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

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.

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.

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.

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 \(n^2\).

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.

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.

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.

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.

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.

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.

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.

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 \(S_1\).

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.

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.

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.

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.

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.

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 \(\frac{n}{2}\).

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.

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.

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.