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

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here, \(824=\frac{16}{2}[2(14)+15d]=8(28+15d)\). Thus, \(28+15d=103\), so \(15d=75\) and \(d=5\). If 4 were used, the sum would be 704, not 824. Exam tip: substitute the given \(a\), \(n\), and \(S_n\) directly into the sum formula to find \(d\).

From (2208=\frac{n}{2}(8+176)), (n=24). Exam tip: use the (a+l) form when first and last terms are known.

(t_8=67), (t_{26}=193), and there are (19) terms, so the sum is (2470). Exam tip: count the selected terms correctly.

For an arithmetic progression, the nth term is \(a_n=S_n-S_{n-1}\). Here, \(a_n=(4n^2+n)-[4(n-1)^2+(n-1)]=8n-3\). Therefore, the difference between consecutive terms is \(a_{n+1}-a_n=8\), so the common difference is 8. Option 7 is not correct; it can result from incorrectly reducing the coefficient of \(n\) by one. Exam tip: first find \(a_n=S_n-S_{n-1}\) from \(S_n\), then subtract consecutive terms.

(S_n=\frac{n}{2}(309-9n)) is positive up to (n=34). Exam tip: solve the inequality and then take the integer limit.

The sum of the first (n) even numbers is (n(n+1)), so (n=25). Exam tip: you can also treat it as the AP (2,4,6,\ldots).

This is the AP (44,55,\ldots,297) with (24) terms. Exam tip: find the number of terms first.

The terms are (112,126,\ldots,350), making (18) terms. Exam tip: choose the first and last valid terms carefully.

These sums give (a=5) and (d=4), so (S_{27}=1539). Exam tip: write (S_n) in terms of (a) and (d).

The two equations give (a=-2) and (d=6), so (S_{20}=1100). Exam tip: convert term sums into (a) and (d).

The last term is (134), so (d=\frac{134-18}{29}=4). Exam tip: connect the average with the average of first and last terms.

The correct answer is C, 6. For an arithmetic progression with first term a = 9, the sum formula is S_n = n/2[2a + (n−1)d]. Hence S_6 = 6/2[18 + 5d] = 54 + 15d, and S_12 = 12/2[18 + 11d] = 108 + 66d. The ratio condition gives (54 + 15d)/(108 + 66d) = 2/7. Cross-multiplying, 7(54 + 15d) = 2(108 + 66d), so 378 + 105d = 216 + 132d. Thus 162 = 27d and d = 6. Substituting d = 6 gives S6 = 144 and S12 = 504, whose ratio is 2:7, confirming the answer.

The selected (12) terms give (6(2a+245)=1602), so (a=11). Exam tip: treat the selected part as a separate AP.

The numbers are (114,133,\ldots,988), and there are (47) terms. Exam tip: find the first and last three-digit multiples.

The total sum is (9870), and the sum of multiples of (7) is (1470), so the answer is (8400). Exam tip: subtract the complementary sum.

(S_{24}=1476) and (S_{25}=1600), so the sum first exceeds (1500) at (n=25). Exam tip: check nearby integers in boundary questions.

From (555=\frac{15}{2}(a+65)), (a=9), and (d=\frac{65-9}{14}=4). Exam tip: use last term and sum to find the first term.

For an AP, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(S_5=35+10d\) and \(S_{15}=105+105d\). Using the given ratio, \(\frac{35+10d}{105+105d}=\frac{1}{7}\). On cross-multiplication, \(245+70d=105+105d\), which gives \(d=4\). If \(d=3\), the required ratio of sums is not obtained. Exam tip: simplify each sum before cross-multiplying a ratio equation.

Equating both sums gives (4n+16=2n+42), so (n=13). Exam tip: when (n) is common, cancel (\frac{n}{2}).

The required sum is (S_{25}-S_{15}=1975-735=1240). Exam tip: use the difference of cumulative sums for middle terms.