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

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.

The first 15 odd natural numbers are 1, 3, 5, ..., 29. The sum of the first n odd natural numbers is n². Therefore, the sum is 15² = 225. A value such as 215 does not give the correct sum for this sequence with common difference 2. Exam tip: Remember that the sum of the first n odd numbers is n².

Here, the first term is \(a=17\), the common difference is \(d=22-17=5\), and \(n=10\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{10}=\frac{10}{2}[2(17)+9(5)]=5(79)=395\). Therefore, option C is correct. An answer such as 390 usually results from an error in finding the 10th term or the common difference. Exam tip: Remember to use \(n-1\) in the AP sum formula.

\(S_n=a_1+a_2+\cdots+a_{n-1}+a_n\), whereas \(S_{n-1}=a_1+a_2+\cdots+a_{n-1}\). Thus, subtracting \(S_{n-1}\) from \(S_n\) leaves only the nth term: \(a_n=S_n-S_{n-1}\). \(S_n/n\) is the average of the first n terms, not generally the nth term. Exam tip: obtain a term from consecutive sums by subtracting the previous sum from the larger sum.

In this AP, the first term is \(a=60\) and the common difference is \(d=54-60=-6\). The first five terms are \(60,54,48,42,36\), and their sum is \(60+54+48+42+36=240\). Therefore, 240 is the correct answer. A result such as 250 can arise from not handling the decreasing terms or the common difference \(-6\) correctly. Exam tip: remember to use a negative sign for \(d\) in a decreasing AP.

Here, \(a=8\), \(d=8\), and \(n=14\). The 14th term is \(a_{14}=a+(n-1)d=8+13\times8=112\). Hence, \(S_{14}=\frac{14}{2}(8+112)=7\times120=840\). Therefore, 840 is the correct answer. While \(14\times8=112\) gives the 14th term, it does not give the sum. Exam tip: Use both the first and last terms in the sum formula.

The numbers of seats form an AP, where \(a=8\), \(d=3\), and \(n=12\). Therefore, \(S_{12}=\frac{12}{2}[2(8)+(12-1)3]=6(16+33)=6\times49=294\). Hence, the correct answer is 294. An answer such as 276 does not correctly account for all \(11\) increases up to the last row. Exam tip: in the AP sum formula, the number of common differences is \(n-1\).

Here, the first term is \(a=5\), the common difference is \(d=4\), and \(n=18\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{18}=\frac{18}{2}[2(5)+17(4)]=9(10+68)=9\times78=702\). Hence, 702 is correct. The option 700 can result from an error in the final multiplication \(9\times78\). Exam tip: for \(n\) terms, the common difference is added \(n-1\) times, so use \((n-1)d\) in the formula.

The number of seats forms an AP with \(a=20\), \(d=2\), and \(n=15\). Thus, \(S_{15}=\frac{15}{2}[2(20)+(15-1)(2)]=\frac{15}{2}(68)=510\). Therefore, the total number of seats is 510. Getting 480 usually results from counting the number of terms or increments incorrectly. Exam tip: always use \(n-1\) with the common difference in the sum formula.

In this AP, the first term is \(a=19\), the common difference is \(d=10\), and \(n=6\). The sixth term is \(a_6=19+(6-1)\times10=69\). Thus, \(S_6=\frac{6}{2}(19+69)=3\times88=264\). Therefore, 264 is correct. An answer such as 254 can result from an error in finding the last term or multiplying. Exam tip: check the last term before substituting it in the sum formula.

Here, the first term is \(a=32\), the common difference is \(d=4\), and the number of terms is \(n=22\). The 22nd term is \(a_{22}=32+(22-1)\times4=116\). Hence, \(S_{22}=\frac{22}{2}(32+116)=11\times148=1628\). Therefore, option C is correct. A value such as 1616 may result from using an incorrect last term or mishandling \(n-1\). In an exam, identify \(a\), \(d\), and \(n\) before applying \(S_n=\frac{n}{2}[2a+(n-1)d]\).