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

This is the sum of the first (6) multiples of (16), so (16\times21=336). For multiples, use (1+2+\cdots+n).

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(310=\frac{10}{2}(a+49)=5(a+49)\). Hence \(a+49=62\), so \(a=13\). If \(a=12\), the sum would be \(305\), not \(310\). Exam tip: when the last term is given, apply \(S_n=\frac{n}{2}(a+l)\) directly.

The tenth term is (30), so (S_{10}=\frac{10}{2}(75+30)=525). The formula for (S_n) does not change in decreasing order.

This is the sum of the first (9) multiples of (7), so (7\times45=315). Treat the number of shelves as the number of terms.

The odd natural numbers form the arithmetic progression 1, 3, 5, ..., 31, with first term 1, common difference 2, and 16 terms. The sum of the first n odd natural numbers is n². Therefore, S₁₆ = 16² = 16 × 16 = 256. The same result follows from the AP formula Sₙ = n/2[2a + (n − 1)d] = 16/2[2(1) + 15(2)] = 8 × 32 = 256. Hence option B is correct. Options A, C, and D result from arithmetic errors or from using an incorrect number of terms.

This is the sum of the first (8) multiples of (9), so (9\times36=324). If (a=d), the multiples method is faster.

The fourteenth term is (61), so (S_{14}=\frac{14}{2}(22+61)=581). Correct calculation of the last term gives the correct sum.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term. Thus, \(S_7=\frac{7}{2}(31+13)=\frac{7}{2}\times44=154\). Therefore, 154 is correct. Since the first term is greater than the last term, the AP may be decreasing, but this does not change the sum formula. Exam tip: When the first term, last term, and number of terms are given, use \(S_n=\frac{n}{2}(a+l)\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=35\), we get \(\frac{35\times36}{2}=35\times18=630\). Therefore, 630 is the correct answer. An option such as 610 may result from an error in multiplication or division. Exam tip: use \(\frac{n(n+1)}{2}\) directly for \(1+2+\cdots+n\).

The twelfth term is (98), so (S_{12}=\frac{12}{2}(10+98)=648). The average of the first and last terms is useful in sums.

The eighteenth term is (92), so (S_{18}=\frac{18}{2}(7+92)=891). Finding the last term correctly is the key step.

For an arithmetic progression, the sum of the first n terms is Sₙ = n/2[2a + (n − 1)d]. Substituting a = 3, d = 7, and n = 20 gives S₂₀ = 20/2[2(3) + 19(7)] = 10[6 + 133] = 10 × 139 = 1390. Therefore option D is correct. As a check, the twentieth term is a₂₀ = 3 + 19 × 7 = 136, and the sum is 20/2 × (first term + last term) = 10 × (3 + 136) = 1390. The other options arise from mishandling 19d or the factor n/2.

Here (a=-5), (d=4), and (n=15), so (S_{15}=345). Do not make a sign error with a negative first term.

(S_{13}=\frac{13}{2}(11+83)=611). When the last term is given, (S_n=\frac{n}{2}(a+l)) is faster.

The fourteenth term is (-5), so (S_{14}=\frac{14}{2}(60-5)=385). In a decreasing progression, the last term may become negative.

Let Sₙ denote the sum of the first n terms of the arithmetic progression. The sum of terms from the 6th through the 10th is obtained by removing the sum of the first five terms from the sum of the first ten terms. Thus, a₆ + a₇ + a₈ + a₉ + a₁₀ = S₁₀ − S₅ = 270 − 85 = 185. Therefore option C is correct. No separate calculation of the first or last term is needed. Options A, B, and D result from an incorrect subtraction or from confusing the requested five terms with another partial sum.

The first term is (3) and the twelfth term is (47), so (S_{12}=\frac{12}{2}(3+47)=300). Use (a_n) to find the first and last terms.

The sixteenth row has (78) seats, so the total is (S_{16}=\frac{16}{2}(18+78)=768). In word problems, treat rows as terms.

The seventeenth term is (149), so (S_{17}=\frac{17}{2}(5+149)=1309). Find the last term first and use the average method.

From (\frac{n(n+1)}{2}=325), (n(n+1)=650), so (n=25). In such questions, look for products of consecutive numbers.