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

Correct answer: D. (21672)

The numbers are (210,224,\ldots,798), and their sum is (21672). Choose the first and last multiples within the limits correctly.

Correct answer: A. (101100)

This is an AP with (a=2200), (d=175), (n=24), and the total is (101100). In word problems, treat each amount as a term.

Correct answer: B. (28832)

The first number is (102), the last is (986), and there are (53) terms, so the sum is (28832). In divisibility questions, choose the first and last values carefully.

Correct answer: C. (900)

The direct answer is option C, 900. An arithmetic progression (AP) changes by the same amount each time. Here the common difference is d=113-120=-7, so the terms decrease by 7. For the first n terms, use S_n=n/2[2a+(n-1)d]. Thus S_25=25/2[2(120)+24(-7)]=25/2(240-168)=25/2(72)=900. Equivalently, the 25th term is 120+24(-7)=-48, so S_25=25/2(120-48)=900. Option A, 850, is too small and results from an incorrect final term or arithmetic. Option B, 875, is also not the calculated total. Option C, 900, matches both correct methods. Option D, 925, is too large. A negative common difference must be retained; changing it to +7 is a common mistake. Memory cue: decreasing AP means d is negative, but the sum formula remains unchanged.

Correct answer: D. (4845)

The notation gives the sum of the first n terms as \\(S_n=5n^2-2n\\). To add terms from the 26th through the 40th, use the total through the 40th term and remove the total through the 25th term. This works because the first 25 terms end immediately before the 26th term, so no required term is lost or counted twice.

Calculate \\(S_{40}=5(40)^2-2(40)=8000-80=7920\\). Next, \\(S_{25}=5(25)^2-2(25)=3125-50=3075\\). Therefore the required sum is \\(S_{40}-S_{25}=7920-3075=4845\\). Hence option D follows. Subtracting \\(S_{26}\\) would be wrong because the 26th term must be included.

Correct answer: A. (3474)

For an arithmetic progression with first term a and common difference d, the sum formula is \\(S_n=\frac{n}{2}[2a+(n-1)d]\\). Using \\(S_{12}=438\\) gives \\(2a+11d=73\\), and using \\(S_{24}=1596\\) gives \\(2a+23d=133\\). Subtracting these equations gives \\(12d=60\\), so \\(d=5\\). Then \\(2a+55=73\\), hence \\(a=9\\).

Now apply the same formula at n=36: \\(S_{36}=\frac{36}{2}[2(9)+35(5)]\\=18(18+175)=18(193)=3474\\). Therefore option A is correct. The important step is to determine the progression from the two given partial sums before calculating the requested sum.

Correct answer: C. (1462)

The direct answer is option C, 1462. The first term is a=14, the common difference is d=23-14=9, and n=17. Use S_n=n/2[2a+(n-1)d]. Therefore S_17=17/2[2(14)+16(9)]=17/2(28+144)=17/2(172)=17(86)=1462. The last term check gives l=14+16(9)=158, and S_17=17/2(14+158)=17/2(172)=1462. Option A, 1411, is not obtained from the AP formula. Option B, 1436, is also numerically incorrect. Option C, 1462, is correct because it uses all 17 terms and their average, 86. Option D, 1488, is too large. An odd number of terms causes no difficulty; the same sum formula works for odd and even n. Exam cue: find a, d, and n carefully before substituting.

Correct answer: A. 33

Use the sum formula \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(1512=\frac{18}{2}[2a+17\times6]=9(2a+102)\). Hence \(2a+102=168\), so \(2a=66\) and \(a=33\). If 30 were used as the first term, the sum would be 1458, so it is not correct. Exam tip: In the formula, use \((n-1)d\), not \(nd\).

Correct answer: B. (16200)

The numbers are (105,135,\ldots,975), and their sum is (16200). In this condition, the new AP has common difference (30).

Correct answer: C. (13780)

The first multiple is (91), the last is (598), and there are (40) terms, so the sum is (13780). Choose the first term according to the range carefully.

Correct answer: D. (540)

The sum from the 10th term through the 18th term can be obtained by removing the first nine terms from the sum of the first eighteen terms. In symbols, the required range is \(a_{10}+a_{11}+\cdots+a_{18}=S_{18}-S_9\). The given values are \(S_{18}=810\) and \(S_9=270\), so substitution gives \(810-270=540\). There are nine terms in this range, but no common difference or individual terms are needed because the two partial sums already contain exactly the required information.

Thus option D, 540, is correct. A frequent mistake is to subtract \(S_8\), which would include only terms 9 through 18, or to subtract \(S_{10}\), which would remove the 10th term as well. Since the requested sequence begins immediately after the first nine terms, \(S_{18}-S_9\) is the precise calculation.

