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

The first term is (3), and the (50)th term is (199), so the sum is (5050). Finding the first and last terms from (a_n) is an easy method.

Direct answer: B, 2280. Average means total sum divided by the number of terms: \(\text{average}=\frac{S_n}{n}\). Rearranging gives \(S_n=n\times\text{average}\). The question gives 30 terms and average 76, so \(S_{30}=30\times76=2280\). This rule works for an AP and, more generally, for any finite list of numbers. Option B is therefore correct. Option A, 2160, would correspond to an average of 72 for 30 terms, not 76. Option C, 2340, would require an average of 78. Option D, 2400, would require an average of 80. No common difference is needed because the average is already supplied. A useful AP check is that the average of the first and last terms also equals 76, but the direct multiplication is enough here. Memory cue: total equals number of entries multiplied by average; do not divide again.

From (S_{10}=5(a+a_{10})), (a=3), then (d=4) and (S_{20}=820). When a last-like term is given, the average method helps.

The notation gives the sum of the first n terms as \\(S_n=n(5n-2)\\). To add the terms from the 16th through the 25th, subtract the sum through the 15th term from the sum through the 25th term. First, \\(S_{25}=25(5\times25-2)=25(123)=3075\\). Next, \\(S_{15}=15(5\times15-2)=15(73)=1095\\). Thus the required sum is \\(S_{25}-S_{15}=3075-1095=1980\\).

Hence option D is correct. Subtracting \\(S_{15}\\), rather than \\(S_{16}\\), is necessary because the 16th term is the first term wanted. The difference removes terms 1 through 15 and leaves exactly terms 16 through 25. The calculated value agrees with the supplied answer, so no correction is needed.

The selected terms are (a_3,a_6,\ldots,a_{30}), and their sum is (485). In position-based questions, form the new AP of selected terms.

The sum of all two-digit numbers is (4905), and the sum of multiples of (4) is (1188), so the answer is (3717). For not divisible, the complement method is fast.

The numbers are (102,114,\ldots,990), and their sum is (40950). In such a condition, a new AP with common difference (12) is formed.

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 that a range sum can be found by subtracting partial sums. The sum from the 11th through the 20th term is S₂₀ − S₁₀. Using Sₙ = n/2[2a + (n − 1)d] with a = 3, we get S₂₀ = 10(6 + 19d) and S₁₀ = 5(6 + 9d). Hence 10(6 + 19d) − 5(6 + 9d) = 465. Expanding gives 60 + 190d − 30 − 45d = 465, so 30 + 145d = 465. Therefore 145d = 435 and d = 3. Option C is correct. A common mistake is to subtract S₁₁ rather than S₁₀, because the terms before the 11th term are precisely the first ten terms. Other options also arise from counting the interval incorrectly or using the wrong coefficient in the sum formula.

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 AP sum formula Sₙ = n/2[2a + (n − 1)d]. From S₅ = 75, we obtain 5/2(2a + 4d) = 75, so a + 2d = 15. From S₁₀ = 275, 10/2(2a + 9d) = 275, giving 2a + 9d = 55. Doubling the first simplified equation gives 2a + 4d = 30. Subtracting, 5d = 25, so d = 5; then a + 10 = 15, hence a = 5. Therefore S₂₀ = 20/2[2(5) + 19(5)] = 10(10 + 95) = 1050. Option C is correct. The other options result from an incorrect common difference, confusing S₁₀ with the sum of terms 6 to 10, or using an incorrect value of n − 1.

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.

Direct answer: B, 14. In an AP, the sum of the first n terms is \(S_n=\frac{n}{2}(a+l)\), where a is the first term and l is the nth, or last, term of those n terms. Here \(S_{20}=740\), \(n=20\), and \(l=60\). Substitute: \(740=\frac{20}{2}(a+60)=10(a+60)\). Divide by 10: \(74=a+60\). Subtract 60: \(a=14\). Thus B is correct. Option A, 10, would give sum \(10(10+60)=700\), not 740. Option C, 18, would give \(10(18+60)=780\). Option D, 22, would give \(10(22+60)=820\). The given 20th term is used as the last term because the sum ends at the 20th term. Memory cue: when sum and last term are known, use \(S_n=\frac n2(a+l)\).

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.