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

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 \(nd\).

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

Here (\frac{S_n}{t_n}=\frac{n+1}{2}), so (n=29). Exam tip: write the ratio in simplified form.

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

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

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.

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.

From (6(12+11d)=4\cdot3(12+5d)), (d=12). Exam tip: write both sums separately and then compare.

(S_2=12) and (S_3=12), while the sum decreases afterward, so the maximum is (12). Exam tip: check nearby integer values around the vertex.

The sum of the first (n) odd numbers is (n^2), so (60^2=3600). Exam tip: remember this result directly.

(S_n=3n(n+6)), and substituting (n=19) gives (1425). Exam tip: options can be checked quickly too.

The correct answer is C, 855. The first term is a = −12, and the common difference is d = −5 − (−12) = 7. Use the sum formula S_n = n/2[2a + (n−1)d]. For n = 18, S_18 = 18/2[2(−12) + 17(7)] = 9[−24 + 119] = 9 × 95 = 855. The negative first term does not make the total negative because the sequence increases by 7 and later terms are positive; the last term is −12 + 17(7) = 107. Options 805, 830 and 880 arise from an incorrect difference, a wrong number of intervals, or an arithmetic error in the bracket.

These give (a=3) and (d=7), so (S_{30}=3135). Exam tip: first find (a) and (d) from the two sums.

The difference between the sums of two equal blocks is (225d), so (d=4). Exam tip: comparing equal-length blocks is a fast method.

(t_5=64) and (t_{20}=4), so the sum is (\frac{16}{2}(64+4)=544). Exam tip: count the selected terms correctly.

The correct answer is C, 225. A term of a sequence represented by partial sums is found through a_n = S_n − S_(n−1). First calculate S_12 = 5(12)(2×12 − 1) = 60(23) = 1380. Next calculate S_11 = 5(11)(2×11 − 1) = 55(21) = 1155. Therefore a_12 = 1380 − 1155 = 225. This method is preferable to trying to identify the first term and common difference separately, because the sum formula already contains all the required information. The other choices differ by 5 or 10 and may result from using 2n instead of 2n−1, subtracting incorrectly, or confusing the 11th and 12th terms.

The required sum is (S_{18}-S_8=2304-464=1840). Exam tip: the sum from the (m)th to (n)th term is (S_n-S_{m-1}).

The equation gives (4n+26=6n+6), so (n=10). Exam tip: simplify the common (\frac{n}{2}) in both sums.

Total seats are (S_{25}=\frac{25}{2}[44+24(3)]=1450). Exam tip: convert real-life situations into an AP.

The daily savings form an arithmetic progression. The first term is \\(a=50\\), the common difference is \\(d=20\\), and the number of terms is \\(n=18\\). The sum formula is \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Substitution gives \\(S_{18}=\\frac{18}{2}[2(50)+17(20)]\\), which is \\(9[100+340]=9\\times440=3960\\) rupees.

The expression correctly includes the saving on the first day and the increase on each of the next 17 days. A common mistake is to multiply the last day's saving by 18, which would not give the sum of all terms. Therefore, option C, 3960 rupees, is correct. The other options result from errors in the number of increases or in adding the progression.