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

This is an application of the sum of an arithmetic progression. The numbers of bricks in successive levels are 3, 6, 9, ..., so the first term is a = 3, the common difference is d = 3, and the number of levels is n = 12. The twelfth level contains 3 + 11 × 3 = 36 bricks. Therefore the total is S_12 = 12/2[2(3) + (12 - 1)(3)] = 6(6 + 33) = 6 × 39 = 234. Another interpretation is that the amounts are 3 times the first 12 natural numbers, so the total is 3 × [12 × 13/2] = 234. Hence option D is correct; the other options result from undercounting levels or miscalculating the final term.

The direct answer is option C, 891. The first term is \(a=7\), the common difference is \(d=12-7=5\), and there are 18 terms. The eighteenth term is \(a_{18}=7+(18-1)5=7+85=92\). Hence \(S_{18}=\frac{18}{2}(7+92)=9\times99=891\). Option A, 861, is too small and usually comes from finding the wrong last term. Option B, 876, is also not obtained from the correct first and last terms. Option C, 891, is correct. Option D, 906, is too large. Since an AP sum equals number of terms times the average of the first and last terms, the check is 18×((7+92)/2)=18×49.5=891. The last term is the key step.

The governing concept is the sum of the first n terms of an arithmetic progression: Sₙ = n/2[2a + (n − 1)d]. Substituting a = 3, d = 7, and n = 20 gives S₂₀ = 20/2[2(3) + 19(7)] = 10[6 + 133] = 10 × 139 = 1390. Hence option D is correct. A second method verifies the result. The twentieth term is a₂₀ = a + 19d = 3 + 19 × 7 = 136. The sum is then the number of terms multiplied by the average of the first and last terms: S₂₀ = 20(3 + 136)/2 = 10 × 139 = 1390. The other choices may arise from using 20d rather than 19d, finding the last term incorrectly, or making a small arithmetic error.

Direct answer: option A, 345. The AP is -5,-1,3,..., so the first term is a=-5 and the common difference is d=4. For 15 terms, the last term is a15=a+14d=-5+14×4=-5+56=51. Now use S_n=n/2(first term+last term): S15=15/2×(-5+51)=15/2×46=15×23=345. Option A, 345, is exactly this result. Option B, 355, usually comes from an arithmetic error in adding the end terms. Option C, 365, may come from mishandling the negative first term or the number of steps. Option D, 375, is also too large and does not satisfy the formula. Although the first terms are negative, later positive terms are larger; the average of the first and last terms is 23, and 15×23=345. Always use n-1 when finding the nth term.

The direct answer is option C, 611. For an arithmetic progression with first term a, last term l, and n terms, the sum is S_n = n/2(a+l). Here a = 11, l = 83, and n = 13. Therefore S_13 = 13/2(11+83) = 13/2 × 94 = 13 × 47 = 611. Option A, 591, results from an incorrect calculation. Option B, 601, is also not the value of the formula. Option C is correct because it uses both end terms and the exact number of terms. Option D, 621, is ten greater than the correct result and may come from careless addition or multiplication. The key memory rule is: when the first term, last term, and number of terms are known, use half the number of terms times the sum of the end terms.

The fourteenth term is (-5), so (S_{14}=\frac{14}{2}(60-5)=385). In a decreasing progression, the last term may become negative.

The first term is (3) and the twelfth term is (47), so (S_{12}=\frac{12}{2}(3+47)=300). Use (a_n) to find the first and last terms.

The sixteenth row has (78) seats, so the total is (S_{16}=\frac{16}{2}(18+78)=768). In word problems, treat rows as terms.

The seventeenth term is (149), so (S_{17}=\frac{17}{2}(5+149)=1309). Find the last term first and use the average method.

From (\frac{n(n+1)}{2}=325), (n(n+1)=650), so (n=25). In such questions, look for products of consecutive numbers.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(456=\frac{12}{2}(8+l)=6(8+l)\). Hence, \(8+l=76\), so \(l=68\). If 66 were used, the sum would be 444, not 456. Exam tip: When the first and last terms are involved, use \(S_n=\frac{n}{2}(a+l)\) directly.

Using (S_n=\frac{n}{2}[4+9(n-1)]), we get (n=12). When finding (n) from a given sum, check the options.

The multiples are from (6) to (96), and there are (16) terms, so the sum is (816). In boundary questions, decide the last term first.

The numbers are (56,63,\ldots,147), with (14) terms, so the sum is (1407). For numbers between limits, choose the first and last terms carefully.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(l\) is the last term. Therefore, \(279=\frac{9}{2}(7+l)\). This gives \(7+l=62\), so \(l=55\). If the last term were \(53\), the sum would be \(270\), not \(279\). Exam tip: When the first term and the sum are given, use \(S_n=\frac{n}{2}(a+l)\) to find the last term directly.

The eleventh term is (-8), so (S_{11}=\frac{11}{2}(32-8)=132). The formula remains the same even with a negative last term.

The fifteenth term is (96), so the total is (S_{15}=\frac{15}{2}(12+96)=810) fruits. Treat the number of days as the number of terms.

The sum of the (9)th to (15)th terms is (S_{15}-S_8=371). For a group of consecutive terms, take the difference of partial sums.

The sum of the first n odd natural numbers is \(n^2\). Therefore, the sum of the first 18 odd numbers is \(18^2=324\), and the sum of the first 10 odd numbers is \(10^2=100\). Hence, the sum of the remaining 8 numbers is \(324-100=224\). A value such as 234 may result from an error in subtraction. Exam tip: when an initial set of terms is removed, subtract its sum from the total sum.

The sum equals the number of terms multiplied by the average of the first and last terms. Hence \(S_n=\frac{n}{2}(a+l)\). Option C gives only the average, not the sum. Exam tip: check the factor \(n\).

Putting (n=12) in (S_n=\frac{n}{2}[2+7(n-1)]) gives (504). Checking options is a quick method.

The tenth deposit is (825) rupees, so the total is (S_{10}=\frac{10}{2}(150+825)=4875) rupees. Do not forget the unit in money problems.

The sequence is (20,30,\ldots,200) with (19) terms, so the sum is (2090). Include both boundary terms.

The twelfth term is (-21), so (S_{12}=\frac{12}{2}(45-21)=144). In decreasing order, keep (d) negative while finding the last term.

To find the sum of the first 8 terms, substitute \(n=8\) in the given formula: \(S_8=3(8)^2+2(8)=3\times64+16=192+16=208\). Therefore, the correct answer is 208. The value 198 may result from an error while evaluating the \(2n\) term. Exam tip: Calculate \(n^2\) first, then perform multiplication and addition carefully.