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

The notation \(S_n=6n^2-5n\) gives the sum of the first \(n\) terms, not the value of the \(n\)th term. To obtain the sum beginning with the 31st term and ending with the 50th term, subtract the sum through the 30th term from the sum through the 50th term. This removes terms 1 to 30 and leaves exactly terms 31 to 50.

Calculate \(S_{50}=6(50)^2-5(50)=15000-250=14750\), and \(S_{30}=6(30)^2-5(30)=5400-150=5250\). Therefore the required range sum is \(S_{50}-S_{30}=14750-5250=9500\). Hence option B is correct. Subtracting \(S_{31}\) instead would wrongly omit the 31st term.

The two partial sums give (a=\frac{17}{2}) and (d=5), so (S_{48}=6048). Determine the AP first.

The direct answer is option D, 1100. The first term is a=140, the common difference is d=132-140=-8, and n=25. Find the last term: l=a+(n-1)d=140+24(-8)=140-192=-52. Now use S_n=n/2(a+l): S_25=25/2(140-52)=25/2(88)=25(44)=1100. Option A, 1040, is too small and does not result from the correct endpoint average. Option B, 1060, is also an incorrect arithmetic result. Option C, 1080, is close but still not the correct sum. Option D, 1100, agrees with the formula. A decreasing AP can cross zero and have a negative final term; that does not make the sum formula invalid. Check the last term carefully: there are 24 differences after the first term, not 25. Memory cue: use l=a+(n-1)d, then average the first and last terms.

The direct answer is option A: 3220. In the AP 19, 30, 41, …, the first term is a=19 and the common difference is d=30−19=11. The number of terms is n=23. Apply \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). First, n−1=22. Then (n−1)d=22×11=242. Also, 2a=2×19=38. Add inside the bracket: 38+242=280. Therefore \\(S_{23}=\\frac{23}{2}×280=23×140=3220\\). Hence option A is correct. Option B, 3197, is wrong because it is 23 less than the correct sum and does not result from the formula. Option C, 3243, is wrong because it is 23 more than the correct sum, suggesting an incorrect extra amount. Option D, 3266, is also wrong because it uses an incorrect final total or arithmetic step. A useful check is to find the last term: \\(a_{23}=19+22×11=261\\). The average of the first and last terms is \\((19+261)/2=140\\), so the sum is 23×140=3220. The fact that n is odd causes no difficulty; the standard formula works for every positive whole number of terms. Exam cue: identify a, d and n before substituting, and do not forget that n−1=22, not 23.

The governing concept is the sum formula for the first n terms of an arithmetic progression: S_n = n/2[2a + (n − 1)d]. Substitute a = −35, d = 12, and n = 28. Thus S_28 = 28/2[2(−35) + 27(12)] = 14[−70 + 324] = 14 × 254 = 3556. Therefore option C is correct. The negative sign in the first term must be preserved when calculating 2a; treating −35 as positive changes the result. The other options are plausible distractors caused by using an incorrect multiplier, mishandling the negative sign, or making an arithmetic error in the final multiplication. The formula directly gives the required total.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(2450=\frac{20}{2}[2a+19\times9]\), so \(2450=10(2a+171)\). Hence \(245=2a+171\), giving \(2a=74\) and \(a=37\). If 39 is used, the sum becomes greater, so it is a close but incorrect distractor. Exam tip: remember to use \((n-1)d\), which is \(19d\) when \(n=20\).

The numbers are (105,147,\ldots,987), and their sum is (12012). In this condition, the new AP has common difference (42).

The first multiple is (128), the last is (896), and there are (49) terms, so the sum is (25088). Choose the first term according to the range carefully.

S₂₂ represents the sum of the first 22 terms, while S₁₁ represents the sum of the first 11 terms. Removing the first 11 terms from the first 22 terms leaves exactly the 12th through 22nd terms, inclusive. Therefore the required sum is S₂₂ − S₁₁ = 1474 − 407 = 1067. There are 11 terms in this range, but it is not necessary to find the first term or common difference because the two given partial sums already provide the required subtraction directly. Hence option C, 1067, is correct. The other options result from addition, an incorrect subtraction, or a small arithmetic error.

To find the nth term of an AP from its sum, use \(a_n=S_n-S_{n-1}\). Here, \(a_n=(4n^2+9n)-[4(n-1)^2+9(n-1)]=8n+5\). Therefore, \(a_{27}=8\times27+5=221\). Hence, 221 is the correct option. The value 223 may result from an error of 2 during simplification. Exam tip: When \(S_n\) is given, obtain a term by subtracting two consecutive partial sums.

