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

Correct answer: B. \(S_n=\frac{n}{2}(a+l)\)

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.

Correct answer: B. 2550

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.

Correct answer: B. 702

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 \(S_n=\frac{n}{2}(a+l)\).

Correct answer: C. 450

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.

Correct answer: C. 168

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.

Correct answer: A. The last row has 46 chairs and there are 270 chairs in total.

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.

Correct answer: A. 55

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.

Correct answer: B. 364

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 \(S_n=\frac{n}{2}[2a+(n-1)d]\).

Correct answer: A. \(S_n=\frac{n}{2}[2a+(n-1)d]\)

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 \(nd\).

Correct answer: B. 270

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.

Correct answer: B. 220

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.

Correct answer: A. \(S_n=\frac{n}{2}[2a+(n-1)d]\)

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.

Correct answer: B. 1550

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.

Correct answer: C. 462

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.

Correct answer: C. 1104

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.

Correct answer: B. 225

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.

Correct answer: C. 395

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.

Correct answer: A. \(a_n=S_n-S_{n-1}\)

\(S_n=a_1+a_2+\cdots+a_{n-1}+a_n\), 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.

Correct answer: B. 240

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.

Correct answer: C. 840

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.

Correct answer: B. 294

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 \(n-1\).

Correct answer: D. 702

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.

Correct answer: C. 510

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.

Correct answer: A. 264

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.

Correct answer: C. 1628

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 \(S_n=\frac{n}{2}[2a+(n-1)d]\).