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

The governing concept is subtraction of partial sums. S₁₅ represents the sum of the first 15 terms, while S₁₀ represents the sum of the first 10 terms. When the latter is subtracted from the former, the first 10 terms cancel and only terms 11 through 15 remain. Therefore, required sum = S₁₅ - S₁₀ = 465 - 240 = 225. Option D is correct. This method does not require finding the first term or the common difference. A, B and C are distractors produced by incorrect subtraction or by counting the block incorrectly. Because the question includes both boundary terms, subtracting S₁₀ is exactly appropriate; subtracting S₁₁, for example, would omit the 11th term.

To find the nth term of an AP from its sum, use \(a_n=S_n-S_{n-1}\). Here, \(S_{18}=2(18)^2+7(18)=774\) and \(S_{17}=2(17)^2+7(17)=697\). Therefore, \(a_{18}=S_{18}-S_{17}=774-697=77\). The values 75, 79, and 81 do not equal this required difference. Exam tip: When \(S_n\) is given, obtain the nth term by calculating \(S_n-S_{n-1}\).

This is an AP with first term \(a=20\) and common difference \(d=4\). The 15th row has \(20+14\times4=76\) seats, while the 35th row has \(20+34\times4=156\) seats. There are \(35-15+1=21\) rows from the 15th to the 35th row, inclusive. Hence, the required sum is \(\frac{21}{2}(76+156)=2436\). Option 2408 may result from counting the rows or finding the last term incorrectly. Exam tip: When summing from one term number to another, include both endpoints by adding \(+1\) to the difference of their positions.

Solving (\frac{n}{2}[14+5(n-1)]=1090) gives (n=20). The number of terms must always be a positive integer.

The governing concept is the AP sum formula written in terms of the first and last terms: Sₙ = n/2 (a + l). The question gives a + l = 150 and Sₙ = 1800. Substitution gives 1800 = n/2 × 150 = 75n. Dividing both sides by 75 gives n = 1800/75 = 24. Thus the AP contains 24 terms, so option D is correct. The common difference is not needed because the sum of the first and last terms is already supplied. The other options would produce totals of 1500, 1650 and 1950 respectively when multiplied by 75, so they do not satisfy the given total of 1800.

The numbers are (11,16,\ldots,96), and the sum of (18) terms is (963). In remainder questions, identifying the first value correctly is important.

The required sum begins with the fourth term and ends with the twenty-fifth term. In an arithmetic progression, the difference between consecutive terms is constant. Here the first term is 6 and the common difference is 4. The first three terms must not be included because the question starts at the fourth term, not at the first term.

The fourth term is 18 and the twenty-fifth term is 102. The number of terms from the fourth through the twenty-fifth is exactly 25 − 4 + 1 = 22. Therefore, their sum is the number of terms multiplied by the average of the first and last terms: \\(S=\frac{22}{2}(18+102)=11\times120=1320\\). Hence option B is correct. Option A would result from an incorrect count or average.

If \(S_n\) is the sum of the first \(n\) terms, the sum from the 21st through the 30th term is \(S_{30}-S_{20}\). The subtraction removes the first 20 terms and leaves exactly terms 21 to 30. Using the given rule, \(S_{30}=3(30)^2+4(30)=2700+120=2820\), while \(S_{20}=3(20)^2+4(20)=1200+80=1280\).

Therefore, the required sum is \(2820-1280=1540\). This matches option C. It is important not to use \(S_{30}-S_{21}\), because that would remove the first 21 terms and begin with the 22nd term. The endpoints are included: the calculation must contain both the 21st and the 30th terms, and subtracting \(S_{20}\) does exactly that.

The given sums give (a=5), (d=5), and (S_{24}-S_{16}=820). First determine the AP from the two partial sums.

The governing concept is the difference between two partial sums. S₁₂ is the total of terms 1 through 12, and S₆ is the total of terms 1 through 6. Subtracting S₆ from S₁₂ removes the first six terms and leaves exactly terms 7 through 12. Hence the required sum is S₁₂ - S₆ = 420 - 150 = 270. Therefore option A is correct. No common difference or individual term needs to be calculated. Options B, C and D are close numerical distractors, but each fails to equal the exact difference of the supplied partial sums. The inclusion of the 7th term is why S₆, rather than S₇, must be subtracted.

From the two terms, (d=4) and (a=6), so (S_{20}=880). First find (a,d), then apply the sum formula.

The governing concept is the general term of an AP, aₙ = a + (n - 1)d, followed by the AP sum formula. Since a = 4, the 8th term is 4 + 7d and the 3rd term is 4 + 2d. The condition gives 4 + 7d = 3(4 + 2d), so 4 + 7d = 12 + 6d and d = 8. Now use S₁₂ = 12/2 [2(4) + 11(8)] = 6(8 + 88) = 6 × 96 = 576. Therefore option C is correct. The other options arise from using an incorrect position difference, treating the third term as 3d, or substituting the wrong common difference into the sum formula.

(S_{17}=986) and (S_{18}=1107), so the sum first exceeds (1000) at (18) terms. In such questions, also check the previous sum.

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.

The sum is (30\times76=2280). In any sequence, total sum is directly found from average and number of terms.

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.