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

Correct answer: A. (2470)

(t_8=67), (t_{26}=193), and there are (19) terms, so the sum is (2470). Exam tip: count the selected terms correctly.

Correct answer: C. 8

For an arithmetic progression, the nth term is \(a_n=S_n-S_{n-1}\). Here, \(a_n=(4n^2+n)-[4(n-1)^2+(n-1)]=8n-3\). Therefore, the difference between consecutive terms is \(a_{n+1}-a_n=8\), so the common difference is 8. Option 7 is not correct; it can result from incorrectly reducing the coefficient of \(n\) by one. Exam tip: first find \(a_n=S_n-S_{n-1}\) from \(S_n\), then subtract consecutive terms.

Correct answer: B. (34)

(S_n=\frac{n}{2}(309-9n)) is positive up to (n=34). Exam tip: solve the inequality and then take the integer limit.

Correct answer: D. (25)

The sum of the first (n) even numbers is (n(n+1)), so (n=25). Exam tip: you can also treat it as the AP (2,4,6,\ldots).

Correct answer: A. (4092)

The direct answer is option A: 4092. These are the multiples of 11 from 44 to 297, so they form an arithmetic progression. Divide the endpoints by 11: 44 = 4 × 11 and 297 = 27 × 11. Thus the sequence is 44, 55, 66, ..., 297, with first term a = 44, common difference d = 11, and last term l = 297. The number of terms is 27 − 4 + 1 = 24. Using the sum formula for an AP, S = n(a + l)/2, we get S = 24(44 + 297)/2 = 12 × 341 = 4092. Option A is correct. Option B, 4212, does not result from the correct AP sum. Option C, 4332, is also an arithmetic error or an incorrect term count. Option D, 4452, likewise does not match the sequence sum. A useful check is 24 terms with average (44 + 297)/2 = 170.5; 24 × 170.5 = 4092. Common mistake: do not count 297 as 27 terms; it is 27 × 11, so the multiplier indices run from 4 through 27.

Correct answer: C. (4158)

The terms are (112,126,\ldots,350), making (18) terms. Exam tip: choose the first and last valid terms carefully.

Correct answer: B. (1539)

These sums give (a=5) and (d=4), so (S_{27}=1539). Exam tip: write (S_n) in terms of (a) and (d).

Correct answer: D. (1100)

The two equations give (a=-2) and (d=6), so (S_{20}=1100). Exam tip: convert term sums into (a) and (d).

Correct answer: A. (4)

The last term is (134), so (d=\frac{134-18}{29}=4). Exam tip: connect the average with the average of first and last terms.

Correct answer: C. 6

Answer: C, d = 6. Use S_n = n/2[2a + (n − 1)d] with a = 9. For six terms, S_6 = 6/2[18 + 5d] = 3(18 + 5d) = 54 + 15d. For twelve terms, S_12 = 12/2[18 + 11d] = 6(18 + 11d) = 108 + 66d. The ratio condition gives (54 + 15d)/(108 + 66d) = 2/7. Cross-multiply: 7(54 + 15d) = 2(108 + 66d), so 378 + 105d = 216 + 132d. Rearranging gives 162 = 27d, hence d = 6. Check: S_6 = 54 + 90 = 144 and S_12 = 108 + 396 = 504; 144:504 reduces by 72 to 2:7. Therefore option C is correct. Options A, B, and D fail this ratio check. The ratio applies to complete partial sums, not directly to individual terms.

Correct answer: B. (11)

The selected (12) terms give (6(2a+245)=1602), so (a=11). Exam tip: treat the selected part as a separate AP.

Correct answer: D. (25897)

The numbers are (114,133,\ldots,988), and there are (47) terms. Exam tip: find the first and last three-digit multiples.

Correct answer: A. (8400)

The total sum is (9870), and the sum of multiples of (7) is (1470), so the answer is (8400). Exam tip: subtract the complementary sum.

Correct answer: C. (25)

(S_{24}=1476) and (S_{25}=1600), so the sum first exceeds (1500) at (n=25). Exam tip: check nearby integers in boundary questions.

Correct answer: B. (4)

From (555=\frac{15}{2}(a+65)), (a=9), and (d=\frac{65-9}{14}=4). Exam tip: use last term and sum to find the first term.

Correct answer: D. \(4\)

For an AP, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(S_5=35+10d\) and \(S_{15}=105+105d\). Using the given ratio, \(\frac{35+10d}{105+105d}=\frac{1}{7}\). On cross-multiplication, \(245+70d=105+105d\), which gives \(d=4\). If \(d=3\), the required ratio of sums is not obtained. Exam tip: simplify each sum before cross-multiplying a ratio equation.

Correct answer: A. (13)

Equating both sums gives (4n+16=2n+42), so (n=13). Exam tip: when (n) is common, cancel (\frac{n}{2}).

Correct answer: C. (1240)

The required sum is (S_{25}-S_{15}=1975-735=1240). Exam tip: use the difference of cumulative sums for middle terms.

Correct answer: B. 13

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(999=\frac{18}{2}[2a+17\times5]=9(2a+85)\). Hence \(2a+85=111\), so \(2a=26\) and \(a=13\). If 15 is used as the first term, the sum will not be 999. Exam tip: in an \(S_n\) question, carefully use \((n-1)d\), not \(nd\).

Correct answer: D. (720)

(S_n=3n(31-n)), and the maximum (720) occurs at (n=15) or (n=16). Exam tip: check integer values near the vertex.

Correct answer: A. (29)

The direct answer is option A, 29. The AP is 7, 14, 21, so its first term is a = 7 and common difference is d = 7. The nth term is t_n = a + (n-1)d = 7 + 7(n-1) = 7n. The sum of the first n terms is S_n = n/2[2a + (n-1)d] = n/2[14 + 7(n-1)] = 7n(n+1)/2. Therefore, S_n/t_n = [7n(n+1)/2]/7n = (n+1)/2. The question says this ratio is 15, so (n+1)/2 = 15, giving n+1 = 30 and n = 29. Option A is correct. Option B, 30, is one more than the required value. Option C, 31, and option D, 32, also result from continuing the arithmetic incorrectly. Memory cue: for an AP beginning 7,14,21, the ratio S_n/t_n is (n+1)/2.

Correct answer: C. (780)

The numbers are (10,16,\ldots,94), and there are (15) terms. Exam tip: choose the first term of the remainder sequence carefully.

Correct answer: B. ( -8 )

From (0=\frac{25}{2}[192+24d]), (d=-8). Exam tip: in zero-sum questions, set the bracket equal to zero.

Correct answer: D. (1386)

There will be (21) terms, and the sum is (\frac{21}{2}(6+126)=1386). Exam tip: when arithmetic means are inserted, the total number of terms increases.

Correct answer: A. (1392)

For an AP with an odd number of terms, the sum is the product of the number of terms and the middle term. Exam tip: remember the middle-term property.