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Exercise 5.2 Class 10 Maths NCERT Solutions

Looking for the Exercise 5.2 Class 10 Maths NCERT Solutions? This page solves all 20 questions of Exercise 5.2 from Chapter 5 (Arithmetic Progressions) of the NCERT Class 10 Mathematics textbook, one careful step at a time. Every question below is reproduced word for word from the textbook, so you can check your homework without keeping the book open.

The exercise trains a single idea — the nth term formula \( a_n = a + (n-1)d \) — used in four ways: finding a term when the first term and common difference are known, finding \( n \) when the term is given, checking whether a number belongs to the AP at all, and rebuilding the whole list from two known terms.

Exercise 5.2 Class 10 Maths NCERT Solutions

Exercise 5.2 (NCERT, pp. 62–63) is the first big workout on the nth term of an arithmetic progression. It expects two facts before you start: the first term \( a \) and the common difference \( d \) — and the one formula \( a_n = a + (n-1)d \). Each of the 20 questions is a different way of using that formula.

Here is the exam-style map of what each question tests:

Skill tested Questions What a full-marks answer needs
Direct nth term computation Q1(i), (v), Q2, Q4, Q11 State \( a \), \( d \), \( n \), substitute, give the term
Solve for \( n \) or \( d \) Q1(ii), (iv), Q5, Q6, Q13, Q14, Q19, Q20 Rearrange the formula; check \( n \) is a positive whole number
Rebuild the AP from two terms Q3, Q7, Q8, Q9, Q16, Q18 Write two equations in \( a \) and \( d \), eliminate, list the AP
Comparing terms or two APs Q10, Q12, Q15 Subtract the two nth-term expressions and simplify
Term counted from the last Q17 Find the total count first, then count from the front

These questions are the exact exercise 5.2 class 10 maths ncert solutions students search for the night before homework is due — the worked answers below follow the pattern a board examiner expects: formula first, substitution shown, result stated plainly.

Only four ideas are needed for this exercise, all from the chapter’s own development:

  • Arithmetic progression (AP): a list in which each term is obtained by adding a fixed number \( d \) to the preceding term, except the first term (NCERT, p. 54). The fixed number \( d \) is the common difference; it can be positive, negative or zero (NCERT, p. 53).
  • Common difference: \( d = a_{k+1} – a_k \) — always the later term minus the earlier one, even when the later term is smaller (NCERT, p. 54).
  • General form: \( a, \; a + d, \; a + 2d, \; a + 3d, \dots \) (NCERT, p. 53).
  • nth term formula: \( a_n = a + (n-1)d \), where \( a_n \) is also called the general term of the AP (NCERT, p. 58).
  • Arithmetic mean: if \( a, b, c \) are in AP then \( b = \frac{a+c}{2} \) (NCERT, p. 24, Note to the Reader) — this is the quick way to fill a single missing box.

The sum formula \( S = \frac{n}{2}[2a + (n-1)d] \) belongs to Section 5.4 and is practised in Exercise 5.3 — it is not needed anywhere in Exercise 5.2. You use only \( a_n = a + (n-1)d \) here.

Question 1: Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP:

\( a \) \( d \) \( n \) \( a_n \)
(i) 7 3 8 …
(ii) -18 … 10 0
(iii) … -3 18 -5
(iv) -18.9 2.5 … 3.6
(v) 3.5 0 105 …

The formula \( a_n = a + (n-1)d \) has four letters. Each row gives you three of them, so every part is a one-line substitution — your job is only to decide which quantity is missing. Part (v) is the special case where \( d = 0 \), which forces every term to equal the first term.

Part (i): \( a = 7, \; d = 3, \; n = 8 \).

Substitute: \( a_8 = 7 + (8-1) \times 3 = 7 + 21 = 28 \).

Part (ii): \( a = -18, \; n = 10, \; a_{10} = 0 \), so \( 0 = -18 + 9d \).

This gives \( 9d = 18 \), hence \( d = 2 \).

Part (iii): \( d = -3, \; n = 18, \; a_{18} = -5 \).

