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

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).

The direct answer is D, 3234. The multiples of 7 from 21 to 210 form an arithmetic progression: 21, 28, 35, ..., 210. The first term is 21, the last term is 210, and the common difference is 7. To count terms, divide the endpoints by 7: the sequence runs from 3 to 30, so the number of terms is 30 - 3 + 1 = 28. Now use the AP sum formula, \\(S_n=\frac{n}{2}(a+l)\\). Therefore, \\(S_{28}=\frac{28}{2}(21+210)=14\times231=3234\\). Option A, 3178, is not the result of this correct calculation. Option B, 3192, is also too small and likely reflects an arithmetic or term-count error. Option C, 3210, is close but still not equal to the calculated sum. Option D, 3234, exactly matches the formula and is correct. Memory cue: for multiples, first count the terms carefully; include both endpoints.

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.

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).

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.