CBSE Class 10 Mathematics Arithmetic Progressions MCQs Set 25

Answer: (b) 40

Explanation:
1. The last term is 49 and the common difference is 3.
2. Counting back, the 4th term from the end is 49 − 3 × 3 = 40.

Teacher's Note:
1. The 4th term from the end is the last term minus three common differences.
2. Reverse the AP: 49, 46, 43, 40, ...

Answer: (b) −45/2

Explanation:
1. a = 15 and d = 10 − 25/2 = −5/2.
2. a₁₆ = 15 + 15 × (−5/2) = 15 − 75/2 = −45/2.

Teacher's Note:
1. Find d from any two consecutive terms.
2. Use aₙ = a + (n − 1)d.

Answer: (a) −152

Explanation:
1. a = −2 and d = −7 − (−2) = −5.
2. a₃₁ = −2 + 30 × (−5) = −152.

Teacher's Note:
1. The common difference is negative, so the terms decrease.
2. Use aₙ = a + (n − 1)d.

Answer: (d) 35

Explanation:
1. a + 2d = 11 and a + 7d = 26. Subtracting: 5d = 15, so d = 3 and a = 5.
2. a₁₁ = 5 + 10 × 3 = 35.

Teacher's Note:
1. Subtract the two equations to eliminate a.
2. Check: a₈ = 5 + 21 = 26.

Answer: (b) 32nd

Explanation:
1. a = 8, d = 8. aₙ = 8 + (n − 1) × 8 = 8n.
2. 8n = 256, so n = 32.

Teacher's Note:
1. Here aₙ = 8n, a multiple of 8.
2. 256 ÷ 8 = 32.

Answer: (c) 7

Explanation:
1. a₄ = a + 3d for both, with the same d.
2. Difference = (−1 + 3d) − (−8 + 3d) = 7.

Teacher's Note:
1. With equal common differences, the difference between corresponding terms stays constant.
2. It always equals the difference of the first terms.

Answer: (c) 13/5

Explanation:
1. For three numbers in AP, the middle term is the mean of the other two.
2. a = (6/5 + 4)/2 = (26/5)/2 = 13/5.

Teacher's Note:
1. The condition is 2a = 6/5 + 4.
2. Compare the differences a − 6/5 and 4 − a.

Answer: (c) 508th

Explanation:
1. d = (9 − 499)/(499 − 9) = −1.
2. a₉ = a + 8d = 499 gives a = 507.
3. aₙ = 507 − (n − 1) = 0, so n = 508.

Teacher's Note:
1. The terms fall by 1 each time.
2. Check: a₅₀₈ = 507 − 507 = 0.

Answer: (c) n²

Explanation:
1. a₁ = 1 and aₙ = 2n − 1.
2. Sₙ = n/2 (1 + 2n − 1) = n².

Teacher's Note:
1. Sum of the first n odd numbers is n².
2. Use Sₙ = n(a + l)/2.

Answer: (c) \( \frac{3ab}{2(b − a)} \)

Explanation:
1. d = b − a. The last term: a + (n − 1)(b − a) = 2a, so n − 1 = a/(b − a) and n = b/(b − a).
2. S = n/2 (a + 2a) = \( \frac{b}{2(b − a)} × 3a = \frac{3ab}{2(b − a)} \).

Teacher's Note:
1. Find n from the last term first.
2. Then apply Sₙ = n(first + last)/2.

Answer: (d) −(p + q)

Explanation:
1. Let the AP have first term a and common difference d. Then Sₚ − S_q = a(p − q) + (d/2)(p² − q² − p + q) = (p − q)[a + (d/2)(p + q − 1)].
2. Also Sₚ − S_q = q − p. Dividing by (p − q) gives a + (d/2)(p + q − 1) = −1, that is 2a + (p + q − 1)d = −2.
3. S₍ₚ₊q₎ = (p + q)/2 × [2a + (p + q − 1)d] = (p + q)/2 × (−2) = −(p + q).

Teacher's Note:
1. Subtract the two given sums and factor out (p − q).
2. This is a standard result worth remembering: if Sₚ = q and S_q = p then S₍ₚ₊q₎ = −(p + q).

