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

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.

The governing AP sum formula is S_n = n/2[2a + (n − 1)d]. Here n = 20, a = 25, and d = −2. Substitution gives S_20 = 20/2[2(25) + 19(−2)] = 10[50 − 38] = 10×12 = 120. Thus option C is correct. The negative common difference means the terms decrease: 25, 23, 21, and so on, so the bracket must contain 50 − 38 rather than 50 + 38. The values 100, 110, and 130 can result from mishandling the negative sign or using the wrong number of difference intervals.

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 \(d\).

To find the nth term from the sum, use \(a_n=S_n-S_{n-1}\). Thus, \(a_n=(4n^2+3n)-[4(n-1)^2+3(n-1)]=8n-1\). Hence, \(a_{25}=8\times25-1=199\). The value 203 usually results from an error while subtracting or expanding the \((n-1)\) expression. Exam tip: when \(S_n\) is given, subtract consecutive sums to obtain a term.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here, \(345=\frac{10}{2}[2(12)+9d]=5(24+9d)\). Thus, \(24+9d=69\), so \(9d=45\) and \(d=5\). If \(d=4\), the sum would be 300, not 345. Exam tip: substitute the given \(a\), \(n\), and \(S_n\) directly into the sum formula to find \(d\).

From (1264=\frac{n}{2}(7+151)), (n=16). Exam tip: use the (a+l) form when first and last terms are known.

This is the sum of (16) terms with (t_5=38) and (t_{20}=128). Exam tip: treat the required middle part as a smaller AP.

The nth term of a sequence is \(a_n=S_n-S_{n-1}\). Substituting \(S_{n-1}=3(n-1)^2-2(n-1)\) gives \(a_n=6n-5\). Hence, \(a_{n+1}-a_n=6\), so the common difference is 6. Option 4 can be tempting because the coefficient of \(n^2\) in \(S_n\) is 3, but the common difference is twice this coefficient: \(2\times3=6\). Exam tip: if \(S_n=An^2+Bn\), the common difference of the AP is \(2A\).

(S_n=\frac{n}{2}(247-7n)) is positive up to (n=35). Exam tip: solve the inequality and then take the integer limit.

The sum of the first (n) odd numbers is (n^2), so (n=24). Exam tip: you can also treat it as the AP (1,3,5,\ldots).

This is the AP (21,28,\ldots,210) with (28) terms. Exam tip: find the number of terms first.

The terms are (54,63,\ldots,198), making (17) terms. Exam tip: choose the first and last valid terms carefully.

These give (a=8) and (d=3), so (S_{36}=2178). Exam tip: you may write (S_n) as a quadratic in (n).

Let the first term be \(a\) and the common difference be \(d\). From \(t_3+t_9=(a+2d)+(a+8d)=70\), we get \(a+5d=35\). Similarly, \(t_5+t_{15}=(a+4d)+(a+14d)=110\) gives \(a+9d=55\). Subtracting gives \(4d=20\), so \(d=5\) and \(a=10\). Therefore, \(S_{20}=\frac{20}{2}[2(10)+19(5)]=575\). The value 1150 can result from incorrectly using \(n\) instead of \(\frac{n}{2}\) in the sum formula. Exam tip: convert the given term relations into equations in \(a\) and \(d\) first.