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

Correct answer: C. (2090)

Using (S_n=\frac{n}{2}[2a+(n-1)d]), the sum is (2090). In fraction-based questions, simplify the bracket first.

Correct answer: A. (10602)

The required sum is (S_{60}-S_{24}=10602). For a middle range, subtract the sum up to the term just before it.

Correct answer: D. (2484)

First (209=-25+(n-1)9) gives (n=27), and the sum is (2484). When the last term is given, find (n) first.

Correct answer: B. (83664)

Subtracting the sum of multiples of (\operatorname{lcm}(18,30)) from the sum of multiples of (18) gives (83664). In a but-not condition, remove the overlap.

Correct answer: C. (400500)

This is an AP with (a=5000), (d=350), (n=36), and the total is (400500). In word problems, treat each amount as a term.

Correct answer: A. (1339659)

The first number is (1036), the last is (9990), and there are (243) terms, so the sum is (1339659). Choose the first and last multiples carefully.

Correct answer: D. (2765)

The direct answer is D, 2765. In the AP, the first term is 300, the common difference is d=287-300=-13, and the number of terms is 35. Use the last-term formula: l=a+(n-1)d=300+34(-13)=300-442=-142. Now use S_n=n/2(a+l). Thus S_35=35/2(300-142)=35/2(158)=35(79)=2765. Option D is correct. Option A, 2715, is too small and results from an arithmetic error. Option B, 2735, is also incorrect because it does not follow the AP sum calculation. Option C, 2755, is close but still misses the correct value, usually because of an error in the last term or multiplication. A decreasing AP can still have a positive sum here because the average of the first and last terms is 79. The exam cue is to keep the negative sign in d and check the last term before adding.

Correct answer: B. (11760)

The expression \(S_n=3n^2+2n\) gives the sum of the first \(n\) terms. To find the sum from the 51st through the 80th term, subtract the sum of the first 50 terms from the sum of the first 80 terms. This leaves exactly the required terms, since all terms before the 51st are removed.

Compute \(S_{80}=3(80)^2+2(80)=3(6400)+160=19360\). Also, \(S_{50}=3(50)^2+2(50)=3(2500)+100=7600\). Therefore the required sum is \(S_{80}-S_{50}=19360-7600=11760\). Hence option B is correct. The subtraction must use \(S_{50}\), not \(S_{51}\), because the 51st term must remain in the answer.

Correct answer: C. (4212)

The given sums give (a=12) and (d=6), so (S_{36}=4212). From two partial sums, determine the AP first.

Correct answer: D. (2280)

The direct answer is D, 2280. Here a=250, d=238-250=-12, and n=30. Find the last term first: l=a+(n-1)d=250+29(-12)=250-348=-98. Then S_n=n/2(a+l), so S_30=30/2(250-98)=15(152)=2280. Therefore option D is correct. Option A, 2200, is wrong because it does not result from the correct average of the first and last terms. Option B, 2240, is also an arithmetic or substitution error. Option C, 2320, is too large and may come from treating the common difference as positive or mishandling the final term. The formula works because the terms of an AP are evenly spaced, so their average is (first term+last term)/2. Multiplying that average by the number of terms gives the sum. Remember: for a decreasing AP, d is negative; find l carefully before using S_n=n/2(a+l).

Correct answer: A. 7409

Option A is correct. The sequence is an arithmetic progression because the difference between consecutive terms is constant: d = 43 − 29 = 14 and 57 − 43 = 14. Here the first term is a = 29 and the number of terms is n = 31. Use S_n = n/2 [2a + (n − 1)d]. Substitution gives S_31 = 31/2 [2(29) + (31 − 1)(14)] = 31/2 [58 + 420] = 31/2 × 478 = 31 × 239 = 7409. As a check, the 31st term is 29 + 30 × 14 = 449, and the average of the first and last terms is (29 + 449)/2 = 239; multiplying this average by 31 again gives 7409. Therefore A is unambiguous. The other values result from arithmetic or formula errors.

