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

For an AP, \(S_n=\frac{n}{2}(a+l)\), where \(l\) is the last term. Thus, \(203=\frac{7}{2}(5+l)\). So \(406=7(5+l)\), giving \(5+l=58\) and hence \(l=53\). Option 58 is the value of \(a+l\), not the seventh term. Exam tip: When the sum and the last term are involved, use \(S_n=\frac{n}{2}(a+l)\).

The twenty-first term is (149), so (S_{21}=\frac{21}{2}(9+149)=1659). For larger (n), write the last term first.

Direct answer: option A, 160. Here a=25, d=-2, and n=16. The sixteenth term is a16=a+(16-1)d=25+15(-2)=25-30=-5. Therefore S16=n/2(a+a16)=16/2×(25-5)=8×20=160. The negative common difference means the terms decrease: 25,23,21,...,-5. Option A, 160, follows exactly from the formula. Option B, 170, can result from a small arithmetic error and is not the required sum. Option C, 180, often comes from using an incorrect last term or ignoring the negative sign. Option D, 190, is also too large; it does not match either the final term or the average. Another check is that the average of the first and last terms is (25+(-5))/2=10, and 16×10=160. Do not turn -2 into +2.

In the twelfth week, (195) books arrive, so the total is (1350) books. Treat the pattern as (a=30), (d=15), (n=12).

The governing concept is the relationship between partial sums. S₇ denotes the sum of the first seven terms, whereas S₁₄ denotes the sum of the first fourteen terms. When S₇ is subtracted from S₁₄, the first seven terms cancel term by term, leaving only the 8th, 9th, ..., 14th terms. Hence the required sum is S₁₄−S₇ = 427−126 = 301. Therefore option C is correct. It is unnecessary to calculate the first term or the common difference because the two given partial sums already isolate the requested block of terms. The other choices do not result from the correct subtraction: 281, 291, and 311 reflect arithmetic or transcription errors rather than a valid AP method. This subtraction rule works for any sequence whenever the relevant partial sums are known.

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.

The governing formula for the sum of an arithmetic progression is Sₙ = n/2[2a+(n−1)d], or equivalently Sₙ = n(a+l)/2, where a is the first term and l is the nth term. Here a=13, the common difference is d=17−13=4, and n=18. The eighteenth term is l=a+(n−1)d=13+17×4=81. Therefore S₁₈=18(13+81)/2=9×94=846. Thus option A is correct. The same result follows from the direct formula: S₁₈=18/2[26+17×4]=9(94)=846. The other options are not consistent with the correct eighteenth term or contain an arithmetic error while multiplying the average of the first and last terms by 18.

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.

Direct answer: Option D, 383. An AP has partial sums written as S_n, where S_n means the sum of the first n terms. The first six terms have sum S_6=111, while the first thirteen terms have sum S_13=494. The terms from the seventh through the thirteenth are exactly the terms included in S_13 but not included in S_6. Therefore, subtract the earlier sum: S_13-S_6=494-111=383. Option A, 353, is not obtained from the given subtraction. Option B, 363, is also an incorrect difference. Option C, 373, is another arithmetic distractor. Option D, 383, is correct because it represents precisely terms 7 through 13, inclusive. Memory cue: for a consecutive block beginning after the first m terms, use the larger partial sum minus the earlier partial sum.

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.

The seventeenth term is (105), so (S_{17}=\frac{17}{2}(9+105)=969). Finding the last term first makes calculation easier.