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

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(456=\frac{12}{2}(8+l)=6(8+l)\). Hence, \(8+l=76\), so \(l=68\). If 66 were used, the sum would be 444, not 456. Exam tip: When the first and last terms are involved, use \(S_n=\frac{n}{2}(a+l)\) directly.

Using (S_n=\frac{n}{2}[4+9(n-1)]), we get (n=12). When finding (n) from a given sum, check the options.

The multiples are from (6) to (96), and there are (16) terms, so the sum is (816). In boundary questions, decide the last term first.

The numbers are (56,63,\ldots,147), with (14) terms, so the sum is (1407). For numbers between limits, choose the first and last terms carefully.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(l\) is the last term. Therefore, \(279=\frac{9}{2}(7+l)\). This gives \(7+l=62\), so \(l=55\). If the last term were \(53\), the sum would be \(270\), not \(279\). Exam tip: When the first term and the sum are given, use \(S_n=\frac{n}{2}(a+l)\) to find the last term directly.

The eleventh term is (-8), so (S_{11}=\frac{11}{2}(32-8)=132). The formula remains the same even with a negative last term.

The fifteenth term is (96), so the total is (S_{15}=\frac{15}{2}(12+96)=810) fruits. Treat the number of days as the number of terms.

The sum of the (9)th to (15)th terms is (S_{15}-S_8=371). For a group of consecutive terms, take the difference of partial sums.

The sum of the first n odd natural numbers is \(n^2\). Therefore, the sum of the first 18 odd numbers is \(18^2=324\), and the sum of the first 10 odd numbers is \(10^2=100\). Hence, the sum of the remaining 8 numbers is \(324-100=224\). A value such as 234 may result from an error in subtraction. Exam tip: when an initial set of terms is removed, subtract its sum from the total sum.

The sum equals the number of terms multiplied by the average of the first and last terms. Hence \(S_n=\frac{n}{2}(a+l)\). Option C gives only the average, not the sum. Exam tip: check the factor \(n\).

Putting (n=12) in (S_n=\frac{n}{2}[2+7(n-1)]) gives (504). Checking options is a quick method.

The tenth deposit is (825) rupees, so the total is (S_{10}=\frac{10}{2}(150+825)=4875) rupees. Do not forget the unit in money problems.

The sequence is (20,30,\ldots,200) with (19) terms, so the sum is (2090). Include both boundary terms.

The first n even natural numbers are 2, 4, 6, ..., 2n. Their sum is 2(1 + 2 + ... + n) = 2[n(n + 1)/2] = n(n + 1). Therefore n(n + 1) = 420. We look for consecutive positive integers whose product is 420: 20 × 21 = 420. Hence n = 20, so option C is correct. Checking directly, the first 20 even numbers end at 40 and their sum is 20 × 21 = 420. The other choices give products 342, 380, and 462, so they do not satisfy the condition.

The twelfth term is (-21), so (S_{12}=\frac{12}{2}(45-21)=144). In decreasing order, keep (d) negative while finding the last term.

To find the sum of the first 8 terms, substitute \(n=8\) in the given formula: \(S_8=3(8)^2+2(8)=3\times64+16=192+16=208\). Therefore, the correct answer is 208. The value 198 may result from an error while evaluating the \(2n\) term. Exam tip: Calculate \(n^2\) first, then perform multiplication and addition carefully.

For an AP, \(S_n=\frac{n}{2}(a+l)\), where \(l\) is the last term. Thus, \(203=\frac{7}{2}(5+l)\). So \(406=7(5+l)\), giving \(5+l=58\) and hence \(l=53\). Option 58 is the value of \(a+l\), not the seventh term. Exam tip: When the sum and the last term are involved, use \(S_n=\frac{n}{2}(a+l)\).

The twenty-first term is (149), so (S_{21}=\frac{21}{2}(9+149)=1659). For larger (n), write the last term first.

The sixteenth term is (-5), so (S_{16}=\frac{16}{2}(25-5)=160). With a negative difference, the last term decreases.

In the twelfth week, (195) books arrive, so the total is (1350) books. Treat the pattern as (a=30), (d=15), (n=12).