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

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

Using (S_n=\frac{n}{2}[2a+(n-1)d]), we get (S_{10}=120). In exams, first identify (a), (d), and (n).

The governing concept is the sum of the first n terms of an arithmetic progression: S_n = n/2[2a + (n - 1)d]. Here the first term is a = 5, the common difference is d = 8 - 5 = 3, and n = 12. Therefore, S_12 = 12/2[2(5) + 11(3)] = 6(10 + 33) = 6 × 43 = 258. Hence option B is correct. Option A may result from using an incorrect final term or number of terms. Options C and D do not follow from the AP-sum formula and are therefore not valid.

For this AP, the first term is \(a=2\), the common difference is \(d=4\), and \(n=15\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{15}=\frac{15}{2}[2(2)+14(4)]=\frac{15}{2}\times60=450\). Hence, 450 is correct. A value such as 420 usually results from an error in finding the last term or using \((n-1)\). Exam tip: write down \(a\), \(d\), and \(n\) separately before substituting in the formula.

Using (S_n=\frac{n}{2}(a+l)), we get (S_{10}=250). When the last term is given, this formula is faster.

The governing result is that the sum of the first n odd natural numbers is n². These numbers form the arithmetic progression 1, 3, 5, ..., whose first term is 1 and common difference is 2. With n = 18, the sum is S_18 = 18² = 18 × 18 = 324. The same result follows from the AP formula: S_18 = 18/2[2(1) + 17(2)] = 9(2 + 34) = 324. Thus option A is correct. Option B is larger than the correct square, while C and D also arise from incorrect counting or substitution.

The sum of the first (n) even numbers is (n(n+1)), so (25\times26=650). For even numbers, start from (2,4,6,\ldots).

Here (a=10), (d=5), and (n=8), so the sum is (220). For small (n), you can also check by finding the last term.

The sum of the first \(n\) terms of an AP equals the number of terms multiplied by the average of the first and last terms. Thus, \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Option C is the formula for the \(n\)th term, while option D gives only the average of the first and last terms. Exam tip: use this form when the last term \(l\) is known.

Use the AP sum formula S_n = n/2[2a + (n - 1)d]. The first term is a = 12, the common difference is d = 10 - 12 = -2, and the number of terms is n = 7. Hence S_7 = 7/2[24 + 6(-2)] = 7/2(24 - 12) = 7/2 × 12 = 42. Equivalently, the seventh term is 12 + 6(-2) = 0, so the average of the first and last terms is 6 and the sum is 7 × 6 = 42. Therefore option B is correct. The negative difference must not be replaced by +2.

The governing concept is the sum of the first n terms of an arithmetic progression. When the first term a, last term l, and number of terms n are known, use Sₙ = n/2(a + l). Here n = 10, a = 6, and l = 60. Substitution gives S₁₀ = 10/2(6 + 60) = 5 × 66 = 330. Therefore option B is correct. The common difference is not needed because both end terms and the number of terms already determine the average term: (6 + 60)/2 = 33, and 10 terms give 10 × 33 = 330. Option A is too small, while C and D arise from incorrect arithmetic or misuse of the formula.