CBSE Class 10 Mathematics Arithmetic Progressions MCQs Set 21

Answer: (b) \( 101\ \text{and}\ \frac{3}{2} \)

Explanation:
1. 7171 = n × 142 / 2 = 71n, so n = 101.
2. 146 = −4 + (101 − 1)d, so 100d = 150.
3. d = 3/2.

Teacher's Note:
1. Use S = n(a + l)/2 to find n.
2. Then use l = a + (n − 1)d.

Answer: (a) 6 and 10, so it is not an AP

Explanation:
1. a₁ = 3, a₂ = 9, a₃ = 19. The differences are 6 and 10, which are not equal, so the sequence is not an AP.

Teacher's Note:
1. Compute the first three terms.
2. An AP has equal consecutive differences.

Answer: (d) 500

Explanation:
1. a = 2, d = 3/2. S₂₅ = 25/2 [4 + 24 × 3/2] = 25/2 × 40 = 500.

Teacher's Note:
1. Find the common difference first.
2. Use S = n/2 [2a + (n − 1)d].

Answer: (c) −465

Explanation:
1. a₁ = 4 and a₁₅ = 9 − 75 = −66. S = 15 × (4 − 66)/2 = −465.

Teacher's Note:
1. Find the first and last of the 15 terms.
2. S = n(first + last)/2.

Answer: (b) 2, 7, 12, 17, ...

Explanation:
1. (a + 12d) − (a + 6d) = 30, so d = 5.
2. a = 32 − 30 = 2.
3. The AP is 2, 7, 12, 17, ...

Teacher's Note:
1. Subtract the two equations to find d.
2. Then substitute to find a.

Answer: (a) 10

Explanation:
1. S = n/2 [2(3) + (n − 1)2] = n² + 2n.
2. n² + 2n − 120 = 0.
3. (n − 10)(n + 12) = 0, so n = 10.

Teacher's Note:
1. a = 3, d = 2.
2. Reject the negative root.

Answer: (d) 3

Explanation:
1. The middle term is the average: 2(3k) = (k + 1) + (4k + 2), so 6k = 5k + 3 and k = 3. The terms are 4, 9, 14.

Teacher's Note:
1. Use 2 × middle = sum of the neighbours.
2. Check the common difference.

Answer: (c) \( n^2 \)

Explanation:
1. The odd numbers 1, 3, 5, ... form an AP with a = 1 and d = 2, so S = n/2 [2 + 2(n − 1)] = n².

Teacher's Note:
1. a = 1, d = 2.
2. Check with n = 3: 1 + 3 + 5 = 9.

Answer: (b) 11, 7, 3, −1, −5, −9

Explanation:
1. The AP has 8 terms with a = 15 and a₈ = −13. So 7d = −28 and d = −4. The means are 11, 7, 3, −1, −5, −9.

Teacher's Note:
1. Six means give 8 terms in all.
2. d = (b − a)/(k + 1).

Answer: (a) 102 terms; 331

Explanation:
1. n = (407 − 3)/4 + 1 = 102. The 20th term from the end is 407 − 19 × 4 = 331.

Teacher's Note:
1. Use n = (l − a)/d + 1.
2. From the end, the common difference is −4.

Answer: (d) \( 2a_m \)

Explanation:
1. a_{m+n} + a_{m−n} = (a + (m + n − 1)d) + (a + (m − n − 1)d) = 2a + 2(m − 1)d = 2a_m.

Teacher's Note:
1. Write each term using a + (k − 1)d.
2. The n terms cancel.

Answer: (c) \( 2S_2 \)

Explanation:
1. S₁ = n(n + 1)/2, S₂ = n², S₃ = n(3n − 1)/2. Then S₁ + S₃ = n(4n)/2 = 2n² = 2S₂.

Teacher's Note:
1. Compute each sum with S = n/2 [2a + (n − 1)d].
2. Add S₁ and S₃.

Answer: (b) 4

Explanation:
1. a + 4d = 0, so a = −4d. Then a₃₃ = a + 32d = 28d and a₁₂ = a + 11d = 7d, so the ratio is 4.

Teacher's Note:
1. Use a = −4d.
2. Substitute in both terms.

Answer: (a) 0

Explanation:
1. S_p/p = A + (p − 1)d/2 where A is the first term. So the expression is Σ [A + (p − 1)d/2](q − r) = A·0 + (d/2) Σ (p − 1)(q − r) = 0 since Σ p(q − r) = 0 and Σ (q − r) = 0.

Teacher's Note:
1. a/p is the average of the first p terms, a linear function of p.
2. A linear function of p gives zero in the cyclic sum.

