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

The two terms give (d=3) and (a=8), so (S_{50}=4075). Finding the common difference from distant terms is the first step.

In an arithmetic progression, each term changes by the same common difference. Here the first term is 12 and the common difference is 5, so the term at position n is found from the rule \(a_n=12+(n-1)5\). The requested terms run from the 21st through the 40th, giving 20 terms in all. Their first term is \(a_{21}=112\), and their last term is \(a_{40}=207\). The sum of an AP can be found by multiplying the number of terms by the average of the first and last terms.

Thus, the required sum is \(20\times\frac{112+207}{2}=10\times319=3190\). Equivalently, one may calculate \(S_{40}-S_{20}\), because subtracting the first 20 terms leaves terms 21 to 40. Therefore option A is correct. A common error is to subtract \(S_{21}\), which would remove the 21st term too and give the wrong range.

From \(S_{15}=S_{21}\), the sum of the 16th to 21st terms is zero. Their sum is \(3[(a+15d)+(a+20d)]\), giving \(2a+35d=0\). Exam tip: subtract equal partial sums to isolate the intervening terms.

The governing concept is finding a range sum through partial sums, together with the AP sum formula. Since the first term is a = 3, S₂₀ = 20/2 [2(3) + 19d] = 10(6 + 19d), and S₁₀ = 10/2 [2(3) + 9d] = 5(6 + 9d). The sum from the 11th through the 20th term is S₂₀ - S₁₀, so 10(6 + 19d) - 5(6 + 9d) = 465. This simplifies to 60 + 190d - 30 - 45d = 465, hence 30 + 145d = 465, 145d = 435 and d = 3. Therefore option C is correct. The other choices result from using the wrong number of terms or subtracting the wrong partial sum.

The given sums give (a=7) and (d=4), so (S_{25}=1375). From two sums, first find the AP values.

Adding sums of multiples of (3) and (5), then subtracting multiples of (15), gives (58418). Avoiding double counting is important.

The numbers are (16,23,\ldots,93), and the sum of (12) terms is (654). A remainder-based AP has common difference equal to the divisor.

The conditions give (a=6) and (d=5), so (S_{12}=402). Convert the given term and sum into two equations.

Putting (a=2) in (S_{15}=6S_5) gives (d=\frac{2}{3}). In ratio questions, apply the sum formula on both sides.

Here, the first term is \(a=x\), the common difference is \(d=x+2\), and \(n=10\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(365=\frac{10}{2}[2x+9(x+2)]\). Thus, \(365=5(11x+18)=55x+90\), so \(x=5\). Substituting \(x=6\) gives a sum of \(420\), so it is incorrect despite being the closest distractor. Exam tip: In AP questions involving variables, identify \(a\), \(d\), and \(n\) before applying the formula.

The numbers are (200,216,\ldots,488), and their sum is (6536). A but-not condition often forms a new AP.

The governing concept is the sum of the first n terms of an arithmetic progression: Sₙ = n/2[2a + (n−1)d]. Using S₅ = 75 gives 5/2(2a + 4d) = 75, so a + 2d = 15. Using S₁₀ = 275 gives 10/2(2a + 9d) = 275, so 2a + 9d = 55. Since 2a + 4d = 30, subtraction gives 5d = 25, hence d = 5 and a = 5. Therefore S₂₀ = 20/2[2(5) + 19(5)] = 10(105) = 1050. Thus option C is correct. Options A, B, and D result from using an incorrect common difference or an incorrect sum formula.

The sum of multiples of (4) is (125500), and the sum of multiples of (20) is (25500), so the answer is (100000). Remove overlap using (\operatorname{lcm}).

When a question gives the sum of the first n terms, the sum of any consecutive block can be found by subtracting the sum before that block. The terms from the 21st through the 30th are obtained by removing the first 20 terms from the first 30 terms. Thus the required expression is \\(S_{30}-S_{20}\\), not \\(S_{30}-S_{21}\\), because the 21st term must be included.

Using \\(S_n=7n^2-4n\\), we get \\(S_{30}=7(30)^2-4(30)=6180\\) and \\(S_{20}=7(20)^2-4(20)=2720\\). Therefore, the required sum is \\(6180-2720=3460\\). Option A is correct. A common mistake is to subtract \\(S_{21}\\), which would leave out the 21st term.

From (740=10(a+60)), (a=14). When the (n)th term is given, use it as the last term for the first (n) terms.

Using (S_n=\frac{n}{2}[2a+(n-1)d]), the sum is (2616). In exams, calculate ((n-1)d) separately.

The required sum is (S_{32}-S_{14}=2889). To find a middle block sum, subtract the previous partial sum.

First, (169=-6+(n-1)7) gives (n=26), and the sum is (2119). When the last term is given, find (n) first.

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

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.