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

The last term is (158), so (d=\frac{158-2}{39}=4). Exam tip: connect the average with the average of first and last terms.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here, \(a=13\), \(d=7\), and \(n=22\). Thus, \(S_{22}=\frac{22}{2}[2(13)+21(7)]=11(26+147)=11\times173=1903\). Therefore, 1903 is correct. The option 1892 can result from an error in evaluating the expression inside the bracket. Exam tip: use \(n-1=21\), not 22, while finding the last term contribution.

Using the sum formula gives (n=9). Exam tip: even in a decreasing AP, take the positive value of (n).

The governing concept is that the nth term of an arithmetic progression can be obtained by subtracting consecutive partial sums: a_n = S_n − S_(n−1). First calculate S_18 = 5(18²) + 4(18) = 5(324) + 72 = 1692. Next calculate S_17 = 5(17²) + 4(17) = 5(289) + 68 = 1513. Therefore a_18 = S_18 − S_17 = 1692 − 1513 = 179. Hence option C is correct. The other values result from an arithmetic or substitution error, such as using the wrong preceding sum or confusing S_18, which is the total of the first 18 terms, with the 18th term itself.

For an AP, \(T_n=a+(n-1)d\). Thus, \(a+8d=46\) and \(a+20d=106\). Subtracting gives \(12d=60\), so \(d=5\) and \(a=6\). Therefore, \(S_{21}=\frac{21}{2}[2a+20d]=\frac{21}{2}[12+100]=1176\). Option 1113 can result from using \(21\times53\), but the average of the first and 21st terms is \(56\), not 53. Exam tip: when two terms are given, form equations first to find \(a\) and \(d\).

The required sum is (S_{20}-S_{10}=1120-310=810). Exam tip: use the difference of cumulative sums for middle terms.

The terms are (84,96,\ldots,252), making (15) terms. Exam tip: choose the first and last valid terms carefully.

From (810=\frac{15}{2}(19+l)), (l=89). Exam tip: when the last term is needed, the (a+l) form is fast.

Solving (\frac{n}{2}[16+5(n-1)]=775) gives (n=17). Exam tip: choose the positive integer root.

Substituting (n=12) in the given formula gives (S_{12}=564). Exam tip: directly substitute (n) in the given (S_n).

The two sums give (a=3) and (d=5), so (S_{18}=819). Exam tip: write (S_n) in terms of (a) and (d) and solve.

When the last term is known, the sum of the first n terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Here, \(336=\frac{12}{2}(a+51)=6(a+51)\). Thus, \(a+51=56\), so \(a=5\). If the first term were 6, the sum would be \(6(6+51)=342\), not 336. Exam tip: use \(S_n=\frac{n}{2}(a+l)\) directly when the first and last terms are involved.

The numbers are (102,119,\ldots,986), and there are (53) terms. Exam tip: find the first and last three-digit multiples.

The last term is (25+19(-2)=-13), so the average is (\frac{25-13}{2}=6). Exam tip: the average is the average of first and last terms.

The \(n\)th term is \(a_n=S_n-S_{n-1}\). Here, \(a_n=6n-1\), so the difference between consecutive terms is \(6\). Hence A is correct. Exam tip: subtract successive sums to identify the sequence.

The governing concept is the relationship between partial sums and the first two terms of an arithmetic progression. Since S_1 is the first term, a = S_1 = 6(1)² + 2(1) = 8. The second term is obtained from a_2 = S_2 − S_1. Here S_2 = 6(2)² + 2(2) = 24 + 4 = 28, so a_2 = 28 − 8 = 20. Thus the common difference is d = a_2 − a = 20 − 8 = 12. Consequently, a + d = 8 + 12 = 20, which is also equal to the second term a_2. Therefore option D is correct; the other choices arise from confusing S_2 with d or failing to subtract the first partial sum.

The sum is (S_n=n(103-3n)), and the maximum (901) occurs at (n=17). Exam tip: check integer values near the vertex.

The total sum is (11325), and the sum of multiples of (6) is (1875), so the answer is (9450). Exam tip: subtract the complementary sum.

For an AP, the sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+t_n)\). Thus, \(777=\frac{14}{2}(a+96)=7(a+96)\). Hence \(a+96=111\), so \(a=15\). If \(17\) were used, the sum would be \(7(17+96)=791\), not the given sum. Exam tip: when \(S_n\) and \(t_n\) are given, use \(S_n=\frac{n}{2}(a+t_n)\) directly.

In an AP, the average of the first and last terms is \(\frac{a+l}{2}\). Multiplying this by the number of terms \(n\) gives \(S_n=\frac{n}{2}(a+l)\). Option B is only the average, not the sum. In exams, use this form when \(l\) is given.