So \( -5 = a + 17(-3) = a – 51 \), giving \( a = 46 \).

Part (iv): \( a = -18.9, \; d = 2.5, \; a_n = 3.6 \).

Solve \( 3.6 = -18.9 + (n-1)(2.5) \): \( 22.5 = 2.5(n-1) \), so \( n-1 = 9 \) and \( n = 10 \).

Part (v): \( a = 3.5, \; d = 0, \; n = 105 \).

Then \( a_{105} = 3.5 + 104 \times 0 = 3.5 \).

Final answers: (i) 28, (ii) \( d = 2 \), (iii) \( a = 46 \), (iv) \( n = 10 \), (v) 3.5.

Common error: in part (ii), students write \( -18 + 9d = 0 \) and then treat it as \( 9d = -18 \). It is \( 9d = 18 \) — bring \( -18 \) across as \( +18 \). And in part (v), a zero common difference is still a valid AP, not a mistake (NCERT, p. 53).

Question 2: Choose the correct choice in the following and justify :

(i) 30th term of the AP: 10, 7, 4, …, is

  • (A) 97
  • (B) 77
  • (C) -77
  • (D) -87

(ii) 11th term of the AP: -3, -1/2, 2, …, is

  • (A) 28
  • (B) 22
  • (C) -38
  • (D) -48 1/2

A multiple-choice AP question is solved by writing \( a \), \( d \) and \( n \) first, then substituting — the options only hide arithmetic slips, mostly from a wrong sign on \( d \).

Part (i): \( a = 10 \), \( d = 7 – 10 = -3 \), \( n = 30 \).

So \( a_{30} = 10 + (30-1)(-3) = 10 – 87 = -77 \).

Correct answer: (C) -77. Option (A) comes from using \( d = +3 \) (ignoring the decreasing pattern); (B) and (D) are arithmetic slips on \( 29 \times 3 \).

Part (ii): \( a = -3 \), \( d = -\tfrac{1}{2} – (-3) = \tfrac{5}{2} \), \( n = 11 \).

So \( a_{11} = -3 + 10 \times \tfrac{5}{2} = -3 + 25 = 22 \).

Correct answer: (B) 22. Option (A) drops the \( -3 \); (C) and (D) carry a sign error into the subtraction.

Checking method: for part (i), list backwards from the end of the working — the AP is decreasing by 3 each step, so a negative answer is expected. If your result is positive for a decreasing AP, your \( d \) has the wrong sign.

Question 3: In the following APs, find the missing terms in the boxes :

  • (i) 2, [ ], 26
  • (ii) [ ], 13, [ ], 3
  • (iii) 5, [ ], [ ], 9 1/2
  • (iv) -4, [ ], [ ], [ ], [ ], 6
  • (v) [ ], 38, [ ], [ ], [ ], -22

When three consecutive terms are shown, the middle term is the arithmetic mean of its neighbours: \( b = \frac{a+c}{2} \). When the boxes are not consecutive, write the general term \( a + (k-1)d \) for two known positions and solve for \( a \) and \( d \) together.

Part (i): middle box \( = \frac{2+26}{2} = 14 \).

The AP is 2, 14, 26.

Part (ii): four terms, so \( a + d = 13 \) and \( a + 3d = 3 \).

Subtract: \( 2d = -10 \), so \( d = -5 \) and \( a = 18 \).

Missing terms are 18 and 8 (list: 18, 13, 8, 3).

Part (iii): \( a = 5 \), and the 4th term is \( 9\tfrac{1}{2} = 5 + 3d \), so \( d = 1.5 \).

Missing terms are 6.5 and 8 (list: 5, 6.5, 8, 9.5).

Part (iv): six terms, \( a = -4 \), \( a_6 = 6 = -4 + 5d \), so \( d = 2 \).

Missing terms are -2, 0, 2, 4.

Part (v): six terms with \( 38 = a + d \) and \( -22 = a + 5d \).

Subtract: \( 4d = -60 \), so \( d = -15 \) and \( a = 53 \).

