CBSE Class 10 Mathematics Arithmetic Progressions MCQs Set 17

Answer: (d) \( \frac{7}{5} \)

Explanation:
1. Three terms a, b, c are in A.P. when 2b = a + c.
2. 2p = \( \frac{4}{5} \) + 2 = \( \frac{14}{5} \), so p = \( \frac{7}{5} \).

Teacher's Note:
1. The middle term is the average of its neighbours.
2. Check: 7/5 − 4/5 = 3/5 = 2 − 7/5.

Answer: (a) Rs. [12000 + 23(5500)]

Explanation:
1. The fund starts with Rs. 12000 at the end of January 2019.
2. Rs. 5500 is added at the start of each month from February 2019 to December 2020, which is 12 + 11 = 23 months.
3. Total = Rs. [12000 + 23(5500)] = Rs. 1,38,500.

Teacher's Note:
1. Count the months carefully: February 2019 to December 2020.
2. The initial Rs. 12000 is added only once.

Answer: (c) Rs. 2525750

Explanation:
1. The wife's Rs. 125000 is the first donation. From the brother onwards the donations form an A.P. with a = 54000 and d = 5250.
2. The remaining 22 donations: \( S_{22} = \frac{22}{2}[2(54000) + 21(5250)] = 11(108000 + 110250) = 2400750 \).
3. Total = 125000 + 2400750 = Rs. 2525750.

Teacher's Note:
1. The wife's donation does not follow the common difference, so add it separately.
2. Count 22 terms of the A.P., not 23.

Answer: (b) 11th year

Explanation:
1. Salaries form an A.P. with a = 5000 and d = 200.
2. \( a_n = 5000 + (n-1)200 = 7000 \), so n − 1 = 10.
3. n = 11, i.e. in the 11th year (after 10 increments).

Teacher's Note:
1. Count the first year as the 1st term.
2. n − 1 is the number of increments.

Answer: (a) 2s

Explanation:
1. Number of terms: from S = \( \frac{n}{2}(p+q) = s \), n = \( \frac{2s}{p+q} \).
2. Common difference d = \( \frac{q-p}{n-1} = \frac{(q-p)(p+q)}{2s-(p+q)} = \frac{q^2-p^2}{2s-(p+q)} \).
3. Comparing with the given expression, r = 2s.

Teacher's Note:
1. Use S = n(p + q)/2 to find n first.
2. Then d = (last − first)/(n − 1).

Answer: (b) \( 2m - \frac{9}{m} \)

Explanation:
1. a = \( 2m + \frac{1}{m} \) and d = \( -\frac{2}{m} \).
2. \( a_6 = a + 5d = 2m + \frac{1}{m} - \frac{10}{m} \) = \( 2m - \frac{9}{m} \).

Teacher's Note:
1. d = second term − first term.
2. a₆ needs 5 common differences.

Answer: (d) Yes, −40

Explanation:
1. Here d = −7 − (−3) = −4.
2. \( a_{30} - a_{20} = (a + 29d) - (a + 19d) = 10d = -40 \).

Teacher's Note:
1. The difference of two terms is (difference in positions) × d.
2. The first term cancels out.

Answer: (c) 23rd

Explanation:
1. a = 17 and d = \( -\frac{4}{5} \).
2. \( a_n = 17 - (n-1)\frac{4}{5} < 0 \), so n − 1 > 21.25, i.e. n − 1 ≥ 22.
3. The first negative term is the 23rd term, \( a_{23} = 17 - \frac{88}{5} = -\frac{3}{5} \).

Teacher's Note:
1. Solve the inequality a_n < 0 and take the smallest whole number n.
2. Check the 22nd term: 17 − 21 × 0.8 = 0.2 > 0.

Answer: (b) 30

Explanation:
1. a = 10 and d = 2, \( a_n = 10 + (n-1)2 = 68 \).
2. (n − 1) × 2 = 58, so n − 1 = 29 and n = 30.

