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

(S_{42}=9975) and (S_{43}=10449), so the sum first exceeds (10000) at (43) terms. Always check the previous sum.

The first term is (3), and the (80)th term is (556), so the sum is (22360). Finding the first and last terms from (a_n) is an easy method.

Using (S_n=\frac{n}{2}[2a+(n-1)d]) gives (n=24). Exam tip: first reduce the equation to a simple quadratic.

From the two terms (d=4) and (a=3). Hence (S_{20}=980); exam tip: find (a) and (d) before applying the sum formula.

Substituting (n=15) gives (S_{15}=3(15)^2+2(15)=705). Exam tip: when (S_n) is given directly, substitute (n) first.

Using (S_n=\frac{n}{2}(a+l)), (516=6(a+75)), so (a=11). Exam tip: when the last term is given, use the (a+l) form.

(S_{25}=\frac{25}{2}[84+24(-3)]), so the sum is (150). Exam tip: handle the negative sign of the common difference carefully.

(a_{11}=S_{11}-S_{10}), so the value is (572-470=102). Exam tip: use (S_n-S_{n-1}) to find a particular term.

From (315=5[36+9d]), (d=3). Exam tip: when (n) is known, solve the sum formula directly for (d).

Solving gives (n=13), so the last term is (5+12(4)=53). Exam tip: verify both the sum and the last term after finding (n).

Solving (\frac{n}{2}[12+7(n-1)]=760) gives (n=16). Exam tip: accept only the positive integer root.

The direct answer is C, 98. The sum formula is S_n=n/2(a+l). The question gives S_18=999, n=18, and a=13. Substitute these values: 999=18/2(13+l)=9(13+l). Divide by 9: 111=13+l. Therefore l=111-13=98. Option C is correct. Option A, 92, would give 9(13+92)=945, not 999. Option B, 96, would give 981, so it is not the required last term. Option D, 101, would give 1026, also incorrect. The formula is based on pairing the first and last terms: each pair has the same sum, a+l, and there are n/2 such pairs. Do not confuse the last term with the common difference or with the number of terms. The fastest exam method here is to divide the given sum by 9 and subtract the first term.

The sum of the first 14 terms includes the first 13 terms plus the fourteenth term. Therefore, subtracting the two consecutive sums isolates that final term: \\(a_{14}=S_{14}-S_{13}\\). Substituting the given values gives \\(a_{14}=497-429=68\\). This property works for any sequence when consecutive partial sums are known, and it is especially useful in arithmetic-progression questions.

The difference is positive, so the fourteenth term is 68, not one of the nearby values caused by an arithmetic subtraction error. There is no need to find the first term or common difference because the required term follows directly from the two sums. Hence option C is correct. Options A, B, and D do not equal the difference between the given consecutive sums.

The two terms give (d=4) and (a=6), so (S_{15}=510). Exam tip: first find (d) from the difference of two given terms.

(a_1=S_1=9) and (a_2=S_2-S_1=13), so (d=4) and (a+d=13). Exam tip: start with (S_1) and (S_2-S_1).

Solving (\frac{n}{2}[162-6(n-1)]=315) gives (n=7). Exam tip: the same sum formula works for decreasing progressions too.

The required sum is (S_{20}-S_{10}=840-220=620). Exam tip: find sums of middle terms by subtracting cumulative sums.

From (901=\frac{n}{2}(3+103)), (n=17). Exam tip: when first and last terms are known, use (S_n=\frac{n}{2}(a+l)).

The last term is (12+17(7)=131), and the average is (\frac{12+131}{2}=71.5). Exam tip: the average of AP terms equals the average of first and last terms.

Let (S_n=\frac{d}{2}n^2+\frac{2a-d}{2}n). The two sums give (a=7), (d=6), so (S_{27}=2295); exam tip: write (S_n) as a quadratic in (n).

From (904=8(4+l)), (l=109). Exam tip: with sum and first term known, the last term can be found quickly.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here \(a=9\), \(d=4\), and \(n=30\). Thus, \(S_{30}=\frac{30}{2}[2(9)+29(4)]=15(18+116)=15\times134=2010\). Therefore, 2010 is correct. An answer such as 1980 can result from an error in using \((n-1)d\) or in multiplication. Exam tip: for \(n\) terms, there are always \(n-1\) common differences.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, 1003=\(\frac{n}{2}[22+6(n-1)]\)=\(n(3n+8)\). Hence, 3n²+8n−1003=0, giving n=17; the other root is negative. Although 19 is a nearby distractor, it does not give a sum of 1003. Exam tip: since n represents the number of terms, accept only a positive integer root.

(S_{15}=\frac{15}{2}[130+14(-5)]=450). Exam tip: handle the negative common difference carefully.

For an AP, \(a_7=a+6d=34\) and \(a_{18}=a+17d=89\). Subtracting the equations gives \(11d=55\), so \(d=5\). Then \(a+30=34\) gives \(a=4\). Hence, \(S_{18}=\frac{18}{2}[2a+17d]=9[8+85]=837\). An answer such as \(819\) can result from using an incorrect coefficient for the last term in the sum formula. Exam tip: when two terms are given, first write them as \(a+(n-1)d\) to find \(a\) and \(d\).