Correct answer: C. (12600)

The formula gives (S_{45}=\frac{45}{2}[-100+660]=12600). Keep the sign of the negative first term carefully.

Correct answer: B. (\frac{55}{2})

From (4290=13[2a+275]), (a=\frac{55}{2}). In unknown-first-term questions, simplify the bracket first.

Correct answer: D. (1027752)

Subtracting the sum of multiples of (48) from the sum of multiples of (24) gives (1027752). In but-not cases, subtract the stricter condition.

Correct answer: A. (37259)

The first multiple is (209), the last is (1197), and there are (53) terms, so the sum is (37259). Choose the first multiple within the range correctly.

Correct answer: B. (910)

The given sums determine the AP, and (S_{30}-S_{20}=910). Before finding a later block sum, determine (a,d).

Correct answer: C. 341

For an AP, the \(n\)th term is \(a_n=S_n-S_{n-1}\). Therefore, \(a_{35}=S_{35}-S_{34}\). Here, \(S_{35}=5(35)^2-4(35)=5985\) and \(S_{34}=5(34)^2-4(34)=5644\). Hence, \(a_{35}=5985-5644=341\). Option 336 does not give the correct difference for \(n=35\). Exam tip: When \(S_n\) is given, find an individual term using \(S_n-S_{n-1}\).

Correct answer: C. 20792

This is an AP with first term 40 and common difference 8. The 30th row has \(40+29\times8=272\) seats, and the 75th row has \(40+74\times8=632\) seats. There are \(75-30+1=46\) rows from the 30th to the 75th row. Therefore, the required total is \(\frac{46}{2}(272+632)=20792\). Taking only the sum up to the 75th row would incorrectly include the first 29 rows as well. Exam tip: for terms from one position to another, count them as \(\text{last position}-\text{first position}+1\).

Correct answer: B. (29)

Solving (\frac{n}{2}[22+9(n-1)]=3973) gives (n=29). The number of terms must be a positive integer.

Correct answer: D. (38157)

The numbers are (111,124,\ldots,995), and their sum is (38157). A remainder-based AP has common difference equal to the divisor.

Correct answer: B. (12006)

The direct answer is B: the sum is 12006. In an arithmetic progression, consecutive terms change by the same common difference. Here the first term is (-40) and the common difference is 9, because (-31)-(-40)=9. The requested terms are the 25th through the 60th, so their number is 60-25+1=36. The 25th term is a_{25}=-40+24(9)=176, and the 60th term is a_{60}=-40+59(9)=491. Thus the sum is \frac{36}{2}(176+491)=18\times667=12006. Equivalently, use S_{60}-S_{24}. Option A, 11826, is too small and results from an arithmetic error. Option B, 12006, matches the correct calculation. Option C, 12186, and option D, 12366, are also incorrect values, usually caused by using the wrong number of terms or an incorrect endpoint. Remember the inclusive count: last term minus first term plus one.

Correct answer: D. (10341)

The given sums give (a=6) and (d=7), so (S_{54}=10341). From two partial sums, determine the AP first.

Correct answer: B. (1806)

The direct answer is B: 1806. A partial sum S_n means the sum of the first n terms of the AP. The terms from the 15th through the 28th are exactly the first 28 terms after removing the first 14 terms. Therefore the required sum is S_{28}-S_{14}=2576-770=1806. Option A, 1778, is not the stated difference and is therefore incorrect. Option B, 1806, follows directly from the definition of partial sums. Option C, 1834, and option D, 1862, are also incorrect; they may come from adding or subtracting an extra term, but neither equals the sum of terms 15 to 28. No common difference or first term is needed here, because both relevant partial sums are already supplied. The key exam step is to check the endpoints carefully: subtract the sum through the term just before the starting term. Memory cue: terms m to n equal S_n-S_{m-1}.

Correct answer: A. (5850)

From the two terms, (d=6) and (a=-2), so (S_{45}=5850). First find (a,d), then apply the sum formula.

Correct answer: B. (1485)

The condition gives (6+13d=3(6+3d)), so (d=3) and (S_{30}=1485). Convert the term condition into an equation first.