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

Correct answer: D. (901)

The sum is (S_n=n(103-3n)), and the maximum (901) occurs at (n=17). Exam tip: check integer values near the vertex.

Correct answer: B. (9450)

The total sum is (11325), and the sum of multiples of (6) is (1875), so the answer is (9450). Exam tip: subtract the complementary sum.

Correct answer: B. \(15\)

For an AP, the sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+t_n)\). Thus, \(777=\frac{14}{2}(a+96)=7(a+96)\). Hence \(a+96=111\), so \(a=15\). If \(17\) were used, the sum would be \(7(17+96)=791\), not the given sum. Exam tip: when \(S_n\) and \(t_n\) are given, use \(S_n=\frac{n}{2}(a+t_n)\) directly.

Correct answer: D. \(\frac{n}{2}(a+l)\)

In an AP, the average of the first and last terms is \(\frac{a+l}{2}\). Multiplying this by the number of terms \(n\) gives \(S_n=\frac{n}{2}(a+l)\). Option B is only the average, not the sum. In exams, use this form when \(l\) is given.

Correct answer: C. (342)

This is the AP (45,42,39,\ldots), and (S_{12}=6[90+11(-3)]=342). Exam tip: treat the decrease as a negative common difference.

Correct answer: D. (5)

Putting the ratio in the sum formula gives (d=5). Exam tip: cross-multiply the ratio to form an equation.

Correct answer: C. (23)

The equation (\frac{n}{2}[4+6(n-1)]=12[2+6(n-1)]) gives (n=23). Exam tip: write (S_n) and (t_n) separately.

Correct answer: B. (936)

The numbers are (12,17,\ldots,97), and there are (18) terms. Exam tip: choose the first term of the remainder AP correctly.

Correct answer: B. (5)

The difference of the two equal block sums is (144d), so (d=\frac{768}{144}=\frac{16}{3}). Exam tip: recheck block-sum formulas carefully.

Correct answer: D. (1600)

(t_6=50) and (t_{25}=126), so the sum is (\frac{20}{2}(50+126)=1760). Exam tip: count the selected terms correctly.

Correct answer: A. (6)

The direct answer is A, 6. The given expression is S_n=9n-3n^2, where n is a positive integer because it counts terms. Calculate successive values: S_1=9(1)-3(1)^2=6; S_2=18-12=6; S_3=27-27=0; S_4=36-48=-12. After this the negative quadratic term makes the values smaller. Another way is to complete the square: S_n=-3n^2+9n=-3(n-3/2)^2+27/4. The real-valued peak is near n=1.5, so among positive integers n=1 and n=2 give the largest integer value, 6. Option A is correct. Option B, 9, is not the value of S_1 or S_2; it is also below neither? Actually the formula gives no 9 for a positive integer n. Option C, 12, is not obtained; S_2 is 6 and S_3 is 0. Option D, 15, is likewise impossible and ignores the negative quadratic term. A common mistake is to use the continuous maximum 27/4 as the answer, but n must be an integer. Memory cue: for a negative quadratic, check nearby integer values around its vertex.

Correct answer: B. (1364)

For an AP with an odd number of terms, the sum is the product of the number of terms and the middle term. Exam tip: remember the middle-term property.

Correct answer: A. \(a_n=-a_1\)

Using \(S_n=\frac{n}{2}(a_1+a_n)\), and since \(n>0\) and \(S_n=0\), we get \(a_1+a_n=0\). Hence, \(a_n=-a_1\). Neither \(d=0\) nor an even \(n\) is necessary. Exam tip: for a zero sum, check whether the first and last terms are opposites.

Correct answer: B. (840)

The numbers are (12,24,\ldots,120), and there are (10) terms. Exam tip: divisible by both means use the least common multiple.

Correct answer: A. (14)

Substituting (n=14) in (S_n=\frac{n}{2}[32+7(n-1)]) gives (1085). Exam tip: options can be checked quickly too.

Correct answer: D. यह (n) पर निर्भर करेगा

The equation gives (d=\frac{18}{n+1}), so no fixed value is determined. Exam tip: check whether a variable remains in the relation.

Correct answer: C. (35)

In (11) terms, the (6)th term is the middle term, so (t_6=\frac{385}{11}=35). Exam tip: use the middle term for an odd number of terms.

Correct answer: B. (4)

Solving the two sum equations gives (a=-2) and (d=4). Exam tip: find (a) and (d) together from two sums.

Correct answer: C. 2688

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here, \(a=15\), \(d=6\), and \(n=28\). Thus, \(S_{28}=\frac{28}{2}[2(15)+27(6)]=14(30+162)=14\times192=2688\). Therefore, option C is correct. Using \(28\) instead of \(27\) would be incorrect because the common difference is added only \(n-1\) times. Exam tip: write \(n-1\) separately before substituting values in the formula.

Correct answer: B. 18

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Substituting the values gives \(1125=\frac{n}{2}[40+5(n-1)]\), so \(n(n+7)=450\). Hence, \(n=18\). Although \(n=-25\) is another algebraic root, the number of terms cannot be negative. Exam tip: for AP sum questions, select only a positive integral value of n.

Correct answer: D. (624)

(S_{24}=\frac{24}{2}[144+23(-4)]=624). Exam tip: handle the negative common difference carefully.

Correct answer: A. \(931\)

For an AP, \(a_6=a+5d=29\) and \(a_{19}=a+18d=94\). Subtracting gives \(13d=65\), so \(d=5\). Hence, \(a=29-5\times5=4\). Now, \(S_{19}=\frac{19}{2}[2a+18d]=\frac{19}{2}[8+90]=931\). Therefore, \(931\) is correct. A value such as \(950\) can result from using an incorrect first term or number of terms. Exam tip: find \(a\) and \(d\) from the given terms before applying the sum formula.

Correct answer: C. 205

Answer: C, 205. To obtain an individual term from a partial-sum formula, use a_n = S_n − S_{n−1}. The first n terms include the first n−1 terms plus the nth term, so subtraction leaves exactly a_n. Compute S_18 = 6(18²) − 5(18) = 6×324 − 90 = 1944 − 90 = 1854. Compute S_17 = 6(17²) − 5(17) = 6×289 − 85 = 1734 − 85 = 1649. Hence a_18 = 1854 − 1649 = 205, so option C is correct. Option A, B, and D do not satisfy the consecutive-partial-sum calculation. The minus sign in −5n must remain attached to the linear term; changing it to plus would change every value. Also, S_18 itself is a total, not the 18th term. Memory cue: subtract the previous cumulative total before interpreting the result as one term.

Correct answer: B. 5

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here, \(824=\frac{16}{2}[2(14)+15d]=8(28+15d)\). Thus, \(28+15d=103\), so \(15d=75\) and \(d=5\). If 4 were used, the sum would be 704, not 824. Exam tip: substitute the given \(a\), \(n\), and \(S_n\) directly into the sum formula to find \(d\).

Correct answer: D. (24)

From (2208=\frac{n}{2}(8+176)), (n=24). Exam tip: use the (a+l) form when first and last terms are known.