This is an AP with first term 30 and common difference 6. The 25th row has \(30+24\times6=174\) seats, and the 50th row has \(30+49\times6=324\) seats. There are \(50-25+1=26\) rows from the 25th through the 50th row. Therefore, the sum is \(\frac{26}{2}(174+324)=6474\). Both the 25th and 50th rows must be included, so the number of terms is 26, not 25. Exam tip: for a sum over an inclusive range, use last position minus first position plus 1 for the number of terms.

Solving (\frac{n}{2}[32+9(n-1)]=2030) gives (n=20). The number of terms must be a positive integer.

The numbers are (13,22,\ldots,94), and the sum of (10) terms is (535). In remainder questions, the common difference equals the divisor.

The direct answer is option C, 7722. Here a=-30 and d=-22-(-30)=8. The 18th term is t_18=a+17d=-30+136=106, and the 50th term is t_50=a+49d=-30+392=362. The number of terms from 18 through 50 is 50-18+1=33. Therefore the sum is 33/2(106+362)=33/2(468)=33(234)=7722. Equivalently, S_50-S_17 gives the same result. Option A, 7656, has an arithmetic or endpoint error. Option B, 7788, is not the sum of these 33 terms. Option C, 7722, correctly uses both endpoints and their average 234. Option D, 7854, is too large. Do not use 32 as the number of terms: inclusive counting gives 33. Also distinguish the first term -30 from the common difference +8.

Answer: A, 13960. A partial sum S_n contains the first n terms. Thus the sum of terms from position r to position s is S_s − S_{r−1}. For positions 41 through 60, use S_60 − S_40, not S_60 − S_41. First, S_60 = 7(60²) − 2(60) = 7(3600) − 120 = 25,200 − 120 = 25,080. Next, S_40 = 7(40²) − 2(40) = 7(1600) − 80 = 11,200 − 80 = 11,120. Therefore, the required sum is 25,080 − 11,120 = 13,960, so option A is correct. Option B can result from subtracting incorrectly, while C and D do not match the correct partial-sum difference. The important idea is the index just before the first required term: 41 begins after the first 40 terms. Memory cue: selected block r–s means subtract S_{r−1} from S_s.

The given sums give (a=5) and (d=4), so (S_{45}=4185). First determine the AP from the two partial sums.

The direct answer is option B: 1104. S24 means the sum of the first 24 terms, while S12 means the sum of the first 12 terms. The first 24 terms consist of the first 12 terms plus terms 13 through 24. Therefore, to remove the first 12 terms, subtract the two partial sums: sum from the 13th term to the 24th term = S24−S12. Substitute the given values: 1560−456=1104. Hence option B is correct. Option A, 1080, is wrong because it comes from an incorrect subtraction or an incomplete interpretation of the requested range. Option B is correct because subtracting the first 12-term total from the first 24-term total leaves exactly terms 13 to 24. Option C, 1128, is wrong because it is not the difference of the supplied partial sums. Option D, 1152, is also wrong for the same reason and may result from reversing or altering the numbers. No value of a or d is needed; the question directly gives the required partial sums. The key idea is that a later block of consecutive terms can be isolated by subtracting earlier partial sums. Memory cue: “from r to s = Ss−S(r−1).” Here r=13, so subtract S12 from S24, not the other way around.

From the two terms, (d=5) and (a=9), so (S_{35}=3290). First find (a,d), then apply the sum formula.

The condition gives (5+11d=4(5+2d)), so (d=5) and (S_{20}=1050). Convert the term condition into an equation first.

(S_{33}=4983) and (S_{34}=5287), so the sum first exceeds (5000) at (34) terms. In such questions, also check the previous sum.

The first term is (3), and the (70)th term is (348), so the sum is (12285). Finding the first and last terms from (a_n) is an easy method.

The sum of the first \(n\) AP terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B gives the \(n\)th term \(a_n\), not the sum. In exams, carefully check the \(n-1\) factor.

From (S_{18}=9(a+99)), (a=14), then (d=5) and (S_{36}=3654). Treat the given (n)th term as the last term.

Here \(S_n=n(6n-1)\) represents the sum of the first \(n\) terms. A range beginning at the 31st term is found by taking the total through the 50th term and removing the total through the 30th term. This works because \(S_{30}\) contains precisely the terms that come before the required range.

Evaluate \(S_{50}=50(6\cdot50-1)=50(299)=14950\). Next, \(S_{30}=30(6\cdot30-1)=30(179)=5370\). Thus the sum from the 31st through the 50th term is \(S_{50}-S_{30}=14950-5370=9580\). Therefore option D is correct. The number of included terms is 20, from 31 to 50 inclusive, which is consistent with this subtraction method.

The selected terms are (a_5,a_{10},\ldots,a_{50}), and their sum is (1875). In position-based questions, form the new AP of selected terms.