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

Correct answer: A. (120)

Using (S_n=\frac{n}{2}[2a+(n-1)d]), we get (S_{10}=120). In exams, first identify (a), (d), and (n).

Correct answer: B. 258

The governing concept is the sum of the first n terms of an arithmetic progression: S_n = n/2[2a + (n − 1)d]. Here the first term is a = 5, the common difference is d = 8 − 5 = 3, and n = 12. Substitution gives S_12 = 12/2[2(5) + 11(3)] = 6[10 + 33] = 6 × 43 = 258. Therefore option B is correct. A useful independent check is to find the twelfth term: a_12 = 5 + 11 × 3 = 38. The sum is then 12(5 + 38)/2 = 6 × 43 = 258. Option A may result from using an incorrect number of intervals, while options C and D do not satisfy the standard AP formula.

Correct answer: C. 450

For this AP, the first term is \(a=2\), the common difference is \(d=4\), and \(n=15\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{15}=\frac{15}{2}[2(2)+14(4)]=\frac{15}{2}\times60=450\). Hence, 450 is correct. A value such as 420 usually results from an error in finding the last term or using \((n-1)\). Exam tip: write down \(a\), \(d\), and \(n\) separately before substituting in the formula.

Correct answer: B. (250)

The direct answer is option B, 250. The sum of n terms of an AP is S_n=n/2(first term+last term). Here n=10, the first term is 7, and the last term is 43. Therefore S_10=10/2×(7+43)=5×50=250. This works because pairing the first and last terms gives equal sums: 7+43=50, and there are five such pairs. Option A, 240, is too small and does not follow the formula. Option B, 250, is correct. Option C, 260, is too large by 10, and option D, 270, is larger still. No common difference is needed because both the number of terms and the two end terms are already given. Exam cue: when first term, last term, and number of terms are known, use S_n=n/2(a+l), not a longer term-by-term addition.

Correct answer: A. 324

The governing concept is the sum of the first n terms of an arithmetic progression. The first 18 odd natural numbers are 1, 3, 5, ..., 35, with first term a = 1 and common difference d = 2. Using the AP sum formula, S_n = n/2[2a + (n - 1)d], we obtain S_18 = 18/2[2(1) + 17(2)] = 9(2 + 34) = 9 × 36 = 324. There is also a useful identity that the sum of the first n odd natural numbers is n²; therefore 18² = 324. Hence option A is correct. Options B, C, and D are larger values caused by incorrect term counting, an incorrect last term, or an arithmetic error.

Correct answer: C. (650)

The sum of the first (n) even numbers is (n(n+1)), so (25\times26=650). For even numbers, start from (2,4,6,\ldots).

Correct answer: B. (220)

Here (a=10), (d=5), and (n=8), so the sum is (220). For small (n), you can also check by finding the last term.

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

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, \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Option C is the formula for the \(n\)th term, while option D gives only the average of the first and last terms. Exam tip: use this form when the last term \(l\) is known.

Correct answer: B. 42

The governing concept is the finite-sum formula for an arithmetic progression, S_n = n/2[2a + (n − 1)d]. The first term is a = 12, and the common difference is d = 10 − 12 = −2 because the terms decrease. With n = 7, S_7 = 7/2[2(12) + 6(−2)] = 7/2[24 − 12] = 7/2 × 12 = 42. This can be checked by finding the seventh term: a_7 = 12 + 6(−2) = 0. The average of the first and last terms is (12 + 0)/2 = 6, so the sum of seven terms is 7 × 6 = 42. Hence option B is correct. Treating the difference as positive would lead to an incorrect distractor.

Correct answer: B. 330

The governing concept is the sum of the first n terms of an arithmetic progression. When the first term a, last term l, and number of terms n are known, use Sₙ = n/2(a + l). Here n = 10, a = 6, and l = 60. Substitution gives S₁₀ = 10/2(6 + 60) = 5 × 66 = 330. Therefore option B is correct. The common difference is not needed because both end terms and the number of terms already determine the average term: (6 + 60)/2 = 33, and 10 terms give 10 × 33 = 330. Option A is too small, while C and D arise from incorrect arithmetic or misuse of the formula.

Correct answer: C. (820)

Here the last term is (79), and (S_{20}=\frac{20}{2}(3+79)=820). Finding the last term can often be easier.

