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

The governing concept is the meaning of a partial sum. S₇ represents the sum of the first seven terms, while S₁₄ represents the sum of the first fourteen terms. When the earlier partial sum is subtracted from the later one, the first seven terms cancel, leaving exactly the 8th through 14th terms. Therefore, the required sum is S₁₄ − S₇ = 427 − 126 = 301. Hence option C is correct. The other choices result from an arithmetic error or from subtracting the wrong quantities; no common difference or first term is needed for this direct partial-sum method.

This is the sum of the first (25) multiples of (4), so (4\times\frac{25\times26}{2}=1300). Use the sum of natural numbers for multiples.

For an arithmetic progression, the sum of n terms is Sₙ = n/2[2a + (n−1)d], or equivalently n/2(a + l), where a is the first term and l is the last term. Here a = 13, d = 17 − 13 = 4, and n = 18. The eighteenth term is l = a + 17d = 13 + 17(4) = 81. Thus S₁₈ = 18/2(13 + 81) = 9 × 94 = 846. Therefore option A is correct. The other options do not follow from the correct last term or contain an arithmetic error in multiplying the average by the number of terms.

The sum of the first (n) odd numbers is (n^2), so (n^2=441) and (n=21). Recognizing perfect squares is useful.

Sum equals average (\times) number of terms, so (42\times11=462). If the average is given, the long formula is not necessary.

Using (S_n=\frac{n}{2}[32+6(n-1)]), checking the options shows none match (640). This question should be corrected before import.

The nineteenth term is (184), so (S_{19}=\frac{19}{2}(4+184)=1786). With a large difference, find the last term first.

The sum of the first n terms is given as S_n=2n^2+5n. Substituting n=15, S_{15}=2(15)^2+5(15)=2×225+75=525. Hence, the correct answer is 525. A value such as 545 usually results from an arithmetic error while calculating 2×225 or 5×15. Exam tip: When S_n is given directly, substitute the required value of n in the expression.

Putting (n=15) in (S_n=\frac{n}{2}[12+9(n-1)]) gives (1035). In such questions, checking the options is fast.

The numbers are (104,112,\ldots,248), and there are (19) terms, so the sum is (3192). In boundary questions, choose the first and last terms carefully.

The sum of the (7)th to (13)th terms is (S_{13}-S_6=383). For a group of consecutive terms, subtract partial sums.

The fifteenth term is (-12), so (S_{15}=\frac{15}{2}(72-12)=450). In a decreasing progression, keep (d) negative.

The remaining sum is (24\times25-9\times10=510), not any listed option. This question should be corrected before import.

The fourteenth term is (153), so (S_{14}=\frac{14}{2}(10+153)=1141). Finding the last term first makes calculation easier.

Since \(t_n=S_n-S_{n-1}\), we get \(S_n-2S_{n-1}+S_{n-2}=t_n-t_{n-1}\), the common difference of an AP. Option B covers only a constant-term AP. Exam tip: test second differences of sums.

Putting (n=15) in (S_n=\frac{n}{2}[16+6(n-1)]) gives (750). In such questions, check the options in the formula.

The numbers are (204,216,\ldots,396), and there are (17) terms, so the sum is (5100). Choose the first and last terms carefully in boundary questions.

The sum of the (10)th to (18)th terms is (S_{18}-S_9=630). For a group of consecutive terms, take the difference of partial sums.

The thirteenth term is (0), so (S_{13}=\frac{13}{2}(96+0)=624). In a decreasing progression, the last term can also be zero.

The eighteenth row has (143) bricks, so the total is (S_{18}=\frac{18}{2}(24+143)=1503). In word problems, treat rows as terms.