Answer: (d) \( 6n - 1 \)

Explanation:
1. a_n = S_n − S_{n−1} = 3n² + 2n − [3(n − 1)² + 2(n − 1)] = 6n − 1.

Teacher's Note:
1. Use a_n = S_n − S_{n−1}.
2. Check a₁ = S₁ = 5.

Answer: (c) 11

Explanation:
1. First mean = 20 + d and last mean = 80 − d.
2. 3(20 + d) = 80 − d, so 4d = 20 and d = 5.
3. 60/(n + 1) = 5, so n = 11.

Teacher's Note:
1. d = (80 − 20)/(n + 1).
2. Form the ratio equation.

Answer: (b) \( 1 + \frac{1}{n} \)

Explanation:
1. Even sum = n(n + 1), odd sum = n², so the ratio is (n + 1)/n = 1 + 1/n.

Teacher's Note:
1. Use n(n + 1) and n².

Answer: (a) \( \frac{n\,[1 + (n-2)\sqrt{x}]}{1 - x} \)

Explanation:
1. a = (1 − √x)/(1 − x); d = 2√x/(1 − x). The third term minus the second is also 2√x/(1 − x).
2. S = n/2 × [2(1 − √x) + 2(n − 1)√x]/(1 − x).
3. S = n[1 + (n − 2)√x]/(1 − x).

Teacher's Note:
1. Find d from the first two terms; check with the third.
2. Put everything over 1 − x.

Answer: (d) 4, 9, 14 (or 14, 9, 4)

Explanation:
1. Take a − d, a, a + d: 3a = 27 so a = 9. Product 9(81 − d²) = 504 gives d² = 25, so d = ±5.

Teacher's Note:
1. Take the numbers as a − d, a, a + d.
2. The sum gives the middle term.

Answer: (c) \( (2m-1) : (2n-1) \)

Explanation:
1. Since S_n/S_m... the condition gives S_n = kn², so a_n = k(n² − (n − 1)²) = k(2n − 1). Hence a_m : a_n = (2m − 1) : (2n − 1).

Teacher's Note:
1. S proportional to n² means S = kn².
2. Use a_n = S_n − S_{n−1}.

Answer: (b) 0

Explanation:
1. S_p = S_q gives (p − q)[2a + (p + q − 1)d]/2 = 0, so 2a + (p + q − 1)d = 0, hence S_{p+q} = (p + q)/2 × 0 = 0.

Teacher's Note:
1. Subtract S_q from S_p and factor out p − q.

Answer: (a) 3

Explanation:
1. a + 7d = 0 so a = −7d. a₃₈ = 30d and a₁₈ = 10d, so the ratio is 3.

Teacher's Note:
1. Express a in terms of d.
2. Substitute for both terms.

Answer: (d) \( (n+1) : n \)

Explanation:
1. Odd positions give n + 1 terms with average equal to the middle term; even positions give n terms with the same average (the middle term). So the sums are in the ratio (n + 1) : n.

Teacher's Note:
1. The middle term is the average of each group.
2. Count the terms in each group.

Answer: (c) 17

Explanation:
1. S₂ = 24 and S₁ = 7, so a₂ = 24 − 7 = 17.

Teacher's Note:
1. a₂ = S₂ − S₁.

Answer: (b) x = 1

Explanation:
1. The terms 1, 4, 7, ..., 28 form 10 terms with sum 10 × 29/2 = 145. So 10x + 145 = 155 and x = 1.

Teacher's Note:
1. Count the terms: 1 + 3(k − 1) = 28 gives k = 10.
2. Add the x terms and the constants separately.

Answer: (a) 2

Explanation:
1. a_n = n² − (n − 1)² = 2n − 1, so d = a₂ − a₁ = 3 − 1 = 2.

Teacher's Note:
1. a_n = S_n − S_{n−1}.

Answer: (d) 1

Explanation:
1. For n = 1 the expression is (a + b)/(1 + 1) = (a + b)/2, the arithmetic mean. For n = 0 it is 2/(1/a + 1/b), the harmonic mean, so n = 1.

Teacher's Note:
1. Test n = 1 and n = 0.
2. The arithmetic mean is (a + b)/2.

Answer: (c) \( \frac{n-1}{\sqrt{a_1}+\sqrt{a_n}} \)

Explanation:
1. 1/(√a_i + √a_{i+1}) = (√a_{i+1} − √a_i)/d.
2. Sum = (√a_n − √a_1)/d, and d = (a_n − a_1)/(n − 1).
3. Sum = (n − 1)(√a_n − √a_1)/(a_n − a_1) = (n − 1)/(√a_n + √a_1).

Teacher's Note:
1. Rationalise each term using d = a_{i+1} − a_i.
2. The sum telescopes.