Arithmetic Progressions Class 10 Previous Year Questions (Chapter 5)

(i) · 2 marks

1. 1

   Step 1: Find the first term

   1 mark (suggested)

   Let the first term be a and the common difference be d. a = S₁ = 4(1)² − 1 = 3.

   Why? By definition, Sₙ is the sum of the first n terms of an AP. For n equal to 1, that sum contains only the first term, so the first term is S₁.

2. 2

   Step 2: Find the common difference

   1 mark (suggested)

   S₂ = 4(2)² − 2 = 14; a₂ = S₂ − S₁ = 14 − 3 = 11; d = a₂ − a₁ = 11 − 3 = 8.

   Why? The sum of the first two terms minus the sum of the first one term leaves exactly the second term. In an AP, the common difference is the difference between a term and its preceding term.

Marks (i): 1 + 1 = 2

(ii) · 1 mark

3. 3

   Step 3: Write the AP

   1 mark (suggested)

   The AP is 3, 11, 19, 27, ...

   Why? The general form of an AP is a, a + d, a + 2d, a + 3d, ... . Writing the terms uses the first term and the common difference already found.

(iii) · 2 marks

4. 4

   Step 4: Set up the equation for the term number

   1 mark (suggested)

   Let 107 be the nth term. Then aₙ = Sₙ − Sₙ₋₁ = (4n² − n) − [4(n−1)² − (n−1)] = 8n − 5, so 8n − 5 = 107.

   Why? The nth term of an AP is the difference between the sum of its first n terms and the sum of its first (n−1) terms. Equating this nth term to 107 converts the question into an equation in n.

5. 5

   Step 5: Solve for n

   1 mark (suggested)

   8n = 112, so n = 14. Hence, the 14th term of the AP is 107.

   Why? The equation is linear in n; isolating n gives the position of the term whose value is 107.

Marks (iii): 1 + 1 = 2

Marks: 2 + 1 + 2 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Form the AP

   1 mark (suggested)

   Let the production in the first year be a TV sets and the fixed yearly increase be d. Then the yearly productions form an AP.

   Why? A fixed yearly increase means each year's production is obtained by adding the same number to the previous year, which is exactly what an arithmetic progression is; naming a and d lets us use the AP formulas.

2. 2

   Step 2: Write the given terms

   1 mark (suggested)

   a₄ = a + 3d = 720 and a₈ = a + 7d = 880.

   Why? The nth term formula tells how many common differences are added to the first term for a given year: the fourth year adds 3d and the eighth year adds 7d.

3. 3

   Step 3: Solve for a and d

   1 mark (suggested)

   Subtracting: (a + 7d) − (a + 3d) = 880 − 720, so 4d = 160, d = 40. Then a = 720 − 3×40 = 600.

   Why? Subtracting the two equations eliminates a, so the common difference is found first; substituting it back gives the first term.

4. 4

   Step 4: Find the 10th year production

   1 mark (suggested)

   a₁₀ = a + 9d = 600 + 9×40 = 960.

   Why? Each step from the first term adds one common difference, so the tenth term is nine common differences after the first term.

5. 5

   Step 5: Find total production in first 7 years

   1 mark (suggested)

   S₇ = 7/2[2×600 + (7−1)×40] = 7/2 × 1440 = 5040. Therefore, production in the tenth year is 960 TV sets and the total production in the first seven years is 5040 TV sets.

   Why? The sum formula adds the first n terms of an AP without listing them; applying it for the first seven terms gives their total production.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Form the equation for S₇

   1 mark (suggested)

   Let the first term be a and the common difference be d. Using Sₙ = n/2[2a + (n − 1)d], S₇ = 7/2[2a + 6d] = 91.

   Why? The sum formula for an AP turns the given sum of the first 7 terms into a linear equation in a and d.

2. 2

   Step 2: Form the equation for S₁₇

   1 mark (suggested)

   S₁₇ = 17/2[2a + 16d] = 561.

   Why? The same sum formula applied to the first 17 terms gives a second equation, so the two unknowns can be found.

