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

Direct answer: 14 terms, so option C is correct. The sequence is an arithmetic progression because consecutive terms differ by 4: 9 − 5 = 4 and 13 − 9 = 4. Thus a = 5 and d = 4. The sum of the first n terms is Sₙ = n/2[2a + (n − 1)d]. Substitution gives 434 = n/2[10 + 4(n − 1)] = n/2(4n + 6) = n(2n + 3). Hence 2n² + 3n − 434 = 0. Factorising gives (n − 14)(2n + 31) = 0, so n = 14 or n = −31/2. A number of terms must be a positive integer, so n = 14. Option A, 12, option B, 13, and option D, 15, do not produce a sum of 434 when substituted into the formula. Verification: the 14th term is 5 + 13×4 = 57, and S₁₄ = 14/2(5 + 57) = 434. Memory cue: identify a and d first, then use the sum formula, and reject nonpositive roots.

Direct answer: A, 3038. For an arithmetic progression, the sum of the first n terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here \(a=14\), \(d=7\), and \(n=28\). First calculate \(n-1=27\). Then \((n-1)d=27\times7=189\). Next, \(2a=2\times14=28\), so the bracket is \(28+189=217\). Finally, \(\frac{28}{2}\times217=14\times217=3038\). Therefore option A is correct. Option B, 3024, does not result from the correct substitution and may come from a small arithmetic or term-counting error. Option C, 3052, is also not the value of the formula. Option D, 3010, is likewise incorrect. Another check uses the last term: \(l=a+(n-1)d=14+189=203\), so \(S_{28}=\frac{28}{2}(14+203)=14\times217=3038\). Memory cue: never use \(nd\); the formula needs \((n-1)d\).

The required sum is (S_{30}-S_{12}=1566). To find a middle block sum, subtract the previous partial sum from the larger sum.

First, (192=12+(n-1)6) gives (n=31), and the sum is (3162). When the last term is given, find the number of terms first.

The numbers are (110,121,\ldots,495), and their sum is (10890). When between is written, check carefully whether endpoints are included.

This is an AP with (a=1500), (d=250), (n=18), and the total is (65250). In word problems, treat each amount as a term.

The first number is (104), the last is (988), and there are (69) terms, so the sum is (37674). In divisibility questions, choose the first and last values carefully.

The governing concept is the sum of a consecutive block of terms in an AP. Here a = 50 and d = 47 − 50 = −3. The sixth term is a_6 = 50 + 5(−3) = 35, while the twentieth term is a_20 = 50 + 19(−3) = −7. The number of included terms is 20 − 6 + 1 = 15, because both endpoints are included. Hence the block sum is 15/2(35 + (−7)) = 15/2 × 28 = 15 × 14 = 210. Thus option C is correct. Forgetting the inclusive +1 would count only 14 terms and produce an incorrect result; using 20 terms would also include terms outside the requested interval.

The governing concept is the difference of partial sums. For any sequence, the sum from the rth term through the sth term is S_s − S_(r−1), because the earlier terms cancel. Thus the required sum is S_20 − S_11. From S_n = 4n² − 3n, S_20 = 4(20²) − 3(20) = 1600 − 60 = 1540, and S_11 = 4(11²) − 3(11) = 484 − 33 = 451. Therefore, the required sum is 1540 − 451 = 1089, so option D is correct. Subtracting S_12 would omit the twelfth term, while subtracting S_10 would include the eleventh term, explaining why careful indexing is essential.

The two sums give (a=7) and (d=4), so (S_{30}=1950). When two partial sums are given, find (a,d) first.

Here (a=80), (d=-5), (n=18), and the sum is (675). Do not forget the negative sign of (d) in a decreasing AP.

Putting (a=11), (d=6), (n=19) in the formula gives (S_{19}=1235). Do not worry about (\frac{n}{2}) when (n) is odd.

In \(S_n=\frac{n}{2}[2a+(n-1)d]\), the coefficient of \(n^2\) is \(\frac d2\). Since \(d\ne0\), \(S_n\) is quadratic, not linear. Exam tip: check the \(n^2\) term first.

The governing concept is the sum of the first n terms of an arithmetic progression. The formula is Sₙ = n/2[2a + (n − 1)d]. Substituting S₁₆ = 1176, n = 16, and d = 7 gives 1176 = 16/2[2a + 15 × 7] = 8(2a + 105). Dividing by 8 gives 147 = 2a + 105. Therefore 2a = 42 and a = 21. Hence option A is correct. Options B, C, and D can arise from mishandling the factor n/2, using n instead of n − 1, or making an arithmetic error while isolating a. Since d is positive, the terms increase, but that fact alone does not determine a; the sum formula is essential.

The numbers are (117,135,\ldots,999), and the sum of (50) terms is (27900). For odd multiples, the common difference is (18).

The AP is (56,63,\ldots,245) with (28) terms, and the sum is (4214). Choose the first and last multiples within the limits correctly.

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 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.

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

The direct answer is option B, 880. In an arithmetic progression, the difference between consecutive terms is constant. Let the first term be a and the common difference be d. The fifth term is a+4d=22, and the fifteenth term is a+14d=62. Subtracting gives 10d=40, so d=4. Then a+16=22, hence a=6. The sum of the first 20 terms is S20=20/2[2a+19d]=10[12+76]=880. Option A, 840, results from an incorrect sum or term count. Option B is correct because it uses both given terms and the AP sum formula. Option C, 920, is too large and does not follow from the calculated values. Option D, 960, is also an incorrect calculation. Memory cue: first find d, then a, and finally use Sn=n/2[2a+(n−1)d].

The governing concepts are the general term and the sum formula for an arithmetic progression. For an AP, aₙ = a + (n − 1)d. With 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 apply S₁₂ = 12/2[2a + 11d]. Thus S₁₂ = 6[2(4) + 11(8)] = 6(8 + 88) = 6 × 96 = 576. Therefore option C is correct. The other values may result from treating the third term as 3d, using an incorrect positional coefficient, or substituting an incorrect 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.