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

Here, the first term is \(a=11\), the common difference is \(d=11\), and \(n=9\). The ninth term is \(a+(n-1)d=11+8\times11=99\). Hence, \(S_9=\frac{n}{2}(a+l)=\frac{9}{2}(11+99)=495\). Therefore, 495 is correct. An answer such as \(484\) can result from using an incorrect last term. Exam tip: verify the \(n\)th term before substituting values in the sum formula.

Here, the first term is \(a=13\), the common difference is \(d=16-13=3\), and \(n=12\). The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(S_{12}=\frac{12}{2}[2(13)+11(3)]=6(26+33)=6\times59=354\). Therefore, option A is correct. \(360\) can result from incorrectly using \(nd\) instead of \((n-1)d\). Exam tip: always use \(n-1\) with the common difference in the AP sum formula.

The \(n\)th term of an AP is \(a+(n-1)d\). The sum of the first \(n\) terms is \(n\) times the average of the first and last terms, giving \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B gives only the \(n\)th term, not the sum. Exam tip: remember that \(a+(n-1)d\) is the term formula, whereas \(S_n\) is the sum formula.

Here, the first term is \(a=18\), the common difference is \(d=21-18=3\), and \(n=20\). Applying \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{20}=\frac{20}{2}[2(18)+19(3)]=10(36+57)=930\). Hence, 930 is the correct answer. Although 915 is a close distractor, it is not the sum produced by the given first term and common difference. Exam tip: the sum formula uses \(n-1\) because there are \(n-1\) gaps from the first term to the nth term.

Here, \(a=40\), \(d=-5\), and \(n=7\). The seventh term is \(a_7=40+(7-1)(-5)=10\). Hence, \(S_7=\frac{7}{2}(40+10)=175\). Therefore, 175 is correct. A result such as 180 may arise from an error while adding the terms. Exam tip: check the last term using \(n-1\) before applying the sum formula.

For an AP with first term \(a\), last term \(l\), and \(n\) terms, the sum is \(S_n=\frac{n}{2}(a+l)\). It equals the number of terms multiplied by the average of the first and last terms, \(\frac{a+l}{2}\). Option D is also a valid general sum formula, but it is used when the common difference \(d\), rather than the last term \(l\), is known. Exam tip: use \(\frac{n}{2}(a+l)\) directly when \(l\) is given.

Here, the first term is \(a=2\), the common difference is \(d=2\), and the number of terms is \(n=50\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{50}=\frac{50}{2}[2\times2+49\times2]=25\times102=2550\). Therefore, 2550 is the correct answer. The value 2500 may result from taking the last term incorrectly instead of \(100\). Exam tip: identify \(a\), \(d\), and \(n\) before substituting into the sum formula.

Here, the first term is \(a=9\), the common difference is \(d=9\), and \(n=12\). The 12th term is \(a_{12}=a+(n-1)d=9+11\times9=108\). Therefore, \(S_{12}=\frac{12}{2}(9+108)=6\times117=702\). Hence, 702 is correct. Exam tip: First find the \(n\)th term, then use \(S_n=\frac{n}{2}(a+l)\).

Here, the first term is \(a=100\), the common difference is \(d=-10\), and \(n=6\). The sixth term is \(a_6=a+5d=100+5(-10)=50\). Hence, \(S_6=\frac{n}{2}(a+l)=\frac{6}{2}(100+50)=450\). Therefore, 450 is correct. An answer such as 440 can result from an error in finding the last term or adding the terms. Exam tip: for a decreasing AP, the common difference \(d\) is negative.

Putting \(n=12\) in the formula gives \(S_{12}=\frac{12}{2}(2\times12+4)\). Hence, \(S_{12}=6(24+4)=6\times28=168\). Therefore, option C is correct. Option 160 may result from an arithmetic error while multiplying 28 by 6. Exam tip: after substitution, simplify the bracket first and then perform the multiplication.