3. 3

   Step 3: Solve for a and d

   1 mark (suggested)

   7/2(2a + 6d) = 91, so a + 3d = 13. 17/2(2a + 16d) = 561, so a + 8d = 33. Subtracting: 5d = 20, so d = 4. Then a + 12 = 13, so a = 1.

   Why? Dividing the equations by 7 and 17 simplifies them. Subtracting eliminates a, and substituting gives d and then a.

4. 4

   Step 4: Find Sₙ

   1 mark (suggested)

   Sₙ = n/2[2(1) + (n − 1)4] = n/2(4n − 2) = n(2n − 1).

   Why? Putting the values of a and d into the sum formula expresses Sₙ as a function of n alone.

5. 5

   Step 5: Find aₙ

   1 mark (suggested)

   Using aₙ = Sₙ − Sₙ₋₁, aₙ = n(2n − 1) − (n − 1)(2n − 3) = 4n − 3. Hence the nth term is 4n − 3.

   Why? The nth term is the difference between the sum of n terms and the sum of n − 1 terms.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Find the common difference

   1 mark (suggested)

   Let a be the first term and d the common difference of the AP. Then d = a₃ − a₂ = 18 − 14 = 4.

   Why? The common difference of an AP is the fixed number obtained by subtracting a term from its preceding term; this difference is the same throughout the AP.

2. 2

   Step 2: Find the first term

   1 mark (suggested)

   a₂ = a + d, so 14 = a + 4, hence a = 10.

   Why? The nth term of an AP is obtained by adding the common difference to the previous term; the second term therefore equals a + d, and this equation can be solved for a once d is known.

3. 3

   Step 3: Write the sum formula with numbers

   1 mark (suggested)

   S₅₁ = (51/2)[2(10) + (51 − 1)(4)]

   Why? The sum of the first n terms of an AP depends only on a, d and n; substituting the known values into the formula gives the expression to be simplified.

4. 4

   Step 4: Simplify

   1 mark (suggested)

   = (51/2)[20 + 200] = (51/2) × 220 = 51 × 110 = 5610.

   Why? Simplify the expression by first evaluating the bracket, then multiplying by the factor outside; the arithmetic follows the usual order of operations.

5. 5

   Step 5: State the answer

   1 mark (suggested)

   The sum of the first 51 terms of the AP is 5610.

   Why? This states the final result asked for in the question.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Set up the sum equation

   1 mark (suggested)

   Let the number of terms be n. Here a = 45 and d = 39 − 45 = −6, so S_n = n/2 [2(45) + (n − 1)(−6)] = 180.

   Why? The sum of the first n terms of an AP is S_n = n/2 [2a + (n − 1)d]; putting the given values turns the question into an equation in n.

2. 2

   Step 2: Form the quadratic equation

   1 mark (suggested)

   180 = n/2 (90 − 6n + 6) = n/2 (96 − 6n) = 48n − 3n², so 3n² − 48n + 180 = 0, i.e., n² − 16n + 60 = 0.

   Why? Expanding the bracket gives a quadratic equation because n appears both outside and inside the bracket; dividing by 3 simplifies it.

3. 3

   Step 3: Solve for n

   1 mark (suggested)

   n² − 16n + 60 = 0, so (n − 6)(n − 10) = 0, hence n = 6 or n = 10.

   Why? A quadratic equation is solved by factorising; both roots are positive integers, so both are admissible numbers of terms.

4. 4

   Step 4: Explain the double answer

   1 mark (suggested)

   The first 6 terms are 45, 39, 33, 27, 21, 15 and their sum is 180. The 7th to 10th terms are 9, 3, −3, −9 and 9 + 3 + (−3) + (−9) = 0, so the sum of the first 10 terms is also 180.

   Why? When the common difference is negative, later terms become negative and cancel earlier positive terms, so two different values of n can give the same sum.

5. 5

   Step 5: State the result

   1 mark (suggested)

   Thus 6 terms or 10 terms of the AP must be taken so that their sum is 180.

   Why? The question asks for the number of terms, so both admissible values are stated in the final answer.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)