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

For an AP, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(441=\frac{18}{2}[2(3)+17d]=9(6+17d)\). Hence \(49=6+17d\), so \(17d=43\) and \(d=\frac{43}{17}\). If \(d=3\), the sum would be \(513\), so it is not correct. Exam tip: first divide by \(\frac{n}{2}\) to simplify the equation.

The sum of an AP equals the number of terms multiplied by the average of its first and last terms. Thus, \(S_n=n\times\frac{a+l}{2}\). Option B misses division by 2. Exam tip: remember “number of terms × average of extreme terms.”

The numbers are (105,110,\ldots,195), and the sum of (19) terms is (2850). The word between often excludes endpoints.

From (\frac{n}{2}[26+5(n-1)]=253), (n=11). Only the correct positive integer can be the number of terms.

Here (a=1), (d=3), (n=30), so (S_{30}=1335). Even with a small first term, the sum can be large.

This is an AP with first term \(a=40\), common difference \(d=3\), and \(n=15\) terms. Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{15}=\frac{15}{2}[2(40)+14(3)]=\frac{15}{2}(122)=915\) cm. Hence, the correct answer is 915 cm. An option such as 900 cm may result from using the common difference or last term incorrectly. Exam tip: always use \(n-1\) in the AP sum formula.

For an AP, \(S_n=\frac n2[2a+(n-1)d]\), so \(S_n\) is a quadratic expression with zero constant term. \(3n^2-2n=n(3n-2)\) has this form, and its second difference is the constant \(6\). \(n^2+n+1\) has a non-zero constant term, so it does not fit the standard AP-sum form. Exam tip: check differences using \(S_1,S_2,S_3\).

From (\frac{n}{2}[34+2(n-1)]=580), (n=20). After forming the equation, you can also check using options.

The last term is (45), so (S_{12}=\frac{12}{2}(100+45)=870). Once the last term is found, the sum is quick.

The first n natural numbers are 1, 2, 3, ..., n, an arithmetic progression with first term 1 and last term n. Its sum is S_n = n(n + 1)/2. Equating this to 465 gives n(n + 1)/2 = 465, so n(n + 1) = 930. Since 30 × 31 = 930, n = 30. The positive value is selected because n counts terms and must be a positive integer. Therefore, option C is correct. The nearby choices 28, 29, and 31 fail the equation: their products with their successors are 812, 870, and 992, respectively, not 930.

First find (n): (83=8+(n-1)5), so (n=16) and the sum is (728). When the last term is given, find the number of terms first.

This is the AP (15,20,25,\ldots), and the sum for (14) days is (665) minutes. In word problems, treat days as the number of terms.

In an AP, the nth term can be found using \(a_n=S_n-S_{n-1}\). Therefore, \(a_{25}=S_{25}-S_{24}=1225-1128=97\). Hence, the correct answer is 97. For example, 96 is incorrect because it is not the difference between \(S_{25}\) and \(S_{24}\). Exam tip: The difference between two consecutive partial sums always gives the newly added term.

The last term is (-21), and (S_{16}=\frac{16}{2}(24-21)=24). Positive and negative terms can greatly reduce the sum.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(390=\frac{12}{2}[2a+11\times5]=6(2a+55)\). Hence \(2a+55=65\), so \(2a=10\) and \(a=5\). If 4 were used as the first term, the sum would be 378, so it is not correct. Exam tip: In the sum formula, use \((n-1)d\), not \(nd\).

The AP is (4,8,\ldots,48) with (12) terms, and the sum is (312). Identifying the last allowed multiple is important.

The original numerical target 425 is inconsistent with this AP because S_n = n/2[2(5) + (n − 1)4] = n(2n + 3), and no listed integer n gives 425. Correcting the target to 434 makes the question valid. For 434, solve n(2n + 3) = 434, or 2n^2 + 3n − 434 = 0. Factoring gives (n − 14)(2n + 31) = 0, so the positive value is n = 14. Verification: the 14th term is 5 + 13(4) = 57, and S_14 = 14/2(5 + 57) = 7 × 62 = 434. Thus option C is correct.

This is the sum of the first (25) multiples of (6), so the total distance is (1950) meters. In sums of multiples, (a=d).

Given \(S_n=5n^2-n\), substitute \(n=9\): \(S_9=5(9)^2-9=5\times81-9=405-9=396\). Hence, 396 is the correct option. 405 is only \(5\times81\); the final subtraction of 9 must also be done. Exam tip: after substituting \(n\), evaluate the power first, then multiply and subtract.

The governing concept is the sum of the first n terms of an arithmetic progression. Here, the first term is a = 7, the common difference is d = 5, and the last term is 82. Using 82 = 7 + (n − 1)5 gives n = 16. Therefore, Sₙ = n/2(first term + last term) = 16/2 × (7 + 82) = 8 × 89 = 712. Hence option B is correct; the other values result from an incorrect term count or sum.