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

Using (S_n=\frac{n}{2}(a+l)), the sum is (350). When the last term is given, this formula is faster.

The expression S_n directly gives the sum of the first n terms of the sequence, so no separate common difference or nth-term calculation is needed. Substitute n = 12: S_12 = 3(12)^2 + 2(12) = 3 × 144 + 24 = 432 + 24 = 456. Therefore, option B is correct. Option A, 452, could result from an arithmetic subtraction error; option C, 460, does not follow from the given expression; and option D, 468, may arise from adding 36 instead of 24. The governing concept is direct substitution into the formula for the sum of the first n terms.

This is an AP with (a=18), (d=3), (n=16), and the sum is (648). In word problems, identify (a,d,n) first.

Here (d=-3), and the formula gives (S_{18}=441). In a decreasing AP, write the common difference as negative.

From (\frac{n}{2}(4+100)=1040), (n=20). When last term and sum are given, use (S_n=\frac{n}{2}(a+l)).

The sum of the first (n) even numbers is (n(n+1)), so (40\times41=1640). Remember it for even-number AP questions.

Solving (\frac{n}{2}[18+6(n-1)]=525) gives (n=10). For (n), a quadratic may appear, so choose the positive value.

From (\frac{n}{2}[4+5(n-1)]=287), (n=11). When finding the number of terms from sum, reject the negative root.

In an AP, the difference between the sums of the first n and first \(n-1\) terms gives the nth term: \(a_n=S_n-S_{n-1}\). Thus, \(a_{10}=S_{10}-S_9=250-207=43\). Therefore, 43 is correct. Getting 42 would mean subtracting the given sums incorrectly. Exam tip: When two consecutive partial sums are given, subtract them directly to obtain the corresponding term.

Solving (\frac{n}{2}[12+5(n-1)]=660) gives (n=15). Simplify the bracket first, then solve the equation.

From (780=10[10+19d]), (d=4). When sum and first term are given, the common difference can be found directly.

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.

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

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.

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 \(n\).

The formula gives (S_{25}=\frac{25}{2}[40-24]=200). In a decreasing AP, the last term may be smaller.

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.

The daily production forms an arithmetic progression because the increase from one day to the next is constant: d = 135 − 120 = 15. Here the first term is a = 120 and the number of days is n = 10. Using S_n = n/2[2a + (n − 1)d], we get S_10 = 10/2[2(120) + 9(15)] = 5[240 + 135] = 5 × 375 = 1875 items. Hence option A is correct. The other values do not result from applying the AP sum formula; simply multiplying the first or last daily production by ten would also be incorrect because production changes each day.

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

Here the last term is (-26), and (S_{20}=50). In a decreasing AP, the sum can become quite small.