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

The sum of all three-digit numbers is (494550), and the sum of multiples of (11) is (44550), so the answer is (450000). For not divisible, the complement method is fast.

Subtracting the sum of multiples of (50) from the sum of multiples of (10) gives (39600). Remove the common multiples of both conditions.

The two terms give (d=4) and (a=13), so (S_{60}=7860). Finding the common difference from distant terms is the first step.

The sequence \(25,33,41,\ldots\) is an arithmetic progression because the difference between consecutive terms is always \(8\). Its first term is \(25\), so the \(n\)th term is \(a_n=25+(n-1)8\). The required range contains terms 40 through 70, giving \(70-40+1=31\) terms.

The 40th term is \(25+39(8)=337\), and the 70th term is \(25+69(8)=577\). The sum of a finite AP is \(\frac{\text{number of terms}}{2}(\text{first term} + \text{last term})\). Hence the required sum is \(\frac{31}{2}(337+577)=\frac{31}{2}(914)=31(457)=14167\). Therefore option B is correct; using 30 instead of 31 terms would incorrectly exclude one endpoint.

Use the AP sum formula S_n = n/2[2a + (n − 1)d]. Here n = 41, d = −11, and S_41 = 0. Substitution gives 0 = 41/2[2a + 40(−11)] = 41/2(2a − 440). Since 41/2 is non-zero, the expression in parentheses must be zero. Therefore 2a − 440 = 0, so 2a = 440 and a = 220. Thus option C is correct. A zero sum does not imply that the first term is zero; it means that the positive and negative terms cancel in total. The other choices may arise from using 41d instead of 40d, ignoring the negative sign, or making a division error. Substitution of a = 220 gives the required zero sum.

Answer: A, d = 6. The terms from the 21st through the 40th positions contain 40 − 21 + 1 = 20 terms. The 21st term is a + 20d = 7 + 20d, and the 40th term is a + 39d = 7 + 39d. The sum of these 20 AP terms is number of terms divided by 2 multiplied by the sum of the first and last terms: 3680 = 20/2[(7 + 20d) + (7 + 39d)] = 10(14 + 59d). Divide by 10: 368 = 14 + 59d. Hence 59d = 354 and d = 6. Therefore option A is correct. Option B, C, and D fail when substituted into the same equation. The common confusion is using 20d for the last term; 20 is the number of selected terms, whereas the 40th term has 39 differences after the first term. A useful check is that S_40 − S_20 gives the same block sum.

The given sums give (a=8) and (d=4), so (S_{40}=3440). From two sums, first find the AP values.

Adding sums of multiples of (5) and (8), then subtracting multiples of (40), gives (150500). Avoiding double counting is important.

The numbers are (17,28,\ldots,94), and the sum of (8) terms is (444). A remainder-based AP has common difference equal to the divisor.

The conditions give (a=8) and (d=7), so (S_{16}=968). Convert the given term and sum into two equations.

Apply the AP sum formula S_n = n/2[2a + (n − 1)d]. Here n = 12, a = x, and d = 3x − 2. Therefore 1128 = 12/2[2x + 11(3x − 2)] = 6[2x + 33x − 22] = 6(35x − 22). Dividing by 6 gives 188 = 35x − 22. Hence 35x = 210 and x = 6, so option B is correct. The important point is that the common difference depends on x and must be substituted as 3x − 2, rather than treated as an independent number. Afterward d = 3(6) − 2 = 16, which confirms consistency. Other choices result from omitting the factor 11 or distributing incorrectly.

Subtracting the sum of multiples of (72) from the sum of multiples of (18) gives (28278). Use the least common multiple to remove overlap.

The given sums give (a=3) and (d=5), so (S_{36}=3258). First determine the AP from the smaller sums.

Subtracting the sum of multiples of (36) from the sum of multiples of (9) gives (93753). Numbers divisible by both are multiples of (\operatorname{lcm}(9,12)).