Missing terms are 53, 23, 8, -7 (list: 53, 38, 23, 8, -7, -22).

Final answers: (i) 14; (ii) 18, 8; (iii) 6.5, 8; (iv) -2, 0, 2, 4; (v) 53, 23, 8, -7.

Common error: in part (v), students treat 38 as the first term. It is the second term, so the equation is \( a + d = 38 \), not \( a = 38 \).

Question 4: Which term of the AP: 3, 8, 13, 18, …, is 78?

The question gives you the value of a term (78) and asks for its position \( n \). Read \( a \) and \( d \) off the list, put 78 into the formula as \( a_n \), and solve for \( n \).

Step 1: \( a = 3 \), \( d = 8 – 3 = 5 \), and \( a_n = 78 \).

\[ 78 = 3 + (n-1) \times 5 \]

Step 2: \( 75 = 5(n-1) \), so \( n – 1 = 15 \) and \( n = 16 \).

Final answer: 78 is the 16th term of the AP.

Checking method: substitute back — \( a_{16} = 3 + 15 \times 5 = 78 \). If the back-substitution does not reproduce 78, the position is wrong.

Question 5: Find the number of terms in each of the following APs :

  • (i) 7, 13, 19, …, 205
  • (ii) 18, 15 1/2, 13, …, -47

Here the last term is known and \( n \) is unknown. Because terms are numbered 1, 2, 3, …, the \( n \) you solve for must come out a positive whole number — that is itself a check on the working.

Part (i): \( a = 7 \), \( d = 13 – 7 = 6 \), \( a_n = 205 \).

Then \( 205 = 7 + (n-1) \times 6 \), so \( 198 = 6(n-1) \), giving \( n – 1 = 33 \) and \( n = 34 \).

Part (ii): \( a = 18 \), \( d = 15\tfrac{1}{2} – 18 = -\tfrac{5}{2} \), \( a_n = -47 \).

Then \( -47 = 18 + (n-1)(-\tfrac{5}{2}) \), so \( -65 = -\tfrac{5}{2}(n-1) \), giving \( n – 1 = 26 \) and \( n = 27 \).

Final answers: (i) 34 terms; (ii) 27 terms.

Common error: in part (ii) the common difference is negative — a common slip is writing \( d = +\tfrac{5}{2} \), which turns the count upside down. Always subtract the earlier term from the later one: \( 15.5 – 18 = -2.5 \).

Question 6: Check whether -150 is a term of the AP: 11, 8, 5, 2 …

To test membership, assume the number is a term, write it as \( a_n \), and solve for \( n \). A positive whole number for \( n \) means the value really appears in the list; a fraction or a negative result proves it does not (this mirrors Example 6, NCERT, p. 59).

Step 1: \( a = 11 \), \( d = 8 – 11 = -3 \).

Let \( -150 = a_n \).

\[ -150 = 11 + (n-1)(-3) \]

  1. Step 1: \( -161 = -3(n-1) \), so \( n – 1 = \frac{161}{3} \) and \( n = 1 + \frac{161}{3} \).
  2. Step 2: \( 1 + \frac{161}{3} = \frac{164}{3} \), which is not a positive integer.

Final answer: -150 is not a term of the AP.

Positive-integer rule (NCERT, p. 59): a term number is always a counting number. If your \( n \) is a fraction such as \( \frac{164}{3} \), stop — the number is not in the list. There is no “164/3rd term” of an AP.

Question 7: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Two given terms give two equations in the two unknowns \( a \) and \( d \). Write each term as \( a + (k-1)d \), eliminate one unknown, then use the recovered \( a \) and \( d \) to build the term the question actually wants.

  1. Step 1: \( a_{11} = a + 10d = 38 \) and \( a_{16} = a + 15d = 73 \).
  2. Step 2: subtract the first equation from the second: \( 5d = 35 \), so \( d = 7 \).

Put \( d = 7 \) into \( a + 10d = 38 \): \( a + 70 = 38 \), so \( a = -32 \).

\[ a_{31} = -32 + (31-1) \times 7 = -32 + 210 = 178 \]

Final answer: the 31st term is 178.