Teacher's Note:
1. Weekly savings form an A.P.
2. Check: 10 + 29 × 2 = 68.

Answer: (d) 100

Explanation:
1. The nth terms are a₁ + (n − 1)d and a₂ + (n − 1)d.
2. Their difference is a₁ − a₂, whatever the value of n.
3. So the difference between the 1000th terms is also 100.

Teacher's Note:
1. Equal common differences keep the gap between corresponding terms fixed.
2. Do not scale the difference with n.

Answer: (c) \( \sqrt{2} \) and \( 3 + 4\sqrt{2} \)

Explanation:
1. Each term exceeds the previous one by √2, so it is an A.P. with d = √2.
2. The next three terms are 3 + 4√2, 3 + 5√2, 3 + 6√2.

Teacher's Note:
1. Check that all consecutive differences are equal before calling it an A.P.
2. Add d to the last term to get the next.

Answer: (a) 128

Explanation:
1. The first three-digit multiple of 7 is 105 (= 7 × 15) and the last is 994 (= 7 × 142).
2. \( 994 = 105 + (n-1)7 \), so n − 1 = 127 and n = 128.

Teacher's Note:
1. Find the first and last multiples inside the range.
2. Equivalent count: 142 − 15 + 1 = 128.

Answer: (c) 5

Explanation:
1. The 10th terms are 2 + 9d and 7 + 9d.
2. Their difference is 7 − 2 = 5.
3. The same difference holds for any pair of corresponding terms.

Teacher's Note:
1. The common difference cancels out.
2. So the difference of any two corresponding terms is 5.

Answer: (a) 234

Explanation:
1. Trees by one section of each class: 1 + 2 + 3 + … + 12 = \( \frac{12 \times 13}{2} \) = 78.
2. Three sections in each class: 3 × 78 = 234.

Teacher's Note:
1. Use n(n + 1)/2 for the sum of the first n natural numbers.
2. Multiply by 3 sections.

Answer: (b) 73

Explanation:
1. The numbers are 11, 15, 19, …, 299, an A.P. with a = 11 and d = 4.
2. \( 299 = 11 + (n-1)4 \), so n − 1 = 72 and n = 73.

Teacher's Note:
1. The first number is 11 (4 × 2 + 3).
2. The last number below 300 is 299.

Answer: (d) 5050

Explanation:
1. \( S_{100} = \frac{100}{2}(1+100) = 50 \times 101 \).
2. S₁₀₀ = 5050.

Teacher's Note:
1. Use S = n(a + l)/2.
2. Gauss pairing: 50 pairs each summing to 101.

Answer: (a) \( \frac{1}{m} + n - 1 \)

Explanation:
1. a = \( \frac{1}{m} \) and d = \( \frac{1+m}{m} - \frac{1}{m} = 1 \).
2. \( a_n = \frac{1}{m} + (n-1)(1) = \frac{1 + (n-1)m}{m} \).

Teacher's Note:
1. Find d by subtracting the first term from the second.
2. Here d = 1, not m.

Answer: (d) 20

Explanation:
1. a = −14 and a₅ = a + 4d = 2, so 4d = 16 and d = 4.
2. \( 62 = -14 + (n-1)4 \), so n − 1 = 19 and n = 20.

Teacher's Note:
1. Use the fifth term to find d first.
2. Then apply the nth term formula to the last term.

Answer: (c) 31st

Explanation:
1. d = 12. The required term exceeds the 21st term by 120, i.e. by 120 ÷ 12 = 10 terms.
2. So it is the (21 + 10) = 31st term.

Teacher's Note:
1. Each step along the A.P. adds d = 12.
2. Check: a₃₁ − a₂₁ = 10 × 12 = 120.

Answer: (b) 18th

Explanation:
1. a = 65 and d = −4.
2. \( a_n = 65 - 4(n-1) < 0 \), so n − 1 > 16.25, hence n = 18.
3. \( a_{18} = 65 - 68 = -3 \).

Teacher's Note:
1. Solve 65 − 4(n − 1) < 0 and take the next whole number.
2. a₁₇ = 1 is still positive.