The governing rule for a consecutive range is that the sum from the rth through the sth term equals S_s − S_{r−1}. Thus the sum from the 51st to the 70th term is S_70 − S_50. Calculate S_70 = 8(70²) − 3(70) = 8(4900) − 210 = 38,990. Next, S_50 = 8(50²) − 3(50) = 8(2500) − 150 = 19,850. The required sum is therefore 38,990 − 19,850 = 19,140. Hence option D is correct. Subtracting S_51 would exclude the 51st term, while using an incorrect square or sign would produce the other choices. The endpoint rule ensures that exactly terms 51 through 70 are included.

When the last term \(l\) is known, the sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(3000=\frac{30}{2}(a+150)=15(a+150)\). Hence \(a+150=200\), so \(a=50\). If \(55\) were used, the sum would be \(15(55+150)=3075\), not the given sum. Exam tip: use \(S_n=\frac{n}{2}(a+l)\) directly when the last term is provided.

(a_{21}+a_{40}=a_1+a_{60}=300), so the sum of (20) terms is (3000). Sums of symmetric terms are equal in an AP.

When S_n represents the sum of the first n terms, the sum from the rth term to the sth term is S_s − S_{r−1}. The required range is 61 through 90, so calculate S_90 − S_60. First, S_90 = 4(90²) + 3(90) = 4(8100) + 270 = 32,670. Next, S_60 = 4(60²) + 3(60) = 4(3600) + 180 = 14,580. Therefore the required sum is 32,670 − 14,580 = 18,090. Option A is correct. Subtracting S_61 instead of S_60 would remove the 61st term, and using the wrong square or coefficient would produce a distractor. The use of r − 1 is the key boundary idea.

Answer: B, 155. The sum S_n is cumulative, not the nth term itself. Since S_n = a_1 + a_2 + ... + a_n and S_{n−1} = a_1 + ... + a_{n−1}, subtracting gives a_n = S_n − S_{n−1}. Calculate S_20 = 20(4×20 − 1) = 20(80 − 1) = 20×79 = 1580. Calculate S_19 = 19(4×19 − 1) = 19(76 − 1) = 19×75 = 1425. Therefore a_20 = 1580 − 1425 = 155. Hence option B is correct. Option A, C, and D do not result from the correct difference of consecutive partial sums. A frequent mistake is to call S_20, which equals 1580, the 20th term; it is actually the total of the first twenty terms. Memory cue: a term is the “new amount” added when moving from S_{n−1} to S_n.

Answer: C, 179. Since S_n is the sum through the nth term, subtracting consecutive partial sums isolates one term: a_n = S_n − S_{n−1}. First evaluate S_18 = 5(18²) + 4(18) = 5(324) + 72 = 1620 + 72 = 1692. Then evaluate S_17 = 5(17²) + 4(17) = 5(289) + 68 = 1445 + 68 = 1513. Therefore a_18 = S_18 − S_17 = 1692 − 1513 = 179. Hence option C is correct. Option A, B, and D result from arithmetic or substitution errors. In particular, 1692 is not the answer because it is the total of the first 18 terms. The method works for any sequence described by partial sums, not only when the sequence is explicitly written term by term. Memory cue: consecutive totals differ by the newly added term.

Answer: D, 20. The first partial sum is the first term, so a = S_1. Also, the second term is a_2 = S_2 − S_1, and in an AP the second term equals a + d. Evaluate S_1 = 6(1²) + 2(1) = 6 + 2 = 8, so a = 8. Next, S_2 = 6(2²) + 2(2) = 6×4 + 4 = 28. Therefore a_2 = S_2 − S_1 = 28 − 8 = 20. Since a_2 = a + d, the required value a + d is 20. Thus option D is correct. Option A, B, and C do not equal the second term obtained from the two partial sums. A common error is to use S_2 = 28 as a + d; S_2 is the sum of the first two terms, not the second term alone. Memory cue: S_1 = a and S_2 − S_1 = a + d.