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

From the two terms, (d=6) and (a=-2), so (S_{45}=5850). First find (a,d), then apply the sum formula.

The condition gives (6+13d=3(6+3d)), so (d=3) and (S_{30}=1485). Convert the term condition into an equation first.

(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.

From \(999=9(13+l)\), \(l=98\). Exam tip: \(S_n=\frac{n}{2}(a+l)\) is the shortest method here.

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.

The governing relationship between consecutive partial sums and an individual term is a_n = S_n − S_{n−1}. First evaluate S_20 using the given expression: S_20 = 20(4×20 − 1) = 20(80 − 1) = 1580. Next evaluate S_19: S_19 = 19(4×19 − 1) = 19(76 − 1) = 19×75 = 1425. Therefore, a_20 = S_20 − S_19 = 1580 − 1425 = 155. Hence option B is correct. A common mistake is to use S_20 itself as the 20th term, while the other choices can arise from an arithmetic error in one of the two evaluations.