Answer: (c) 400

Explanation:
1. 7a + 21d = 49, so a + 3d = 7. 17a + 136d = 289, so a + 8d = 17.
2. Subtracting: 5d = 10, so d = 2 and a = 1.
3. \( S_{20} = \frac{20}{2}[2(1) + 19(2)] = 10 \times 40 = 400 \).

Teacher's Note:
1. Form two equations in a and d.
2. Here Sₙ = n², so S₂₀ = 400.

Answer: (a) 19

Explanation:
1. Coins per day are 1, 2, 3, …, so \( \frac{n(n+1)}{2} = 190 \).
2. n² + n − 380 = 0, i.e. (n − 19)(n + 20) = 0, so n = 19.

Teacher's Note:
1. Reject the negative root.
2. Total money saved = 190 × Rs. 5 = Rs. 950.

Answer: (b) 38

Explanation:
1. \( S_n = \frac{n}{2}(a + a_n) = \frac{n}{2}(21) = 399 \).
2. n = 399 × 2 ÷ 21 = 38.

Teacher's Note:
1. Use Sₙ = n(a + l)/2 when the last term is given.
2. Check: 38 × 21/2 = 399.

Answer: (d) 137

Explanation:
1. d = 4 − (−3) = 7.
2. \( a_{21} = -3 + 20 \times 7 = 137 \).

Teacher's Note:
1. The 21st term uses 20 common differences.
2. Do not use 21 differences.

Answer: (d) 25

Explanation:
1. \( a_{18} - a_{13} = 5d \) (5 terms apart).
2. = 5 × 5 = 25.

Teacher's Note:
1. Difference of terms = (gap in positions) × d.
2. 18 − 13 = 5.

Answer: (a) 7

Explanation:
1. The 4th terms are −1 + 3d and −8 + 3d.
2. Their difference is (−1) − (−8) = 7.

Teacher's Note:
1. The common difference cancels.
2. The difference equals the difference of the first terms.

Answer: (c) −320

Explanation:
1. a = 10, d = −4.
2. \( S_{16} = \frac{16}{2}[2(10) + 15(-4)] = 8(20 - 60) = 8(-40) = -320 \).

Teacher's Note:
1. Substitute d with its sign.
2. Check using a₁₆ = 10 − 60 = −50: 8(10 − 50) = −320.

Answer: (b) 33

Explanation:
1. a + d = 13 and a + 4d = 25, so 3d = 12 and d = 4, a = 9.
2. a₇ = 9 + 6 × 4 = 33.

Teacher's Note:
1. Subtract the two equations to eliminate a.
2. Check: a₇ = a₅ + 2d = 25 + 8 = 33.

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

Explanation:
1. Sum of first n odd numbers = n².
2. Sum of first n even numbers = n(n + 1).
3. n² = k × n(n + 1), so \( k = \frac{n}{n+1} \).

Teacher's Note:
1. Use 1 + 3 + … = n² and 2 + 4 + … = n(n + 1).
2. Cancel n from both sides.

Answer: (a) \( \sqrt{112} \)

Explanation:
1. √28 = 2√7 and √63 = 3√7, so the terms are √7, 2√7, 3√7.
2. d = √7, so the next term is 4√7 = √112.

Teacher's Note:
1. Simplify the surds first.
2. Express every term as a multiple of √7.

Answer: (c) \( (q-p)\left[d - \frac{1}{pq}\right] \)

Explanation:
1. First A.P.: second term = \( \frac{1}{q} - (p-2)d \). Second A.P.: second term = \( \frac{1}{p} - (q-2)d \).
2. Difference = \( \frac{1}{q} - \frac{1}{p} + (q-p)d = \frac{p-q}{pq} + (q-p)d \).
3. = \( (q-p)\left[d - \frac{1}{pq}\right] \).

Teacher's Note:
1. Write each second term using a₂ = a_k − (k − 2)d.
2. Take (q − p) common at the end.

