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

The governing concept is the sum of a finite arithmetic progression. The first term is a = 7, and the common difference is d = 15 − 7 = 8. The fifth term is a₅ = a + (5 − 1)d = 7 + 4 × 8 = 39. Therefore, the five terms are 7, 15, 23, 31, and 39. Adding them gives 7 + 15 + 23 + 31 + 39 = 115. The formula provides an independent check: Sₙ = n/2[2a + (n − 1)d], so S₅ = 5/2[14 + 32] = 5/2 × 46 = 115. Thus option C is correct. Options A, B, and D arise from incomplete addition, using an incorrect fifth term, or making an arithmetic error; none agrees with the direct sum and the formula.

Direct answer: option D, 155. The first term is a = 2, the common difference is d = 5−2 = 3, and n = 10. For an arithmetic progression, Sₙ = n/2[2a + (n−1)d]. Therefore S₁₀ = 10/2[2(2) + 9(3)] = 5[4 + 27] = 5×31 = 155. Verify by finding the last term: a₁₀ = a + 9d = 2 + 9×3 = 29. The average of the first and last terms is (2 + 29)/2 = 15.5, and 10×15.5 = 155. Thus D is correct. A, B, and C do not satisfy the AP sum calculation. A common error is using 10d instead of (10−1)d when finding the tenth term, or counting eleven terms. Remember that ten terms contain nine equal steps after the first term.

Direct answer: option C, 164. In an arithmetic progression (AP), each term changes by the same common difference. Here the first term is a = 3, the common difference is d = 8 − 3 = 5, and the number of terms is n = 8. Use Sₙ = n/2[2a + (n − 1)d]. Thus S₈ = 8/2[2(3) + 7(5)] = 4(6 + 35) = 164. A useful check is to find the eighth term: a₈ = 3 + 7(5) = 38. The average of the first and last terms is (3 + 38)/2 = 20.5, so the sum is 8 × 20.5 = 164. A (156) is too small, B (160) does not result from the formula, and D (168) is too large. Remember: for n terms, use n − 1 differences, not n differences.

Direct answer: B, 555. In the AP, the first term is a=2, the common difference is d=7−2=5, and the number of terms is n=15. Use Sₙ=n/2[2a+(n−1)d]. Therefore S₁₅=15/2[2(2)+14(5)]=15/2(4+70)=15/2×74=15×37=555. A second check uses the last term: a₁₅=a+(15−1)d=2+14×5=72. In an AP, the average of the first and last terms is (2+72)/2=37, so the sum is 15×37=555. A, 545, C, 560, and D, 570, do not agree with either calculation. The common error is using 15d instead of 14d for the fifteenth term; there are fourteen equal gaps from the first to the fifteenth term.

The governing concept is the sum formula for the first n terms of an arithmetic progression: Sₙ = n/2[2a + (n − 1)d]. Here n = 12, the first term a = 5, and the common difference is d = 9 − 5 = 4. Therefore S₁₂ = 12/2[2(5) + 11(4)] = 6[10 + 44] = 6 × 54 = 324. A second method confirms this result. The twelfth term is a₁₂ = 5 + 11(4) = 49, so S₁₂ = 12(5 + 49)/2 = 6 × 54 = 324. Hence option B is correct. The other values may result from using the wrong number of intervals, making an arithmetic error inside the brackets, or confusing the final term with the total sum.

The governing concept is the sum of the first n terms of an arithmetic progression. The formula is Sₙ = n/2[2a + (n − 1)d], where a is the first term and d is the common difference. Here a = 8, d = 14 − 8 = 6, and n = 9. Therefore, S₉ = 9/2[2(8) + 8(6)] = 9/2[16 + 48] = 9/2 × 64 = 288. Hence option B is correct. A useful independent check is to find the ninth term: a₉ = 8 + 8(6) = 56. The average of the first and last terms is (8 + 56)/2 = 32, so the total is 9 × 32 = 288. The other options can arise from an incorrect common difference, miscounting the eight intervals, or an arithmetic mistake.

Use the arithmetic-progression sum formula Sₙ = n/2[2a + (n − 1)d]. The first term is a = 4, the common difference is d = 9 − 4 = 5, and the number of terms is n = 16. Substitution gives S₁₆ = 16/2[2(4) + 15(5)] = 8[8 + 75] = 8 × 83 = 664. Therefore option B is correct. The result can also be checked by calculating the last term: a₁₆ = 4 + 15(5) = 79. In an arithmetic progression, the mean of the first and last terms is (4 + 79)/2 = 41.5, and multiplying this mean by 16 gives 16 × 41.5 = 664. The alternatives may result from using 14 instead of 15 intervals, finding the wrong last term, or making a minor calculation error.

Direct answer: option A, 468. An arithmetic progression (AP) is a sequence in which the difference between consecutive terms remains constant. Here, the first term is a = 6, the common difference is d = 11 − 6 = 5, and the number of terms is n = 13. Use the sum formula Sₙ = n/2[2a + (n − 1)d]. Therefore, S₁₃ = 13/2[2(6) + (13 − 1)5] = 13/2[12 + 60] = 13/2 × 72 = 13 × 36 = 468. A second method confirms this: the thirteenth term is a₁₃ = a + 12d = 6 + 60 = 66, so the average term is (6 + 66)/2 = 36 and the sum is 13 × 36 = 468. A, 468, is correct because both methods agree. B, 474, is too large and does not result from the formula. C, 486, and D, 492, also fail the same calculation. A useful memory cue is: for n terms, there are n − 1 common-difference steps; do not use n instead of n − 1. Also, the nth-term formula finds one term, whereas the sum formula finds all n terms together.

