Finding the sum of the first $n$ terms of an AP

Correct answer: A. (1060)

The selected terms are (a_4,a_8,\ldots,a_{40}), and their sum is (1060). In position-based questions, form the new AP of selected terms.

Correct answer: B. (439200)

The sum of all three-digit numbers is (494550), and the sum of multiples of (9) is (55350), so the answer is (439200). For not divisible, the complement method is fast.

Correct answer: C. (33368)

Subtracting the sum of multiples of (24) from the sum of multiples of (8) gives (33368). In a but-not condition, subtract the complement.

Correct answer: D. (3120)

The two terms give (d=4) and (a=0), so (S_{40}=3120). Finding the common difference from distant terms is the first step.

Correct answer: A. (8021)

The sequence is an arithmetic progression with first term 18 and common difference 7. Its nth term is \(a_n=18+(n-1)7\). The requested terms are the 30th through the 55th, so their number is \(55-30+1=26\). The first term of this block is \(a_{30}=221\), and the last is \(a_{55}=396\). The sum of consecutive AP terms equals the number of terms multiplied by the average of the first and last terms.

Hence the sum is \(26\times\frac{221+396}{2}=13\times617=8021\). The same result follows from \(S_{55}-S_{29}\), because subtracting through the 29th term leaves precisely terms 30 to 55. Therefore option A is correct. Subtracting \(S_{30}\) would incorrectly omit the 30th term, so the endpoint handling is important.

Correct answer: B. 135

Direct answer: 135, so option B is correct. Use Sₙ = n/2[2a + (n − 1)d]. Here n = 31, d = −9, and S₃₁ = 0. Therefore 0 = 31/2[2a + 30(−9)] = 31/2(2a − 270). Since 31/2 is not zero, the bracket must be zero: 2a − 270 = 0, so 2a = 270 and a = 135. Option B is correct. Option A, 126, option C, 144, and option D, 153, would leave the bracket nonzero and hence would not make S₃₁ equal to zero. A conceptual check is useful: 31 is odd, so the 16th term is the middle term. In an odd-numbered AP, symmetric terms have sums twice the middle term; for the total to be zero, the middle term must be zero. Thus a + 15d = 0, giving a − 135 = 0 and a = 135. Memory cue: for an odd number of AP terms with zero sum, the middle term is zero.

Correct answer: C. 4

Direct answer: d = 4, so option C is correct. The terms from the 16th through the 30th are counted inclusively, so their number is 30 − 16 + 1 = 15. The first term of this block is t₁₆ = a + 15d = 5 + 15d, and the last is t₃₀ = a + 29d = 5 + 29d. Apply the AP sum formula to these 15 terms: 1395 = 15/2[(5 + 15d) + (5 + 29d)] = 15/2(10 + 44d) = 75 + 330d. Hence 330d = 1320 and d = 4. Option C is correct. Option A, 2, option B, 3, and option D, 5, do not satisfy the given sum when substituted. A common mistake is to count 14 rather than 15 terms by forgetting the +1 in 30 − 16 + 1. Another valid approach is to write the required sum as S₃₀ − S₁₅. Memory cue: whenever both endpoints are included, number of terms = last index − first index + 1.

Correct answer: D. (1925)

The given sums give (a=4) and (d=3), so (S_{35}=1925). From two sums, first find the AP values.

Correct answer: A. (87850)

Adding sums of multiples of (4) and (7), then subtracting multiples of (28), gives (87850). Avoiding double counting is important.

Correct answer: B. (795)

The numbers are (11,17,\ldots,95), and the sum of (15) terms is (795). A remainder-based AP has common difference equal to the divisor.

Correct answer: C. (707)

The conditions give (a=18) and (d=5), so (S_{14}=707). Convert the given term and sum into two equations.

Correct answer: D. 4

Answer: D, x = 4. For an arithmetic progression, the sum of the first n terms is S_n = n/2[2a + (n − 1)d], where a is the first term and d is the common difference. Here n = 10, a = x, and d = 2x + 1. Substitute these values: 445 = 10/2[2x + 9(2x + 1)] = 5[2x + 18x + 9] = 5(20x + 9). Dividing by 5 gives 89 = 20x + 9, so 20x = 80 and x = 4. Check: when x = 4, a = 4 and d = 9; hence S_10 = 5[8 + 81] = 445. Option A gives S_10 = 225, option B gives 335, and option C gives 555, so none of them works. Memory cue: for n terms, use (n − 1)d, not nd.

Correct answer: A. (26460)

Subtracting the sum of multiples of (36) from the sum of multiples of (12) gives (26460). Use the least common multiple to remove overlap.

Correct answer: B. (1404)

The given sums give (a=1) and (d=5), so (S_{24}=1404). First determine the AP from the smaller sums.

