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

The supplied answer is not consistent with the AP calculation, so the correct decision is revise. Let the first term be a = 6 and the common difference be d. The sum formula is S_n = n/2[2a + (n-1)d]. Thus S_12 = 12/2[12 + 11d] = 6(12 + 11d), while S_6 = 6/2[12 + 5d] = 3(12 + 5d). The condition S_12 = 4S_6 gives 6(12 + 11d) = 12(12 + 5d). Dividing by 6 gives 12 + 11d = 24 + 10d, so d = 12. Therefore option C is mathematically supported, not merely by the supplied explanation’s displayed equation. However, the supplied explanation writes 4·3(12+5d), which is correct, and solving it actually gives d = 12. Option A, 8, does not satisfy the equation; option B, 10, also fails; option C, 12, satisfies both sums; option D, 14, fails. The concern is not the answer but that the displayed intermediate equation should be explained more clearly. Memory cue: calculate S_12 and S_6 separately before substituting.

(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 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.

From (S_{10}=0), (2a+9d=0), and from (S_{20}=600), (d=6), so (a=-27). Exam tip: use the zero sum to form a relation.

In (17) terms, the (9)th term is the middle term, so the sum is (17(36)=612). Exam tip: use the middle term for an odd number of AP terms.

The numbers are (52,60,\ldots,244), and there are (25) terms. Exam tip: convert the condition into the correct AP.