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

This is the sum of the first (9) multiples of (5), so (5\times45=225). In word problems, treat levels as terms.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term. Thus, \(S_{16}=\frac{16}{2}(11+71)=8\times82=656\). Therefore, 656 is correct. The value 646 would result from an arithmetic error in addition or multiplication. Exam tip: When the first and last terms are given, use \(\frac{n}{2}(a+l)\) directly.

This is the sum of the first (11) multiples of (8), so (8\times66=528). For multiples, use the sum of natural numbers.

(\frac{28\times29}{2}=406), so the sum is (406). The natural-number sum formula gives a quick answer.

The sixth term is (60), so (S_6=\frac{6}{2}(90+60)=450). In decreasing order, calculate the last term carefully.

The sum of the first n AP terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). It also equals \(\frac{n}{2}(a+l)\), where \(l=a+(n-1)d\). Option B gives only the nth term. Exam tip: distinguish the sum formula from the nth-term formula.

The tenth term is (51), so the total is (S_{10}=\frac{10}{2}(6+51)=285) plants. The same sum formula works in real situations.

The governing concept is the sum formula for an arithmetic progression. The first term is a = 1 and the common difference is d = 6 − 1 = 5. To reach the eighteenth term from the first, there are 18 − 1 = 17 equal intervals, so a_18 = a + 17d = 1 + 17 × 5 = 86. The sum is therefore S_18 = n(a + l)/2 = 18(1 + 86)/2 = 9 × 87 = 783. Thus option B is correct. Using the other form gives S_18 = 18/2[2(1) + 17(5)] = 9(2 + 85) = 783. The nearby alternatives generally result from using 18 instead of 17 intervals, finding the wrong last term, or making an arithmetic error.

Here \(a=14\), \(l=84\), and \(n=15\). Therefore, \(S_{15}=\frac{15}{2}(14+84)=\frac{15}{2}\times98=15\times49=735\). Hence, 735 is correct. Getting 745 would result from an error in addition or multiplication. Exam tip: add \(a+l\) first and simplify by 2 when the sum is even.

Answer: 462, so option B is correct. The first 21 even natural numbers are 2, 4, 6, ..., 42. They form an arithmetic progression with first term a=2, last term l=42, and n=21 terms. The last term is 2×21=42 because the kth even number is 2k. Apply the AP sum formula S_n=n(a+l)/2: S_21=21(2+42)/2=21×44/2=21×22=462. There is also a useful shortcut: the sum of the first n even natural numbers is n(n+1). For n=21, this gives 21×22=462. Option A, 452, is 10 less; option C, 472, is 10 more; and option D, 482, is 20 more. These do not satisfy the formula. Be careful not to start with 0, because the first positive even natural number in this sequence is 2 and the question asks for the first 21 even numbers.

Answer: 402, so option B is correct. The first term is a=17 and the common difference is d=20-17=3. To use the sum formula, find the twelfth term first. The nth-term formula is a_n=a+(n-1)d, so a_12=17+(12-1)3=17+33=50. Now apply S_n=n(a+l)/2: S_12=12(17+50)/2=12×67/2=6×67=402. Another equivalent calculation is S_12=12/2[2(17)+(12-1)3]=6[34+33]=6×67=402. The factor 11 appears because there are eleven equal jumps from the first term to the twelfth term, not twelve. Option A, 392, C, 412, and D, 422, do not result from the correct formula. A useful check is that the average of the first and last terms is 33.5, and 12×33.5=402.

To find the average of the first 9 terms, divide their total sum by the number of terms: \(\frac{198}{9}=22\). Therefore, the correct answer is 22. Getting 20 would be incorrect because 198 divided by 9 is not 20. Exam tip: use \(\text{average}=\frac{\text{sum}}{\text{number of terms}}\) in average-based questions.

The seventh term is (53), so (S_7=\frac{7}{2}(5+53)=203). With an odd number of terms, you can also check using the middle term.

The partial sum \\(S_4=44\\) contains the first four terms, while \\(S_9=189\\) contains the first nine terms. Subtracting the first sum from the second removes the common first four terms and leaves exactly the fifth, sixth, seventh, eighth, and ninth terms. Therefore, the required sum is \\(S_9-S_4=189-44=145\\).

This subtraction works for any sequence when partial sums are defined in the usual way, and it does not require finding the first term or common difference. Hence option C is correct. A value such as 140 could result from an arithmetic subtraction error, while adding the sums would count the first four terms twice and would not represent the requested block of terms.

This is the sum of the first (6) multiples of (16), so (16\times21=336). For multiples, use (1+2+\cdots+n).

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(310=\frac{10}{2}(a+49)=5(a+49)\). Hence \(a+49=62\), so \(a=13\). If \(a=12\), the sum would be \(305\), not \(310\). Exam tip: when the last term is given, apply \(S_n=\frac{n}{2}(a+l)\) directly.

The tenth term is (30), so (S_{10}=\frac{10}{2}(75+30)=525). The formula for (S_n) does not change in decreasing order.

This is the sum of the first (9) multiples of (7), so (7\times45=315). Treat the number of shelves as the number of terms.

The governing concept is the sum of the first n odd natural numbers, which is n², or equivalently the sum formula for an arithmetic progression. The first 16 odd numbers are 1, 3, 5, ..., 31, so a = 1, d = 2, and n = 16. Using the identity, the sum is 16² = 256. The AP formula confirms this: S₁₆ = n/2[2a + (n − 1)d] = 16/2[2(1) + 15(2)] = 8(2 + 30) = 8 × 32 = 256. Therefore option B is correct. Option A, C, or D could result from miscounting the terms, using an incorrect last term, or making an arithmetic error. The result is also consistent with the familiar pattern that the sum of the first n odd numbers forms the square n².

This is the sum of the first (8) multiples of (9), so (9\times36=324). If (a=d), the multiples method is faster.

The fourteenth term is (61), so (S_{14}=\frac{14}{2}(22+61)=581). Correct calculation of the last term gives the correct sum.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term. Thus, \(S_7=\frac{7}{2}(31+13)=\frac{7}{2}\times44=154\). Therefore, 154 is correct. Since the first term is greater than the last term, the AP may be decreasing, but this does not change the sum formula. Exam tip: When the first term, last term, and number of terms are given, use \(S_n=\frac{n}{2}(a+l)\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=35\), we get \(\frac{35\times36}{2}=35\times18=630\). Therefore, 630 is the correct answer. An option such as 610 may result from an error in multiplication or division. Exam tip: use \(\frac{n(n+1)}{2}\) directly for \(1+2+\cdots+n\).

The twelfth term is (98), so (S_{12}=\frac{12}{2}(10+98)=648). The average of the first and last terms is useful in sums.

The governing concept is the relationship between partial sums. S₁₀ represents the sum of terms 1 through 10, while S₅ represents the sum of terms 1 through 5. Subtracting the latter from the former removes the first five terms and leaves exactly terms 6 through 10: a₆ + a₇ + a₈ + a₉ + a₁₀ = S₁₀ − S₅ = 270 − 85 = 185. Therefore option C is correct. No first term, common difference, or individual term needs to be found. The other choices result from an incorrect subtraction, reversing the order, or confusing the requested five-term segment with another partial sum.