Correct answer: C. (66336)

Subtracting the sum of multiples of (30) from the sum of multiples of (6) gives (66336). Numbers divisible by both (6) and (15) are multiples of (30).

Correct answer: D. 6765

Answer: D, 6765. S_n denotes the sum of the first n terms. To add terms from the rth through the sth term, subtract the sum before the rth term: S_s − S_{r−1}. Therefore, the required sum is S_45 − S_30, because the 31st term must be included and the first 30 terms must be removed. Calculate S_45 = 6(45²) + 45 = 6(2025) + 45 = 12,150 + 45 = 12,195. Next, S_30 = 6(30²) + 30 = 6(900) + 30 = 5,400 + 30 = 5,430. Thus the required sum is 12,195 − 5,430 = 6,765. Option A, B, and C do not equal this difference. A common boundary mistake is to use S_45 − S_31; that removes the 31st term as well, giving the wrong range. Memory cue: for terms r to s, always write S_s − S_{r−1}.

Correct answer: A. (17)

The direct answer is option A, 17. For n terms of an AP, S_n=n/2(first term+last term). The given last term is the 25th term, so 1625=25/2(a+113). Multiply by 2: 3250=25(a+113). Divide by 25: 130=a+113. Hence a=17. Option A, 17, satisfies the equation exactly. Option B, 15, would give S_25=25/2(128)=1600, not 1625. Option C, 19, would give 25/2(132)=1650, not 1625. Option D, 21, would give 25/2(134)=1675, also incorrect. The important idea is that the sum of an AP equals the number of terms multiplied by the average of the first and last terms. Since the 25th term is the last term among the first 25 terms, it must be paired with the first term. Memory cue: for S_n, use t_n as the last term, not t_{n+1}.

Correct answer: B. (2100)

(a_{11}+a_{30}=a_1+a_{40}=210), so the sum of (20) terms is (2100). Sums of symmetric terms are equal in an AP.

Correct answer: C. (3523)

The direct answer is option C: 3523. For an arithmetic progression, the sum of the first n terms is \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Here the first term is a=23, the common difference is d=9, and the number of terms is n=26. First calculate n-1=25. Then calculate (n-1)d=25×9=225. Next calculate 2a=2×23=46. Add inside the bracket: 46+225=271. Finally, multiply by n/2: \\(S_{26}=\\frac{26}{2}×271=13×271=3523\\). Therefore option C is correct. Option A, 3487, is wrong because it results from an arithmetic or substitution error. Option B, 3505, is also incorrect because the bracketed total should be 271, not the value needed to produce 3505. Option C matches the complete calculation. Option D, 3541, is wrong because it is larger than the correctly calculated sum and usually comes from adding an extra amount or using the wrong difference. Another check is possible: the last term is \\(a_{26}=23+25×9=248\\); the average of first and last terms is \\((23+248)/2=135.5\\), and 26×135.5=3523. Exam cue: calculate (n-1)d separately and keep every bracket clear.

Correct answer: A. (3116)

The required sum is (S_{36}-S_{17}=3116). To find a middle block sum, subtract the previous partial sum.

Correct answer: D. (2494)

First, (184=-12+(n-1)7) gives (n=29), and the sum is (2494). When the last term is given, find (n) first.

Correct answer: B. (16218)

The numbers are (153,170,\ldots,748), and their sum is (16218). Choose the first and last multiples within the limits correctly.

Correct answer: C. (193875)

This is an AP with (a=3200), (d=225), (n=30), and the total is (193875). In word problems, treat each amount as a term.

Correct answer: A. (25897)

The first number is (114), the last is (988), and there are (47) terms, so the sum is (25897). In divisibility questions, choose the first and last values carefully.

Correct answer: D. (1441)

The direct answer is option D: 1441. The sequence 160, 151, 142, … is an arithmetic progression because each term decreases by 9, so a=160, d=151−160=−9, and n=22. Use \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). First, n−1=21. Next, (n−1)d=21×(−9)=−189. Also, 2a=2×160=320. Therefore the bracket is 320−189=131. Now \\(S_{22}=\\frac{22}{2}×131=11×131=1441\\). Thus option D is correct. Option A, 1397, is wrong because it does not follow from the correct negative difference calculation. Option B, 1419, is also incorrect and may arise from a small subtraction error. Option C, 1463, is wrong because it is too large; using d=+9 instead of d=−9 would incorrectly increase the sum. Option D is correct because a decreasing AP must use a negative common difference. A second check uses the last term: \\(a_{22}=160+21(−9)=−29\\). The average of the first and last terms is \\((160−29)/2=65.5\\), and 22×65.5=1441. Remember: when terms decrease, d is negative, even though the terms themselves may begin with a large positive number.