Answer: (d) \( \sqrt{1500} \)

Explanation:
1. The terms are √15, 2√15, 3√15, 4√15, …, so a = d = √15.
2. \( a_{10} = 10\sqrt{15} = \sqrt{100 \times 15} = \sqrt{1500} \).

Teacher's Note:
1. Spot that each term is a multiple of √15.
2. Take 10 inside the root as √100.

Answer: (b) (12x + 6), (8x − 4), (3x + 8), (7x + 4), …

Explanation:
1. Option (a): differences are 2x + 2 each time. Option (b): differences are 10x + 10. Option (c): differences are 3x + 5.
2. Option (d): (8x − 4) − (12x + 6) = −4x − 10, but (3x + 8) − (8x − 4) = −5x + 12. These are not equal.
3. So the series in option (d) is not an A.P.

Teacher's Note:
1. Compare every consecutive difference.
2. Even one unequal pair is enough to reject an A.P.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Multiply a, b, c by \( \frac{1}{abc} \): we get \( \frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab} \), which is an A.P. by the Reason. Assertion is true.
2. Multiplying all terms of an A.P. by a non-zero constant multiplies d by that constant, so the sequence stays an A.P. Reason is true.
3. Reason explains Assertion.

Teacher's Note:
1. The Reason is the exact rule used.
2. Multiply by 1/(abc) to see the result.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. 2(3k + 3) = (2k + 1) + (5k − 1), so 6k + 6 = 7k and k = 6. Assertion is true.
2. The formula for Sₙ is correct, so Reason is true.
3. But the sum formula is not used to find k, so Reason does not explain Assertion.

Teacher's Note:
1. The condition 2b = a + c is what is used.
2. Both true, but unrelated: choose (b).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( S_{38} = \frac{38}{2}(17 + 350) = 19 \times 367 = 6973 \). Assertion is true.
2. Reason is the standard formula and it is exactly what is used, so it explains the Assertion.

Teacher's Note:
1. Use Sₙ = n(a + l)/2 when both end terms are known.
2. 19 × 367 = 6973.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. a, b, c are in A.P. ⇔ b − a = c − b ⇔ 2b = a + c. Assertion is true.
2. 1 + 3 + 5 + … + (2n − 1) = n². Reason is true.
3. The Reason does not explain the Assertion.

Teacher's Note:
1. Check each statement on its own first.
2. Related to A.P., but unrelated to the condition 2b = a + c.

Answer: (c) A is true, but R is false.

Explanation:
1. The common difference is \( \frac{5}{2} \) throughout, so the list is an A.P. Assertion is true.
2. The terms of an A.P. can be positive, negative or zero (this very A.P. has both signs), so Reason is false.

Teacher's Note:
1. Check the common difference for the Assertion.
2. An A.P. with positive d passes through negative numbers to positive numbers.

Answer: (c) A is true, but R is false.

Explanation:
1. Take a = −√2 and d = √2: the terms are −√2, 0, √2, …; the 2nd term 0 is rational. So Assertion is true.
2. The sum of a rational and an irrational number is always irrational, so Reason is false.

Teacher's Note:
1. If a rational + irrational were rational, the irrational would itself be rational.
2. Here the rational term appears because a is irrational and a + (n − 1)d cancels it.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( \frac{S_n}{S_n'} = \frac{2a + (n-1)d}{2a' + (n-1)d'} = \frac{5n+1}{7n+27} \). Put n = 2m − 1 to get \( \frac{a_m}{a_m'} \).
2. \( \frac{5(2m-1)+1}{7(2m-1)+27} = \frac{10m-4}{14m+20} = \frac{5m-2}{7m+10} \). Assertion is true.
3. Reason is the sum formula that the method depends on, so it explains the Assertion.

Teacher's Note:
1. Replace n by 2m − 1 to turn the ratio of sums into the ratio of mth terms.
2. Sₙ/Sₙ′ = (2a + (n − 1)d)/(2a′ + (n − 1)d′) after cancelling n/2.