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

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

The sum is \(S_n=\frac{n}{2}[2a+(n-1)d]\), since the last term is \(a+(n-1)d\). Option B gives only the \(n\)th term, not the sum. Exam tip: always check the \((n-1)d\) factor.

Correct answer: A. 480

Here, the first term is \(a=4\), the common difference is \(d=4\), and \(n=15\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{15}=\frac{15}{2}[2(4)+14(4)]=\frac{15}{2}(64)=480\). Therefore, 480 is correct. A value such as 420 can result from using the wrong number of terms or last term. Exam tip: write down \(a\), \(d\), and \(n\) before applying the AP sum formula.

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

The \(n\)th term of an AP is \(l=a+(n-1)d\). Hence, \(S_n=\frac{n}{2}(a+l)=\frac{n}{2}[2a+(n-1)d]\). Option B multiplies the last term by \(n\), so it is incorrect. Exam tip: find \(l\) first to check the formula.

Correct answer: C. 625

Here, the first term is \(a=1\), the common difference is \(d=2\), and \(n=25\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{25}=\frac{25}{2}[2+24\times2]=\frac{25}{2}\times50=625\). Choosing 600 would result from an error in finding the last term or the number of terms. Exam tip: the sum of the first \(n\) odd numbers \(1,3,5,\ldots\) is directly \(n^2\).

Correct answer: B. 780

For this AP, the first term is \(a=10\), the common difference is \(d=10\), and \(n=12\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{12}=\frac{12}{2}[2(10)+11(10)]=6(130)=780\). Therefore, 780 is correct. A value such as 760 usually results from taking the number of terms or the last term incorrectly. Exam tip: always use \(n-1\) in the sum formula before substituting values.

Correct answer: C. 385

Here, the first term is \(a=7\), the common difference is \(d=7\), and \(n=10\). The tenth term is \(a_{10}=7+(10-1)\times7=70\). Therefore, \(S_{10}=\frac{10}{2}(7+70)=5\times77=385\). Hence, 385 is the correct option. Choosing 365 would result from using an incorrect number of terms or last term. Exam tip: identify \(a\), \(d\), \(n\), and the last term before applying the sum formula.

Correct answer: B. 225

When the first and last terms are known, the sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(S_9=\frac{9}{2}(5+45)=\frac{9}{2}\times 50=225\). Hence, 225 is correct. A value such as 215 can result from an addition or multiplication error. Exam tip: do not forget to multiply by the number of terms \(n\) in the sum formula.

Correct answer: A. 1395

For this AP, the first term is \(a=3\), the common difference is \(d=3\), and \(n=30\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{30}=\frac{30}{2}[2(3)+29(3)]=15(93)=1395\). Hence, 1395 is correct. An answer such as 1350 can result from using an incorrect last term or number of terms. Exam tip: write down \(a\), \(d\), and \(n\) before applying the sum formula.

Correct answer: B. 104

For this AP, the first term is \(a=20\), the common difference is \(d=-2\), and \(n=8\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_8=\frac{8}{2}[2(20)+7(-2)]=4(40-14)=104\). Therefore, the correct answer is 104. An option such as 108 can result from not handling the negative common difference correctly. Exam tip: identify \(a\), \(d\), and \(n\) before substituting in the sum formula.

Correct answer: A. 319

Here, the first term is \(a=9\), the common difference is \(d=13-9=4\), and \(n=11\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{11}=\frac{11}{2}[2(9)+10(4)]=\frac{11}{2}\times58=319\). Therefore, option A is correct. A nearby value such as option B may result from using an incorrect value for \((n-1)d\) or making a multiplication error. Exam tip: identify \(a\), \(d\), and \(n\) separately before applying the formula.

Correct answer: C. 623

Here, the first term is \(a=12\), the common difference is \(d=17-12=5\), and the number of terms is \(n=14\). Applying \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{14}=\frac{14}{2}[2(12)+13(5)]=7(89)=623\). Hence, 623 is correct. 615 is incorrect because a 14-term AP has \((n-1)=13\) common-difference intervals. Exam tip: identify \(a\), \(d\), and \(n\) before substituting in the sum formula.