Answer: (c) 4n + 3

Explanation:
1. aₙ = Sₙ − Sₙ₋₁ = 2n² + 5n − [2(n − 1)² + 5(n − 1)].
2. = 4n − 2 + 5 = 4n + 3.

Teacher's Note:
1. The nth term is the difference of consecutive sums.
2. Check n = 1: S₁ = 7 = 4 + 3.

Answer: (b) 27th

Explanation:
1. aₙ = Sₙ − Sₙ₋₁ = 6n + 2.
2. 6n + 2 = 164, so n = 27.

Teacher's Note:
1. First find the general term.
2. Check: 6 × 27 + 2 = 164.

Answer: (c) 9

Explanation:
1. Sum of interior angles = (n − 2) × 180°. Sum of the AP = n/2 (240 + 5(n − 1)).
2. (n − 2) × 180 = n(235 + 5n)/2, giving n² − 25n + 144 = 0, so n = 9 or 16.
3. n = 16 is impossible since the largest angle would be 195° (more than 180°). So n = 9.

Teacher's Note:
1. An interior angle of a convex polygon is less than 180°.
2. Reject the root that gives an angle above 180°.

Answer: (b) 10

Explanation:
1. a = 23, d = −2, last term 5.
2. 5 = 23 + (n − 1)(−2), so n − 1 = 9 and n = 10.

Teacher's Note:
1. List the terms to check: 23, 21, 19, 17, 15, 13, 11, 9, 7, 5.
2. The common difference is −2.

Answer: (b) 2005

Explanation:
1. 5000 + 200(n − 1) = 7000, so n − 1 = 10 and n = 11.
2. The 11th year starting from 1995 is 2005.

Teacher's Note:
1. The first year (n = 1) is 1995, so the nth year is 1994 + n.
2. 1994 + 11 = 2005.

Answer: (d) 10th

Explanation:
1. 5 + 1.75(n − 1) = 20.75.
2. 1.75(n − 1) = 15.75, so n − 1 = 9 and n = 10.

Teacher's Note:
1. Set up aₙ = a + (n − 1)d.
2. Check: 5 + 15.75 = 20.75.

Answer: (c) 12th

Explanation:
1. a + 2d + a + 14d = (a + 5d) + (a + 10d) + (a + 12d).
2. 2a + 16d = 3a + 27d, so a + 11d = 0.
3. a + 11d is the 12th term.

Teacher's Note:
1. Collect a on one side and d on the other.
2. The nth term is a + (n − 1)d.

Answer: (b) 2, 4, 6

Explanation:
1. Let the numbers be 4 − d, 4, 4 + d (the sum is 12).
2. Sum of cubes: 3 × 64 + 6 × 4 × d² = 192 + 24d² = 288, so d² = 4 and d = 2.
3. The numbers are 2, 4, 6.

Teacher's Note:
1. Choosing the middle term as the centre simplifies the algebra.
2. Check: 8 + 64 + 216 = 288.

Answer: (b) −10

Explanation:
1. S₆ = 600 + 15d. S₁₂ = 1200 + 66d, so the next six terms sum to S₁₂ − S₆ = 600 + 51d.
2. 600 + 15d = 5(600 + 51d), so 600 + 15d = 3000 + 255d.
3. −240d = 2400, d = −10.

Teacher's Note:
1. Next six terms means terms 7 to 12.
2. Check: S₆ = 450 and the next six sum to 90.

Answer: (a) 178

Explanation:
1. 5d = 73 − 38 = 35, so d = 7.
2. t₃₁ = t₁₆ + 15d = 73 + 105 = 178.

Teacher's Note:
1. Find d from the difference of the two known terms.
2. Move from t₁₆ to t₃₁ by adding 15d.

Answer: (b) 999

Explanation:
1. The numbers are 13, 18, 23, ..., 98, an AP with d = 5.
2. 98 = 13 + (n − 1)5, so n = 18.
3. Sum = 18/2 × (13 + 98) = 999.

Teacher's Note:
1. The first two-digit number with remainder 3 is 13.
2. Use Sₙ = n(a + l)/2.

Answer: (d) 3

Explanation:
1. d = (−6 − 9)/3 = −5.
2. a = 4 and b = −1, so a + b = 3.

Teacher's Note:
1. In an AP, the sum of terms equidistant from the ends is constant.
2. a + b = 9 + (−6) = 3.