Checking method: confirm the two given terms from your \( a \) and \( d \): \( a_{11} = -32 + 70 = 38 \) and \( a_{16} = -32 + 105 = 73 \). If both match, the answer is safe.

Question 8: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

The phrase “50 terms” tells you the last term is \( a_{50} \). That gives two equations in \( a \) and \( d \) again: one from the 3rd term, one from the last. Once \( a \) and \( d \) are known, any term of the AP can be written directly.

  1. Step 1: \( a_3 = a + 2d = 12 \) and \( a_{50} = a + 49d = 106 \).
  2. Step 2: subtract: \( 47d = 94 \), so \( d = 2 \).

Then \( a + 4 = 12 \), so \( a = 8 \).

\[ a_{29} = 8 + (29-1) \times 2 = 8 + 56 = 64 \]

Final answer: the 29th term is 64.

Common error: treating 106 as the 50th term is right only because the AP has exactly 50 terms — do not assume “the last term” is \( a_{50} \) in questions that do not state a total count.

Question 9: If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Set up \( a \) and \( d \) from the two given terms, then solve \( a_n = 0 \) for \( n \). The result must again be a positive whole number — that is how you know the zero really sits at some position.

  1. Step 1: \( a_3 = a + 2d = 4 \) and \( a_9 = a + 8d = -8 \).
  2. Step 2: subtract: \( 6d = -12 \), so \( d = -2 \).

Then \( a – 4 = 4 \), so \( a = 8 \).

\[ 0 = 8 + (n-1)(-2) \; \Rightarrow \; -8 = -2(n-1) \; \Rightarrow \; n – 1 = 4 \; \Rightarrow \; n = 5 \]

Final answer: the 5th term of this AP is zero.

Checking method: list the early terms from \( a = 8 \), \( d = -2 \): 8, 6, 4, 2, 0 — zero is indeed the 5th term, and the 3rd term is 4 as given.

Question 10: The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

This question never tells you \( a \) — and you do not need it. Write \( a_{17} \) and \( a_{10} \) in terms of \( a \) and \( d \), subtract, and the unknown first term cancels itself out.

\[ a_{17} – a_{10} = [a + 16d] – [a + 9d] = 7d \]

Step: the difference is given as 7, so \( 7d = 7 \), giving \( d = 1 \).

Final answer: the common difference is 1.

Why this works: between the 10th and 17th terms there are exactly 7 jumps of size \( d \), so the gap is always \( 7d \). The first term never enters the calculation because it appears in both expressions identically.

Question 11: Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?

First compute the 54th term, add 132 to get the target value, then find which position \( n \) holds that value. Equivalently, the target is \( n \) jumps of size \( d \) beyond the 54th term.

Step 1: \( a = 3 \), \( d = 15 – 3 = 12 \).

Then \( a_{54} = 3 + 53 \times 12 = 3 + 636 = 639 \).

Step 2: the target value is \( 639 + 132 = 771 \).

Solve \( 771 = 3 + (n-1) \times 12 \): \( 768 = 12(n-1) \), so \( n – 1 = 64 \) and \( n = 65 \).

Final answer: the 65th term is 132 more than the 54th term.

Quick check: from the 54th to the 65th term there are 11 jumps; \( 11 \times 12 = 132 \), exactly the excess asked for.

Question 12: Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Write both APs with the same \( d \), calling their first terms \( a \) and \( b \). Because the common difference is shared, it cancels whenever you subtract the two nth terms — so the gap between the two lists never changes, no matter which term you compare.

\[ a_{100} – b_{100} = (a + 99d) – (b + 99d) = a – b = 100 \]

\[ a_{1000} – b_{1000} = (a + 999d) – (b + 999d) = a – b = 100 \]

Final answer: the difference between their 1000th terms is still 100.

Why this works: both APs grow by the same step size, so the horizontal gap between them is fixed forever from the first term onward. The term number does not matter at all.

Question 13: How many three-digit numbers are divisible by 7?

