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

The two terms give (d=4) and (a=13), so (S_{60}=7860). Finding the common difference from distant terms is the first step.

The sequence \(25,33,41,\ldots\) is an arithmetic progression because the difference between consecutive terms is always \(8\). Its first term is \(25\), so the \(n\)th term is \(a_n=25+(n-1)8\). The required range contains terms 40 through 70, giving \(70-40+1=31\) terms.

The 40th term is \(25+39(8)=337\), and the 70th term is \(25+69(8)=577\). The sum of a finite AP is \(\frac{\text{number of terms}}{2}(\text{first term} + \text{last term})\). Hence the required sum is \(\frac{31}{2}(337+577)=\frac{31}{2}(914)=31(457)=14167\). Therefore option B is correct; using 30 instead of 31 terms would incorrectly exclude one endpoint.

The sum of the first n terms of an AP is Sₙ = n/2[2a + (n − 1)d]. For n = 41, d = −11, and S₄₁ = 0, substitute the values: 0 = 41/2[2a + 40(−11)] = 41/2(2a − 440). Since 41/2 is non-zero, the bracket must equal zero. Thus 2a − 440 = 0, so 2a = 440 and a = 220. Therefore option C is correct. The zero total does not mean that a is zero; it means the positive and negative terms cancel overall. The other choices come from mishandling the 40d term or dividing incorrectly.

The sum from the 21st to the 40th term contains 20 terms. The first term of this block is a + 20d = 7 + 20d, and the last is a + 39d = 7 + 39d. Using the AP sum formula for these 20 terms, 3680 = 20/2[(7 + 20d) + (7 + 39d)] = 10(14 + 59d). Therefore 368 = 14 + 59d, so 59d = 354 and d = 6. Equivalently, S₄₀ − S₂₀ = 3680 gives the same equation. Hence option A is correct. The common mistake is to use 20d as the last-term multiplier instead of 39d, forgetting that the original sequence starts at the first term.

The given sums give (a=8) and (d=4), so (S_{40}=3440). From two sums, first find the AP values.

Adding sums of multiples of (5) and (8), then subtracting multiples of (40), gives (150500). Avoiding double counting is important.

The numbers are (17,28,\ldots,94), and the sum of (8) terms is (444). A remainder-based AP has common difference equal to the divisor.

The conditions give (a=8) and (d=7), so (S_{16}=968). Convert the given term and sum into two equations.

Use Sₙ = n/2[2a + (n − 1)d] with n = 12, a = x, and d = 3x − 2. Then 1128 = 12/2[2x + 11(3x − 2)] = 6[2x + 33x − 22] = 6(35x − 22). Dividing by 6 gives 188 = 35x − 22, so 35x = 210 and x = 6. Therefore option B is correct. After finding x, the common difference would be 3(6) − 2 = 16, which is consistent with the equation but is not needed separately. The other options result from forgetting the factor 11, distributing incorrectly, or treating d as independent of x.

Subtracting the sum of multiples of (72) from the sum of multiples of (18) gives (28278). Use the least common multiple to remove overlap.

The given sums give (a=3) and (d=5), so (S_{36}=3258). First determine the AP from the smaller sums.

Subtracting the sum of multiples of (36) from the sum of multiples of (9) gives (93753). Numbers divisible by both are multiples of (\operatorname{lcm}(9,12)).

The governing concept is that the sum of consecutive terms from the rth term through the sth term is S_s − S_{r−1}. Here the required range is from the 51st through the 70th term, so calculate S_70 − S_50. We have S_70 = 8(70)² − 3(70) = 8(4900) − 210 = 39,200 − 210 = 38,990. Similarly, S_50 = 8(50)² − 3(50) = 20,000 − 150 = 19,850. Therefore, the required sum is 38,990 − 19,850 = 19,140. Hence option D is correct. Option A, B, and C result from incorrect substitution or subtraction; using S_51 would also include the wrong range boundary.

When the last term \(l\) is known, the sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(3000=\frac{30}{2}(a+150)=15(a+150)\). Hence \(a+150=200\), so \(a=50\). If \(55\) were used, the sum would be \(15(55+150)=3075\), not the given sum. Exam tip: use \(S_n=\frac{n}{2}(a+l)\) directly when the last term is provided.

(a_{21}+a_{40}=a_1+a_{60}=300), so the sum of (20) terms is (3000). Sums of symmetric terms are equal in an AP.

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

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

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

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.

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.