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

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.

Here (d=-7), and the formula gives (S_{25}=900). In a decreasing AP, write the common difference as negative.

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.

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.

The governing concept is the sum of a finite arithmetic progression. Here the first term is a = 95, the common difference is d = 89 − 95 = −6, and n = 20. First find the twentieth term: a20 = a + (n − 1)d = 95 + 19(−6) = 95 − 114 = −19. Now use S_n = n/2(a + l): S20 = 20/2(95 + (−19)) = 10 × 76 = 760. Hence option B is correct. The negative common difference only makes the sequence decrease; it does not make the sum negative. The other values result from an arithmetic or last-term error.

Here (a=14), (d=9), (n=17), and the sum is (1462). The formula applies directly even when (n) is odd.

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 = −20, d = 9 and n = 30. Then S30 = 30/2[2(−20) + 29(9)] = 15[−40 + 261] = 15 × 221 = 3315. Equivalently, the last term is l = −20 + 29 × 9 = 241, so S30 = 30/2(−20 + 241) = 15 × 221 = 3315. Therefore option D is correct. The negative first term must be retained; dropping its sign produces a nearby but incorrect result.

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 \(nd\).

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

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.

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.

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 \(S_n\).

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.

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

The governing formula for the sum of n terms of an arithmetic progression is S_n = n/2 × (first term + last term), or S_n = n/2(a + l). Here, a + l = 260 and S_n = 4160. Substituting these values gives 4160 = n/2 × 260 = 130n. Therefore, n = 4160/130 = 32. Hence option D is correct. The common difference and the individual first or last term are unnecessary because their sum is already provided. Options A, B, and C do not produce the given total when multiplied by 130.

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

The required sum is (S_{45}-S_{19}=4810). When starting from the (20)th term, subtract the sum up to (19) terms.

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.

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

The governing idea is that a partial sum S_k contains the first k terms. Therefore, subtracting S₈ from S₁₆ removes the first eight terms and leaves exactly the terms from the 9th through the 16th. Thus, required sum = S₁₆ − S₈ = 880 − 280 = 600. Hence option A is correct. This method avoids finding the first term or common difference separately. A common mistake is to use S₁₆ + S₈, which would count the first eight terms again, or to subtract in the reverse order and obtain a negative value. Options B, C, and D do not equal the correct difference.