Divisible-by-7 numbers make an AP with \( d = 7 \). The skill here is identifying the correct first and last three-digit multiples — pick the smallest and largest, then count the steps between them with the nth term formula.

Step 1: the first three-digit multiple of 7 is 105 (since \( 100 \div 7 \) leaves a remainder); the largest is 994 (since \( 7 \times 142 = 994 \)).

So the AP is 105, 112, …, 994.

Step 2: \( a = 105 \), \( d = 7 \), \( a_n = 994 \).

Then \( 994 = 105 + (n-1) \times 7 \), so \( 889 = 7(n-1) \), giving \( n – 1 = 127 \) and \( n = 128 \).

Final answer: there are 128 three-digit numbers divisible by 7.

Common error: starting at 98 (a two-digit number) or ending at 1001 (four-digit). “Three-digit” means 100 to 999 — check both endpoints before counting.

Question 14: How many multiples of 4 lie between 10 and 250?

Same structure as Q13 with \( d = 4 \). The phrase “between 10 and 250” excludes the endpoints themselves, so start at the first multiple above 10 and stop at the last multiple below 250.

Step 1: the first multiple of 4 after 10 is 12; the last one before 250 is 248.

The list is 12, 16, 20, …, 248.

Step 2: \( a = 12 \), \( d = 4 \), \( a_n = 248 \).

Then \( 248 = 12 + (n-1) \times 4 \), so \( 236 = 4(n-1) \), giving \( n – 1 = 59 \) and \( n = 60 \).

Final answer: 60 multiples of 4 lie between 10 and 250.

Checking method: 10 itself is not a multiple of 4, so it cannot be counted. If you accidentally started at 8, you would be one term wrong — always test the boundary numbers.

Question 15: For what value of n, are the nth terms of two APs: 63, 65, 67, … and 3, 10, 17, … equal?

Write the nth term of each AP with its own \( a \) and \( d \), set the two expressions equal, and solve for \( n \). The common \( n \) is the single position where both lists hit the same value.

\[ \text{AP 1: } 63 + (n-1) \times 2 \qquad \text{AP 2: } 3 + (n-1) \times 7 \]

\[ 63 + 2(n-1) = 3 + 7(n-1) \; \Rightarrow \; 60 = 5(n-1) \; \Rightarrow \; n – 1 = 12 \; \Rightarrow \; n = 13 \]

Final answer: the nth terms are equal at \( n = 13 \).

Checking method: \( a_{13} \) of the first AP is \( 63 + 12 \times 2 = 87 \); the second is \( 3 + 12 \times 7 = 87 \). Both give 87, so the position is correct.

Question 16: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

The “exceeds by 12” sentence hides a second equation: \( a_7 – a_5 = 12 \). Since \( a_7 – a_5 = 2d \), this immediately gives \( d \) without touching \( a \) — then the 3rd term pins down the first term.

  1. Step 1: \( a_3 = a + 2d = 16 \), and \( a_7 – a_5 = (a + 6d) – (a + 4d) = 2d = 12 \), so \( d = 6 \).
  2. Step 2: \( a + 2 \times 6 = 16 \), so \( a = 4 \).

Final answer: the AP is 4, 10, 16, 22, …

Why it works: the 5th and 7th terms are two jumps apart, so their difference is always \( 2d \) — the 7th exceeds the 5th by exactly \( 2d \) no matter what \( a \) is.

Question 17: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

A term “from the last” is counted backwards, so first find how many terms the AP has altogether, then convert the backward position into a forward one. The kth term from the last is the \((n – k + 1)\)th term from the front (mirrors Example 8, NCERT, p. 60).

Step 1: \( a = 3 \), \( d = 5 \), \( a_n = 253 \).

Solve \( 253 = 3 + (n-1) \times 5 \): \( 250 = 5(n-1) \), so \( n – 1 = 50 \) and \( n = 51 \).

The AP has 51 terms.

Step 2: the 20th from the last is the \( (51 – 20 + 1) = 32 \)nd term from the front.

\[ a_{32} = 3 + (32-1) \times 5 = 3 + 155 = 158 \]

Final answer: the 20th term from the last is 158.