Direct answer: option B, 276. In an arithmetic progression, each term changes by the same common difference. The first term is a = 10, the common difference is d = 17 − 10 = 7, and there are n = 8 terms. The sum formula is Sₙ = n/2[2a + (n − 1)d]. Substitution gives S₈ = 8/2[2(10) + (8 − 1)7] = 4[20 + 49] = 4 × 69 = 276. We can check by finding the last term: a₈ = a + (8 − 1)d = 10 + 7 × 7 = 59. The average of the first and last terms is (10 + 59)/2 = 34.5, and 8 × 34.5 = 276. Therefore B is correct. A, 268, is 8 less than the correct result; C, 284, and D, 292, also do not satisfy the sum formula or the average check. The common mistake is using 8d instead of 7d for the eighth term: eight terms contain only seven equal jumps from the first term. Remember: the nth term uses n − 1 steps, and the sum formula then multiplies the average by n.

The governing concept is the sum of the first n terms of an arithmetic progression. For an AP, S_n = n/2[2a + (n − 1)d], where a is the first term, d is the common difference, and n is the number of terms. Here a = 3, d = 5, and n = 20. Therefore, S_20 = 20/2[2(3) + 19(5)] = 10[6 + 95] = 10 × 101 = 1010. Hence option A is correct. As a check, the twentieth term is 3 + 19 × 5 = 98, so the average of the first and last terms is (3 + 98)/2 = 50.5; multiplying by 20 again gives 1010. The other options result from arithmetic or substitution errors.

The direct answer is option B,
\(S_{15}=-30\). An arithmetic progression changes by the same amount each time. Here the first term is \(a=12\), the common difference is \(d=-2\), and there are \(n=15\) terms. Use \(S_n=\frac n2[2a+(n-1)d]\). Thus \(S_{15}=\frac{15}{2}[24+14(-2)]=\frac{15}{2}(24-28)=\frac{15}{2}(-4)=-30\). Option A, 30, has the wrong sign. Option B, -30, is correct. Option C, 45, does not follow the sum formula. Option D, -45, also has an incorrect magnitude. The negative difference makes the terms decrease: 12, 10, 8, ..., -16. Memory cue: substitute the negative sign carefully before simplifying.

The sum of the first (n) odd numbers is (n^2), so (25^2=625). This result is useful for quick calculation.

The AP is (7,14,\ldots,140) with (20) terms, and its sum is (1470). Finding (n) from the last term is an easy method.

Here (a=-5), (d=3), (n=30), so the sum is (1155). The same formula works even with a negative first term.

Using (S_n=\frac{n}{2}(a+l)), the sum is (350). When the last term is given, this formula is faster.

The governing concept is direct evaluation of a formula for the sum of the first n terms. Since S_n is already given, substitute n = 12 directly; there is no need to find the individual nth term or the common difference. Thus S_12 = 3(12²) + 2(12) = 3 × 144 + 24 = 432 + 24 = 456. Therefore, option B is correct. A useful check is that the quadratic part contributes 432 and the linear part contributes 24, whose sum is 456. Option A may come from an arithmetic subtraction or addition mistake, option C is not obtained from the given expression, and option D can result from incorrectly evaluating the linear term. The answer follows solely from the stated partial-sum formula.

This is an AP with (a=18), (d=3), (n=16), and the sum is (648). In word problems, identify (a,d,n) first.

Direct answer: Option A, 441. The AP is 50, 47, 44, and so on. Its first term is a = 50, common difference is d = 47 - 50 = -3, and n = 18. Use the sum formula \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Substitution gives \\(S_{18}=\\frac{18}{2}[2(50)+17(-3)]\\) = 9[100 - 51] = 9 × 49 = 441. As a check, the eighteenth term is \\(a_{18}=50+17(-3)=-1\\); the average of the first and last terms is \\((50+(-1))/2=24.5\\), and 18 × 24.5 = 441. Option A is correct. Option B, 459, results from an arithmetic error or mishandling the negative difference. Option C, 468, is not obtained from the correct formula and may come from using an incorrect last term. Option D, 450, is also an incorrect calculation. Common mistake: in a decreasing AP, d must be written as -3, not +3.

From (\frac{n}{2}(4+100)=1040), (n=20). When last term and sum are given, use (S_n=\frac{n}{2}(a+l)).

The sum of the first (n) even numbers is (n(n+1)), so (40\times41=1640). Remember it for even-number AP questions.

Solving (\frac{n}{2}[18+6(n-1)]=525) gives (n=10). For (n), a quadratic may appear, so choose the positive value.

From (\frac{n}{2}[4+5(n-1)]=287), (n=11). When finding the number of terms from sum, reject the negative root.

In an AP, the difference between the sums of the first n and first \(n-1\) terms gives the nth term: \(a_n=S_n-S_{n-1}\). Thus, \(a_{10}=S_{10}-S_9=250-207=43\). Therefore, 43 is correct. Getting 42 would mean subtracting the given sums incorrectly. Exam tip: When two consecutive partial sums are given, subtract them directly to obtain the corresponding term.

Solving (\frac{n}{2}[12+5(n-1)]=660) gives (n=15). Simplify the bracket first, then solve the equation.

From (780=10[10+19d]), (d=4). When sum and first term are given, the common difference can be found directly.