This forms an AP with first term \(a=14\), common difference \(d=4\), and \(n=9\) terms. The last row has \(a+(n-1)d=14+8\times4=46\) chairs. Therefore, the total is \(S_9=\frac{9}{2}(14+46)=270\). Option D incorrectly counts 9 common differences; 9 rows have only 8 gaps between them. Exam tip: use \(n-1\) in the formula for the last term.

The sum of the five given terms is 3 + 7 + 11 + 15 + 19 = 55. Hence, the correct answer is 55. It can also be checked using the AP sum formula: \(S_5=\frac{5}{2}[2(3)+4(4)]=55\). Although 57 is close, it is not the sum of the given terms. Exam tip: for a small number of terms, direct addition is a quick way to verify the answer.

Here, the first term is \(a=21\), the common difference is \(d=28-21=7\), and \(n=8\). Thus, \(S_8=\frac{8}{2}[2(21)+(8-1)\times7]=4(42+49)=364\). Therefore, 364 is correct. The value 371 may result from incorrectly combining the eighth term, \(70\), with the sum; it is not the sum of the first 8 terms. Exam tip: Identify \(a\), \(d\), and \(n\), then apply \(S_n=\frac{n}{2}[2a+(n-1)d]\).

The \(n\)th term of an AP is \(l=a+(n-1)d\). Therefore, the sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+l)=\frac{n}{2}[2a+(n-1)d]\). Option B gives only the \(n\)th term, not the sum. Exam tip: ensure that the sum formula contains \((n-1)d\), not \(nd\).

Here, the first term is \(a=50\), the common difference is \(d=-5\), and \(n=9\). The ninth term is \(a_9=a+(n-1)d=50+8(-5)=10\). Hence, \(S_9=\frac{n}{2}(a+l)=\frac{9}{2}(50+10)=270\). Therefore, 270 is the correct answer. A value such as 280 can result from using an incorrect last term or number of terms. In exams, find the \(n\)th term first to verify the sum.

Here, the first term is \(a=0\), the common difference is \(d=4\), and \(n=11\). The 11th term is \(a_{11}=a+(n-1)d=0+10\times4=40\). Hence, \(S_{11}=\frac{n}{2}(a+l)=\frac{11}{2}(0+40)=220\). The value \(240\) can result from counting the terms or the last term incorrectly. Exam tip: before using the sum formula, use \(n-1\) while finding the \(n\)th term.

The \(n\)th term of an AP is \(a+(n-1)d\). Therefore, the average of the first and last terms is \(\frac{a+[a+(n-1)d]}{2}\). Multiplying this by \(n\) gives \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B is only the \(n\)th term, while option C incorrectly uses \(nd\) instead of \((n-1)d\). Exam tip: always check the \((n-1)d\) factor for the last term.

Here, the first term is \(a=14\), the common difference is \(d=18-14=4\), and the number of terms is \(n=25\). The 25th term is \(a_{25}=a+(n-1)d=14+24\times4=110\). Hence, \(S_{25}=\frac{n}{2}(a+l)=\frac{25}{2}(14+110)=1550\). Therefore, 1550 is the correct answer. A close option such as 1575 may result from an error in finding the last term or the number of terms. Exam tip: always check the \(n-1\) factor while finding the last term.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term. Here, \(n=11\), \(a=12\), and \(l=72\). Therefore, \(S_{11}=\frac{11}{2}(12+72)=\frac{11}{2}\times84=462\). Hence, 462 is the correct answer. Although 452 is close, the average of the first and last terms is \(42\), and \(42\times11=462\), not 452. Exam tip: When \(a\), \(l\), and \(n\) are given, apply this sum formula directly.

Here, the first term is \(a=24\), the common difference is \(d=30-24=6\), and \(n=16\). The 16th term is \(a_{16}=a+(16-1)d=24+15\times6=114\). Thus, \(S_{16}=\frac{16}{2}(24+114)=8\times138=1104\). Therefore, 1104 is correct. An option such as 1094 can result from an error in addition or multiplication. Exam tip: You can also apply \(S_n=\frac{n}{2}[2a+(n-1)d]\) directly.