Common error: students write \( a_{20} \) instead of \( a_{32} \) — the 20th from the end is not the 20th term. Count backwards: the 51st is 253, the 50th is 248, …, so the 20th from the last is the 32nd.

Question 18: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Translate each sentence into an equation in \( a \) and \( d \), simplify each sum, and solve the pair by elimination. The question asks only for the first three terms, so write \( a \), \( a + d \), \( a + 2d \) with the recovered values.

  1. Step 1: \( a_4 + a_8 = (a+3d) + (a+7d) = 2a + 10d = 24 \), so \( a + 5d = 12 \).
  2. Step 2: \( a_6 + a_{10} = (a+5d) + (a+9d) = 2a + 14d = 44 \), so \( a + 7d = 22 \).
  3. Step 3: subtract: \( 2d = 10 \), so \( d = 5 \) and \( a = 12 – 25 = -13 \).

Final answer: the first three terms are -13, -8, -3.

Checking method: with \( a = -13 \), \( d = 5 \): \( a_4 = 2 \), \( a_8 = 22 \), sum 24; \( a_6 = 12 \), \( a_{10} = 32 \), sum 44. Both given conditions hold.

Question 19: Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

His yearly salaries form an AP: 5000, 5200, 5400, … Find which position holds ₹ 7000, then translate the term number back into a calendar year — the 1st term is 1995.

Step 1: \( a = 5000 \), \( d = 200 \), \( a_n = 7000 \).

Solve \( 7000 = 5000 + (n-1) \times 200 \): \( 2000 = 200(n-1) \), so \( n – 1 = 10 \) and \( n = 11 \).

Step 2: the 11th year of work — 1995 was his 1st year, so 1995 + 10 = 2005.

Final answer: his income reached ₹ 7000 in the year 2005.

Common error: adding 11 to 1995 instead of 10. The 1st term is 1995 itself, so the 11th term is 1995 + (11 – 1), not 1995 + 11.

Question 20: Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Weekly savings form an AP with first term ₹ 5 and common difference ₹ 1.75. You are told the nth week’s saving and asked for its position — exactly the “find n when a_n is known” pattern seen throughout this exercise.

Step: \( a = 5 \), \( d = 1.75 \), \( a_n = 20.75 \).

Solve:

\[ 20.75 = 5 + (n-1) \times 1.75 \; \Rightarrow \; 15.75 = 1.75(n-1) \; \Rightarrow \; n – 1 = 9 \; \Rightarrow \; n = 10 \]

Final answer: \( n = 10 \), so her weekly savings reached ₹ 20.75 in the 10th week.

Checking method: feed \( n = 10 \) back in — \( a_{10} = 5 + 9 \times 1.75 = 5 + 15.75 = 20.75 \). The back-substitution reproduces the given value, so the count is right.

All 20 questions of Exercise 5.2 are now solved. The next step in the chapter is the sum of the first \( n \) terms of an AP, which Exercise 5.3 develops — you can find the full chapter treatment in the complete set of Class 10 Maths notes.

How to Verify Your nth Term Answers

Before you hand in the homework, run every answer through one of these four checks built only from this chapter’s own tools. Each takes seconds and catches the slips students actually make.

  1. Substitute back. After finding a term, plug your \( n \) into \( a_n = a + (n-1)d \) and confirm it reproduces the value. For Q7, \( a_{31} = -32 + 210 = 178 \) brings back the given 11th and 16th terms, so the answer is safe.
  2. Positive-integer test. When checking membership, \( n \) must come out a positive whole number. In Q6 you get \( n = 1 + \frac{161}{3} \), a fraction — so -150 is not a term (NCERT, p. 59). If \( n \) is a fraction or negative, the number is not in the list.
  3. Adjacent-difference test. To confirm a list is an AP, check \( a_{k+1} – a_k \) is the same for consecutive \( k \) (NCERT, p. 54). This is the safest guard against misreading a common difference, especially a negative one.
  4. Arithmetic-mean shortcut. For the missing-box questions, when three consecutive terms appear, the middle term is \( \frac{a+c}{2} \) (NCERT, p. 24). In Q3(i), \( \frac{2+26}{2} = 14 \) — instant verification.