Correct answer: C. 5

The first term of a sequence equals the sum of its first one term, so \(a=S_1\). Substituting \(n=1\) in the given expression gives \(S_1=3(1)^2+2(1)=5\). Therefore, the first term is 5. The number \(3\) is the coefficient of \(n^2\), not the first term. Exam tip: when \(S_n\) is given, find the first term directly by calculating \(S_1\).

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

The last term is \(l=a+(n-1)d\). Substituting it in \(S_n=\frac{n}{2}(a+l)\) gives option A. Option B gives only the last term, not the sum. Exam tip: check the \((n-1)d\) factor carefully.

Correct answer: B. 234

Here, \(a=6\), \(d=11-6=5\), and \(n=9\). The ninth term is \(a_9=a+(n-1)d=6+8\times5=46\). Hence, \(S_9=\frac{n}{2}(a+a_9)=\frac{9}{2}(6+46)=234\). Therefore, 234 is the correct answer. A result such as 238 can occur if the last term or the number of terms is taken incorrectly. Exam tip: write down \(a\), \(d\), and \(n\) before using the AP sum formula.

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

The nth, or last, term of the AP is \(l=a+(n-1)d\). Substituting this in \(S_n=\frac{n}{2}(a+l)\) gives \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B is only the formula for the nth term, not the sum, while option C incorrectly uses \(nd\) in place of \((n-1)d\). Exam tip: always check both the factor \(n\) and \((n-1)d\) in the sum formula.

Correct answer: A. 272

The first 16 even natural numbers are 2, 4, 6, ..., 32. They form an AP with \(a=2\), \(d=2\), and \(n=16\). Therefore, \(S_{16}=\frac{16}{2}[2(2)+(16-1)\cdot2]=272\). Hence, 272 is correct. Although 256 equals \(16^2\), the sum of the first 16 even natural numbers is given by \(n(n+1)\). Exam tip: use \(n(n+1)\) directly for the sum of the first \(n\) even natural numbers.

Correct answer: C. 975

Here, the first term is \(a=15\), the common difference is \(d=25-15=10\), and \(n=13\). Thus, \(S_{13}=\frac{13}{2}[2(15)+(13-1)\times10]=\frac{13}{2}(150)=975\). Therefore, 975 is correct. \(1040\) results if \(n\) common differences are incorrectly used in place of \((n-1)\). Exam tip: Always check the \((n-1)\) term in the AP sum formula.

Correct answer: B. 155

Substituting \(n=10\) gives \(S_{10}=\frac{10}{2}(3\times10+1)=5\times31=155\). Hence, 155 is correct. A value such as 145 may result from evaluating \(3\times10+1\) incorrectly. Exam tip: simplify the expression inside the brackets before multiplying by \(\frac{n}{2}\).

Correct answer: C. 165

For this AP, the first term is \(a=30\), the common difference is \(d=27-30=-3\), and \(n=10\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{10}=\frac{10}{2}[2(30)+9(-3)]=5(60-27)=165\). Therefore, 165 is correct. If you get 160, recheck the common difference or the number of terms. Exam tip: always retain the negative sign of \(d\) in a decreasing AP.

Correct answer: A. If the first term and last term are known, then \(S_n=\frac{n}{2}(a+l)\).

The average of the first and last terms of an AP is \(\frac{a+l}{2}\). Therefore, the sum of the first \(n\) terms is the number of terms multiplied by this average: \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Hence, option A is correct. Option C gives only the average of the first and last terms, not the sum. Exam tip: \(l\) denotes the last term, whereas \(d\) denotes the common difference.

Correct answer: B. 495

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.

Correct answer: A. 354

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.

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

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.

Correct answer: C. 930

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.

Correct answer: C. 175

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.