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

The AP has first term a = 14 and common difference d = 17 - 14 = 3. For n = 10, its tenth term is l = a + (n - 1)d = 14 + 9 × 3 = 41. The sum formula using the first and last terms is S_n = n/2(a + l), so S_10 = 10/2(14 + 41) = 5 × 55 = 275. Therefore option B is correct. A direct calculation with S_n = n/2[2a + (n - 1)d] gives the same result: 5[28 + 27] = 275. Other choices reflect arithmetic or term-counting errors.

The sum of the first \(n\) terms of an AP is the number of terms multiplied by the average of the first and last terms. Hence, \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Option B does not divide by \(2\), while option C gives only the average of the first and last terms, not the sum. Exam tip: use this formula directly when the last term \(l\) is known.

The ninth term is (10), so (S_9=\frac{9}{2}(50+10)=270). Because the difference is negative, the last term will be smaller.

The sum of the (7)th to (12)th terms is (S_{12}-S_6=135). Find the sum of middle terms by subtracting partial sums.

This is the sum of the first (8) multiples of (7), so (7\times36=252). For multiples, use the sum of natural numbers.

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, \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Therefore, option B is correct. Option A misses division by 2, while option C gives only the average of the first and last terms. Exam tip: use this formula directly when the last term \(l\) is known.

The fourteenth term is (41), so (S_{14}=\frac{14}{2}(2+41)=301). Finding the last term correctly is the main step.

Here (a=100), (d=50), and (n=6), so the total saving is (1350) rupees. In money questions, keep the unit in mind.

The ninth term is (40), so (S_9=\frac{9}{2}(16+40)=252). The average of the first and last terms is the middle term.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Substituting the given values, \(S_7=\frac{7}{2}(25+5)=\frac{7}{2}\times30=105\). Therefore, the correct answer is 105. Option 110 is incorrect because \(\frac{7}{2}\) times 30 equals 105, not 110. Exam tip: first add the values inside the brackets, then simplify with 2 before multiplying whenever possible.

This is the sum of the first (5) multiples of (11), so (11\times15=165). For multiples, the sum of the first (n) natural numbers is useful.

The sum of the first 10 terms includes the sum of the first 5 terms. Therefore, the sum of the 6th to 10th terms is 145 - 45 = 100. Obtaining 95 would result from an incorrect subtraction. Exam tip: To find the sum of a consecutive block of terms, subtract the smaller partial sum from the larger partial sum.

The first term is a = 18 and the common difference is d = 24 - 18 = 6. The seventh term is a_7 = a + (7 - 1)d = 18 + 6 × 6 = 54. Using the average-of-end-terms method, S_7 = 7/2(18 + 54) = 7/2 × 72 = 7 × 36 = 252. Therefore option B is correct. The sum formula gives the same result: 7/2[2(18) + 6(6)] = 7/2(72) = 252. The nearby values are distractors from incorrect multiplication or an incorrect final term.

This is a word problem involving the sum of an arithmetic progression. The numbers of bricks in the levels are 3, 6, 9, ..., 36, with first term a = 3, common difference d = 3, and n = 12. The total is S_12 = 12/2[2(3) + (12 - 1)(3)] = 6(6 + 33) = 6 × 39 = 234. Equivalently, these are 3 times the first 12 natural numbers, so the total is 3 × [12 × 13/2] = 234. Thus option D is correct; the other choices undercount or miscalculate the final level.

Using \(\text{average}=\frac{\text{sum}}{\text{number of terms}}\), the sum is \(18\times 8=144\). Therefore, 144 is the correct option. A value such as 148 would result from using an incorrect average or number of terms. Exam tip: When the average and the number of terms are given, multiply them to find the sum.

The tenth term is (13), so (S_{10}=\frac{10}{2}(40+13)=265). In a decreasing progression, the last term is smaller.

This is the sum of the first (50) natural numbers, which is (1275). When (d=1) and (a=1), use the natural-number sum formula.

These are the (3)rd to (8)th multiples of (7), or directly (a=21), (d=7), (n=6) gives (231). Look carefully at the first term.

The eighth term is (59), so (S_8=\frac{8}{2}(17+59)=304). Finding the last term first makes the sum easier.

Here (a=12), (d=4), and (n=11), so there will be (352) books. In word problems, treat the given pattern as an arithmetic progression.