Here is one full check with original numbers. Take the AP 4, 9, 14, 19, … and ask whether 89 is a term. Here \( a = 4 \), \( d = 5 \); putting \( a_n = 89 \) into the formula gives \( 89 = 4 + (n-1) \times 5 \), so \( 85 = 5(n-1) \) and \( n = 18 \). Since \( n \) is a positive integer, 89 is the 18th term. Back-substitute to confirm: \( a_{18} = 4 + 17 \times 5 = 4 + 85 = 89 \). Both tests agree.

Mistake Correct rule How to check your answer
Finding \( d \) the wrong way round (e.g. \( 3 – 6 \) instead of \( 6 – 3 \)) Always \( d = a_{k+1} – a_k \), the later term minus the earlier one (NCERT, p. 54) Write the first three consecutive differences — all three must be identical
Treating a negative \( d \) as positive Keep \( d \) as it is; a term can be smaller than the one before it (NCERT, p. 53) Substitute your \( n \) back into the formula and see if the term matches
Calling a number a term when \( n \) comes out a fraction \( n \) must be a positive whole number (NCERT, p. 59) A fraction such as \( \frac{164}{3} \) proves the number is not in the list
Reading “20th from the last” as \( a_{20} \) Count from the front: position = (total terms – 20 + 1) Count backwards through the last few terms to match
Assuming \( d = 0 \) means “no AP” \( d = 0 \) is a valid AP where every term is equal (NCERT, p. 53) Confirm \( a_{k+1} – a_k = 0 \) for every \( k \)

Revising the whole chapter? The Class 10 notes for every subject, the main notes index, and the build-up chapter on quadratic equations (where you first met solving two equations together) are all one click away.

Frequently Asked Questions

Why must n be a positive whole number when checking whether a value is a term of an AP?

Because terms are numbered 1st, 2nd, 3rd, … — there is no “164/3rd term.” The formula always gives \( n = 1 + \frac{a_n – a}{d} \); when that comes out a fraction or a negative value, the number simply does not appear in the list. This is exactly the test used in Example 6 (NCERT, p. 59) and in Question 6 above.

How do you find the term from the last of an AP, like question 17?

Count from the front. First find the total number of terms \( n \) using the last term, then the kth term from the last is the \((n – k + 1)\)th term. In Question 17, the AP has 51 terms, so the 20th from the last is the \( (51 – 20 + 1) = 32 \)nd term, and \( a_{32} = 158 \)). The method mirrors Example 8 (NCERT, p. 60).

How do you fill missing boxes when both the first term and the common difference are unknown?

Write the nth term formula for two different known positions — that gives two equations in \( a \) and \( d \) — then eliminate one unknown, as in Questions 7, 8 and 18. When three consecutive terms are shown, the middle one is simply the average of its neighbours, \( b = \frac{a+c}{2} \) (NCERT, p. 24), which solves Question 3(i) instantly.

What does a common difference of zero mean in an AP?

Every term equals the first term, so the list is constant — like 3.5, 3.5, 3.5, … It is still a perfectly valid AP (NCERT, p. 53), and the formula still works: \( a_n = a + (n-1) \times 0 = a \). In Question 1 part (v), the 105th term is simply 3.5.

Do I need the sum formula S = n/2(2a + (n-1)d) for Exercise 5.2?

No. Exercise 5.2 asks only for individual terms, so the only formula you need is \( a_n = a + (n-1)d \). The sum of the first \( n \) terms is developed in Section 5.4 and practised in Exercise 5.3 — you will meet it as the next step in the chapter, covered in the Class 10 Maths chapter notes.

Every question solved on this page is taken word for word from the official NCERT Class 10 Mathematics textbook, so you can check the original wording of any Exercise 5.2 question against the source PDF page by page.

Reference: NCERT Class 10 Mathematics textbook, chapter Arithmetic Progressions.

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