to first (n − 1) terms
Used in questions 24 27
Common difference of an Arithmetic Progression (AP)
Used in question 5
Common difference of an AP
Used in question 13
General form of an AP
Used in question 17
Arithmetic mean of three numbers in AP
Used in question 21
Arithmetic Progression
Used in question 23

How to use these questions

Try each question on paper first. Then read the steps: the mark beside each step shows what the examiner looks for. If a step is new to you, revise Section 5.2, Section 5.3 and Section 5.4 of the NCERT chapter.

To revise the lessons behind these questions, read the NCERT chapter, exactly as NCERT publishes it on its own website.

Official papers behind this page

Every question on this page comes from an official CBSE Class 10 Mathematics board paper. The board publishes the question papers on cbse.gov.in and the marking schemes on cbse.gov.in, where you can see how marks are given.

Year	Exam	Course	Question paper	Marking scheme
2026	Main	Standard (041)	Open CBSE's official question paper for the 2026 main exam (one ZIP file of about 61.9 MB, Class 10 Maths Standard and Basic together) to read the full paper.	Open CBSE's official marking scheme for the 2026 main exam (one ZIP file of about 22.8 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Main	Standard (041)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 54.5 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Main	Basic (241)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 49.9 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Compartment	Standard (041)	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 3.4 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 8.1 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Compartment	Basic (241)	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 2.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 18.7 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Main	Standard (041)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 35.3 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Main	Basic (241)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 43.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.8 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Compartment	Basic (241)	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 2.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 5.6 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Main	Standard (041)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 44 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 17.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Basic (241)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 49.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 26 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Compartment	Standard (041)	Open CBSE's official question paper for the 2023 compartment exam (one ZIP file of about 7.6 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 compartment exam (one ZIP file of about 3.3 MB, Class 10 Maths Standard (041)) to see how marks are given.

Sources and session

Each question is reproduced as printed in its CBSE paper, with its year, paper code and question number; only the layout is tidied. The same question printed in several paper sets counts once, and the card says how often it was printed in 2023–2026. These counts describe past papers; they are not predictions.

Every solution is original and follows NCERT Mathematics Class X (reprint 2026-27). The mark beside each step follows the official CBSE marking scheme where the scheme shows it; a mark labelled “(suggested)” follows the board’s usual pattern for that type of question.

Reference: CBSE Class 10 Mathematics board question papers via cbse.gov.in; all solutions are original. learncbse.net is an independent website and is not affiliated with CBSE or NCERT.

1-mark questions on Arithmetic Progressions

All 32 Arithmetic Progressions 1-mark questions

Q1. If the sum of the first \(n\) terms of an A.P be \(3n^2 + n\) and its common difference is 6, then its first term is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q2. The common difference of an A.P., if \(a_{23} - a_{19} = 32\) , is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q3. Which term of the A.P. -29, -26, -23, ..., 61 is 16 ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q4. In an A.P., if \(d = -4\) and \(a_7 = 4\), then the first term 'a' is equal to

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q5. The common difference of the A.P.\(\frac{1}{2x}, \frac{1-4x}{2x}, \frac{1-8x}{2x}, \dots\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q6. The number of terms in the A.P. 3, 6, 9, 12, ..., 111 is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q7. The next (4\(^{th}\)) term of the A.P. \(\sqrt{18}\) , \(\sqrt{50}\) , \(\sqrt{98}\) , ... is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q8. If the first term of an A.P. is 'a' and its common difference is 'b', then its 10\(^{th}\) term is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q9. The 6\(^{th}\) term of the AP \(\sqrt{27}\) , \(\sqrt{75}\) , \(\sqrt{147}\) , ... is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q10. If \(n^{\text{th}}\) term of an A.P. is \(5n - 6\), then its common difference is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q11. The common difference of the AP : \(\sqrt{2}\), \(2\sqrt{2}\), \(3\sqrt{2}\), \(4\sqrt{2}\), ... is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q12. If \(x, 2x + 9, 4x + 3\) are three consecutive terms of an A.P., then the value of \(x\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q13. If 2x, x + 10, 3x + 2 are three consecutive terms of an A.P., then the value of x is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q14. If \(x + 1\), \(3x\) and \(4x + 2\) are three consecutive terms of an A.P., then the value of \(x\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q15. The seventh term of an A.P. whose first term is 28 and common difference \(-4\) , is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q16. The seventh term of an A.P. whose first term is 28 and common difference – 4, is

Step-by-step solution

Step 1: Choose the option

Q5. The common difference of the A.P.
\(\frac{1}{2x}, \frac{1-4x}{2x}, \frac{1-8x}{2x}, \dots\) is :