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

Using (S_n=\frac{n}{2}(a+l)), (S_{18}=\frac{18}{2}\times144=1296). If (a+l) is directly given, use it immediately.

The numbers are (12,16,\ldots,96), and there are (22) terms, so the sum is (1188). Apply the two-digit limit carefully.

The multiples are (100,125,\ldots,275), and there are (8) terms, so the sum is (1500). Decide the first and last terms according to the limit.

The first term is (9) and the twentieth term is (104), so (S_{20}=\frac{20}{2}(9+104)=1130). Use (a_n) to find the first and last terms.

The first term is (-2) and the eighteenth term is (117), so (S_{18}=1035). Put (n=1) and the final (n) in (a_n).

The first term is (13) and the sixteenth term is (58), so (S_{16}=568). Use (a_n) to find (a) and (l), then apply the sum formula.

The first term is (48) and the fourteenth term is (22), so (S_{14}=\frac{14}{2}(48+22)=490). The same sum formula works for a decreasing progression.

The twelfth deposit is (265) rupees, so the total is (S_{12}=\frac{12}{2}(45+265)=1860) rupees. Keep the unit in mind in money questions.

The tenth row has (395) chairs, so the total is (S_{10}=\frac{10}{2}(80+395)=2375). Treat rows as the number of terms.

The eighteenth row has (229) bricks, so the total is (S_{18}=\frac{18}{2}(25+229)=2286). In word problems, convert the pattern into an arithmetic progression.

The governing concept is the sum of the first n terms of an arithmetic progression. The first five terms are 2, 7, 12, 17, and 22, because the common difference is 5. Direct addition gives S₅ = 2 + 7 + 12 + 17 + 22 = 60, so option B is correct. The standard formula confirms this: Sₙ = n/2[2a₁ + (n − 1)d], hence S₅ = 5/2[2(2) + 4(5)] = 5/2(24) = 60. Option A, 55, omits 5 from the total; option C adds an extra 5; and option D is larger still because it does not represent the sum of exactly the first five terms.

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 a_5=a+(5-1)d=7+4(8)=39. Therefore the first five terms are 7, 15, 23, 31 and 39, whose sum is 7+15+23+31+39=115. Equivalently, S_n=n/2[2a+(n-1)d]=5/2[14+32]=5/2(46)=115. Thus option C is correct. The other choices result from incomplete addition or an incorrect fifth term; none equals the verified sum by either method.

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, a = 5, and d = 9 − 5 = 4. Therefore S₁₂ = 12/2[2(5) + 11(4)] = 6[10 + 44] = 6 × 54 = 324. Thus option B is correct. A common alternative check is to find the last term: a₁₂ = 5 + 11 × 4 = 49, then use S₁₂ = 12(5 + 49)/2 = 324. The other choices result from arithmetic errors in the bracket or from using the wrong number of intervals rather than the number of terms.

Use the arithmetic-progression sum formula Sₙ = n/2[2a + (n − 1)d]. The first term is a = 8, the common difference is d = 14 − 8 = 6, and n = 9. Hence S₉ = 9/2[2(8) + 8(6)] = 9/2[16 + 48] = 9/2 × 64 = 288. Therefore, option B is correct. As a verification, the ninth term is a₉ = 8 + 8 × 6 = 56, and pairing the first and last terms gives 9(8 + 56)/2 = 288. Options A, C, and D can arise from using an incorrect difference, miscounting the eight intervals, or making an addition or multiplication error.

Apply Sₙ = n/2[2a + (n − 1)d]. Here a = 4, d = 9 − 4 = 5, and n = 16. Thus S₁₆ = 16/2[2(4) + 15(5)] = 8[8 + 75] = 8 × 83 = 664. Therefore, option B is correct. A second check uses the last term: a₁₆ = 4 + 15 × 5 = 79, so the average of the first and last terms is (4 + 79)/2 = 41.5 and the sum is 16 × 41.5 = 664. The other choices reflect small arithmetic mistakes, such as using 14 instead of 15 intervals or miscalculating the bracket.

The governing concept is the sum of the first n terms of an arithmetic progression. Use S_n = n/2[2a + (n−1)d]. Here a = 3, d = 5, and n = 20. First calculate 2a = 6 and (n−1)d = 19 × 5 = 95. Their sum is 101, and multiplying by n/2 gives S_20 = 20/2 × 101 = 10 × 101 = 1010. Thus option A is correct. A quick check uses the last term: l = a + 19d = 98, so the average of the first and last terms is (3+98)/2 = 50.5; 20 × 50.5 also equals 1010. The other values result from arithmetic errors.

This is a decreasing AP, and the formula gives (S_{15}=-30). Do not forget the sign of the negative common difference.

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.