Correct answer: A. 134

To find the nth term from the sum of an AP, use \(a_n=S_n-S_{n-1}\). Thus, \(a_{22}=S_{22}-S_{21}\). Here, \(S_{22}=3(22)^2+5(22)=1562\) and \(S_{21}=3(21)^2+5(21)=1428\). Therefore, \(a_{22}=1562-1428=134\). Option 131 is not correct because it does not result from the correct difference between \(S_{22}\) and \(S_{21}\). Exam tip: When \(S_n\) is given, obtain the nth term by subtracting \(S_{n-1}\) from \(S_n\).

Correct answer: B. 4250

This is an AP with first term \(a=25\) and common difference \(d=5\). The 18th row has \(a_{18}=25+17\times5=110\) seats, and the 42nd row has \(a_{42}=25+41\times5=230\) seats. There are \(42-18+1=25\) rows from the 18th through the 42nd row. Hence, the sum is \(\frac{25}{2}(110+230)=4250\). An option such as 4175 can result from incorrectly counting the rows without including both end rows. Exam tip: for an inclusive range, use \(\text{last index}-\text{first index}+1\) for the number of terms.

Correct answer: C. (21)

Solving (\frac{n}{2}[22+7(n-1)]=1701) gives (n=21). The number of terms must be a positive integer.

Correct answer: A. (660)

The numbers are (11,19,\ldots,99), and the sum of (12) terms is (660). In remainder questions, the common difference equals the divisor.

Correct answer: B. (4810)

The direct answer is option B, 4810. Here a=-4, d=2-(-4)=6. We need terms 20 through 45, inclusive, so there are 45-20+1=26 terms. The 20th term is t_20=a+19d=-4+19(6)=110, and the 45th term is t_45=a+44d=-4+44(6)=260. Their sum is 26/2(110+260)=13(370)=4810. Equivalently, calculate S_45-S_19; subtracting through the 19th term removes exactly the first 19 terms. Option A, 4720, has an arithmetic error. Option B, 4810, uses the correct 26 terms and their average 185. Option C, 4900, and option D, 4990, are both too large and do not follow the endpoint calculation. Do not count 45-20 as the number of terms; add 1 for both endpoints.

Correct answer: C. (4515)

The expression \\(S_n=4n^2+n\\) gives the sum of the first \\(n\\) terms of the arithmetic progression. The sum from the 31st term through the 45th term is obtained by subtracting the sum through the 30th term from the sum through the 45th term. Thus the required quantity is \\(S_{45}-S_{30}\\), because this removes terms 1 through 30 and leaves terms 31 through 45.

Calculate \\(S_{45}=4(45)^2+45=4(2025)+45=8145\\). Also, \\(S_{30}=4(30)^2+30=4(900)+30=3630\\). Hence the required sum is \\(8145-3630=4515\\). Therefore option C is correct. The subtraction uses \\(S_{30}\\), not \\(S_{31}\\), because the 31st term must remain included.

Correct answer: D. (2295)

The given sums give (a=4) and (d=5), so (S_{30}=2295). First determine the AP from the two partial sums.

Correct answer: B. (2235)

From the two terms, (d=5) and (a=2), so (S_{30}=2235). First find (a,d), then apply the sum formula.

Correct answer: C. (1050)

The condition gives (6+8d=2(6+3d)), so (d=3) and (S_{25}=1050). Convert the term condition into an equation first.

Correct answer: D. (26)

(S_{25}=2800) and (S_{26}=3029), so the sum first exceeds (3000) at (26) terms. In such questions, also check the previous sum.

Correct answer: A. (11280)

The first term is (11), and the (60)th term is (365), so the sum is (11280). Finding the first and last terms from (a_n) is an easy method.

Correct answer: A. The middle term is zero.

For an AP with an odd number of terms, let the middle term be \(m\). The sum of the first \(n\) terms is \(S_n=n m\). Since \(n\neq0\) and \(S_n=0\), \(m=0\). The common difference need not be zero. Exam tip: for odd terms, relate the sum directly to the middle term.

Correct answer: C. (2100)

From (S_{15}=\frac{15}{2}(a+68)), (a=12), then (d=4) and (S_{30}=2100). Treat the given (n)th term as the last term.

Correct answer: D. (3345)

Here the sum of the first n terms is given directly by \\(S_n=n(4n+3)\\). For a range beginning with the 21st term, subtract the sum of the first 20 terms from the sum of the first 35 terms. This leaves exactly terms 21 through 35, including both endpoints.

Calculate \\(S_{35}=35[4(35)+3]=35(140+3)=35(143)=5005\\). Also, \\(S_{20}=20[4(20)+3]=20(80+3)=20(83)=1660\\). Therefore the required sum is \\(S_{35}-S_{20}=5005-1660=3345\\). Hence option D is correct. The subtraction uses 20, not 21, because the 20th term is the last term that must be excluded.