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

Correct answer: A. (12600)

This is an AP with (a=500), (d=100), (n=12), and the total is (12600). In real-life questions, treat each amount as a term.

Correct answer: B. (882)

Putting (a=15), (d=4), (n=18) in the formula gives (S_{18}=882). Always check (d) from the first three terms.

Correct answer: A. (60984)

The AP is (108,117,\ldots,999) with (100) terms, so the sum is (55350), not (60984). Find the last term and number of terms carefully.

Correct answer: A. 168

To find the sum of the first 8 terms, substitute \(n=8\) in the given expression: \(S_8=2(8)^2+5(8)=2\times64+40=168\). Hence, the correct answer is 168. The value 160 may result from an incorrect addition after calculating \(2\times64\). Exam tip: When \(S_n\) is given, directly substitute the required number of terms for \(n\).

Correct answer: B. (200)

The direct answer is option B, \(S_{25}=200\). In an AP, use \(S_n=\frac n2[2a+(n-1)d]\). The given values are \(a=20\), \(d=-1\), and \(n=25\). First calculate \(2a=40\) and \((n-1)d=24(-1)=-24\). Therefore \(S_{25}=\frac{25}{2}(40-24)=\frac{25}{2}\times16=200\). Option A, 195, is too small and may result from an arithmetic error. Option B, 200, follows exactly from the formula. Option C, 205, uses the wrong sign or adds one extra unit. Option D, 210, is also inconsistent with the calculated bracket. The last term is \(a+24d=20-24=-4\), so a positive sum of 200 is still possible because the early positive terms are larger. Memory cue: count \(n-1\) differences, not \(n\).

Correct answer: B. 602

Here, the first term is \(a=4\), the common difference is \(d=10-4=6\), and \(n=14\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{14}=\frac{14}{2}[2(4)+13(6)]=7(86)=602\). Therefore, 602 is correct. Although 604 is close, it does not result from correct substitution in the formula. Exam tip: identify \(a\), \(d\), and \(n\) before applying the sum formula.

Correct answer: A. 1875

The governing concept is applying the AP sum formula to a real-life sequence. Daily production increases by a constant amount: d = 135 − 120 = 15, so the first term is a = 120 and the number of terms is n = 10. Using S_n = n/2[2a + (n − 1)d], we obtain S_10 = 10/2[2(120) + 9(15)] = 5[240 + 135] = 5 × 375 = 1875 items. Thus option A is correct. Equivalently, the tenth-day production is 120 + 9 × 15 = 255, and the average daily production is (120 + 255)/2 = 187.5; multiplying by 10 gives 1875. Simply multiplying the first or last daily value by ten would ignore the changing production.

Correct answer: C. (14)

From (\frac{n}{2}(11+71)=574), (n=14). When the last term is given, the common difference is not needed.

Correct answer: A. (50)

The direct answer is A, 50. For the AP, the first term is approximately written as 31, the common difference is \\(d=-3\\), and the number of terms is \\(n=20\\). First find the twentieth term: \\(a_{20}=a+(20-1)d=31+19(-3)=31-57=-26\\). Now use \\(S_n=\\frac{n}{2}(a+l)\\), where \\(l\\) is the last term. Thus \\(S_{20}=\\frac{20}{2}(31+(-26))=10\\times5=50\\). Option A is correct. Option B, 55, would result from an arithmetic error in the final addition or multiplication. Option C, 60, is also not the value obtained from the AP formula. Option D, 65, is another incorrect calculation. A decreasing AP can have a small positive sum because its later negative terms partly cancel its earlier positive terms. Exam cue: always find the last term carefully, then apply the sum formula.

Correct answer: B. \(\frac{43}{17}\)

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.

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

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.”

Correct answer: A. (3150)

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

Correct answer: D. (11)

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

Correct answer: C. (1335)

Direct answer: Option C, 1335. The AP is 1, 4, 7, and so on. Therefore a = 1, d = 3 and n = 30. Use \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Thus \\(S_{30}=\\frac{30}{2}[2(1)+29(3)]\\) = 15[2 + 87] = 15 × 89 = 1335. Another check is useful: the thirtieth term is \\(a_{30}=1+29(3)=88\\); the average of the first and last terms is \\((1+88)/2=44.5\\), and 30 × 44.5 = 1335. Option C is correct. Option A, 1325, is ten less and may result from an addition or multiplication slip. Option B, 1330, is also not the result of the formula and may come from mishandling the average. Option D, 1340, is five more and is likewise an arithmetic error. The fact that the first term is only 1 does not keep the total small, because there are 30 terms and the terms increase by 3 each time. Exam cue: identify a, d and n before substituting.

Correct answer: B. 915 cm

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.

Correct answer: A. \(3n^2-2n\)

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 \(S_1,S_2,S_3\).

Correct answer: C. (20)

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

Correct answer: A. (870)

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

Correct answer: C. 30

The governing concept is the sum of the first n natural numbers, which form an AP beginning with 1 and ending with n. The formula is S_n = n(n + 1)/2. Setting the sum equal to 465 gives n(n + 1)/2 = 465, so n(n + 1) = 930. Since 30 × 31 = 930, the positive integer solution is n = 30. Option C is therefore correct. The positivity condition matters because n represents a number of terms and cannot be negative. Checking nearby choices confirms the result: 28 × 29 = 812, 29 × 30 = 870, and 31 × 32 = 992, so none of those produces 930. This is also an AP-sum problem because 1, 2, ..., n have a common difference of 1.

Correct answer: B. (728)

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.

Correct answer: C. (665)

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.

Correct answer: C. 97

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.

Correct answer: C. (24)

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

Correct answer: B. 5

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 \(nd\).

Correct answer: C. (312)

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