Correct answer: C. (215)

This is an arithmetic progression with (a=8), (d=3), and (n=10), so there will be (215) chairs. In word problems, convert the pattern into a progression.

Correct answer: C. 125

The sum of the first 10 terms is \(S_{10}=175\), while the sum of the first 5 terms is \(S_5=50\). Hence, the sum from the 6th to the 10th term is \(S_{10}-S_5=175-50=125\). Subtracting \(S_5\) removes exactly the first five terms; 120 does not follow from the correct partial-sum relation. Exam tip: To find the sum of consecutive terms, subtract the smaller partial sum from the larger partial sum.

Correct answer: C. (90)

The first (6) terms go from (20) to (10), and the average is (15), so the sum is (90). The average of equally spaced terms is useful.

Correct answer: A. 217

The governing concept is direct and careful substitution in the arithmetic-progression sum formula S_n = n/2[2a + (n - 1)d]. The supplied values are a = 9, d = 1, and n = 14. Therefore S_14 = 14/2[2(9) + (14 - 1)(1)] = 7[18 + 13] = 7 × 31 = 217. Thus option A is correct. The expression n - 1 is essential because the distance from the first term to the nth term contains n - 1 common differences. Replacing n - 1 by n would give 224, one of the distractors. The other values result from arithmetic or substitution mistakes, not from the given formula.

Correct answer: B. (176)

Here (a=1), (d=3), and (n=11), so (S_{11}=176). For an odd number of terms, you can also check using the middle term.

Correct answer: C. (225)

Sum equals average (\times) number of terms, so (25\times9=225). When the average is given, the long formula is not needed.

Correct answer: B. (429)

The sum of n terms of an arithmetic progression can be found from the first term, last term, and number of terms using \(S_n=\frac{n}{2}(a+l)\). This works because the terms can be paired from the two ends, and every pair has the same total. Equivalently, the sum equals the number of terms multiplied by the average of the first and last terms.

Here \(n=13\), the first term is \(a=15\), and the last term is \(l=51\). Therefore, \(S_{13}=\frac{13}{2}(15+51)=\frac{13}{2}\times66=13\times33=429\). The average term is \((15+51)/2=33\), and 13 times 33 also gives 429. Hence option B, 429, is correct. The number of terms must be 13, not the number of gaps between them.

Correct answer: C. (100)

The first (5) terms are (30,25,20,15,10), so the sum is (100). In easy questions, writing the terms can also verify the answer.

Correct answer: B. (192)

Here (a=5), (d=2), and (n=12), so (192) questions are solved. Treat days as the number of terms.

Correct answer: B. (48)

The first (6) terms are (3,5,7,9,11,13), whose sum is (48). When (a_n) is given, find the first and last terms.

Correct answer: B. 664

Here, the first term is \(a=4\), the common difference is \(d=9-4=5\), and \(n=16\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{16}=\frac{16}{2}[2(4)+15(5)]=8(83)=664\). Therefore, 664 is the correct answer. A value such as 672 usually results from an addition or multiplication error. Exam tip: calculate \((n-1)d\) first, then substitute carefully in the formula.

Correct answer: B. \(35\)

The sum formula for the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Here, \(S_7=140\), \(n=7\), and \(a=5\). Thus, \(140=\frac{7}{2}(5+l)\), which gives \(40=5+l\) and hence \(l=35\). Option \(40\) is not correct because it is the value of \(a+l\), not the last term. Exam tip: in questions involving the last term, first calculate \(a+l=\frac{2S_n}{n}\).

Correct answer: C. (330)

The direct answer is C: 330. The AP is 6, 12, 18, ..., so the first term is \(a=6\), the common difference is \(d=6\), and the number of terms is \(n=10\). The tenth term is \(a_{10}=a+9d=6+9(6)=60\). Now use the sum formula \(S_n=\frac n2[2a+(n-1)d]\): \(S_{10}=\frac{10}{2}[2(6)+9(6)]=5(12+54)=5\times66=330\). Option A, 300, is too small and does not use the correct final term. Option B, 320, is also not the result of the AP sum. Option C, 330, is correct. Option D, 360, is too large; it can arise from an incorrect average or term count. A quick check is first term plus last term, divided by 2: average \((6+60)/2=33\), then \(33\times10=330\).

Correct answer: D. (465)

(\frac{30\times31}{2}=465), so the correct sum is (465). Remember the direct formula for natural numbers.