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

The thirteenth term is (86), so (S_{13}=\frac{13}{2}(2+86)=572). Use ((n-1)d) when finding the last term.

The governing concept is the meaning of partial sums in an arithmetic progression. S₁₁ represents the sum of terms 1 through 11, while S₅ represents the sum of terms 1 through 5. Subtracting the latter from the former cancels the first five terms and leaves exactly the sixth through eleventh terms. Therefore, the required sum is S₁₁ − S₅ = 242 − 65 = 177. Hence option B is correct. There is no need to determine the first term or common difference. Options A, C, and D result from incorrect subtraction or from including or excluding the wrong endpoint terms.

The sum of the first (n) even numbers is (n(n+1)), so (17\times18=306). Do not confuse (n) with the last even number.

The seventh term is (30), so (S_7=\frac{7}{2}(60+30)=315). The same sum formula works for a decreasing progression.

The formula for average is \(\text{average}=\frac{\text{sum of terms}}{\text{number of terms}}\). Hence, \(\frac{126}{6}=21\), so 21 is correct. Values such as 20 or 22 result from not dividing the total sum correctly by 6. Exam tip: for an average question, divide the total sum by the number of terms first.

The sixteenth term is (88), so (S_{16}=\frac{16}{2}(13+88)=808). For larger (n), find the last term first.

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. The eighteenth term is a₁₈ = a + 17d = 1 + 17 × 5 = 86. Therefore, S₁₈ = n(a + l) ÷ 2 = 18(1 + 86) ÷ 2 = 9 × 87 = 783. Thus option B is correct. The factor 17 is used because the eighteenth term is reached through 18 − 1 intervals. The other choices usually result from using 18 differences, miscomputing the 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.

The governing concept is the sum of an arithmetic progression. The first 21 even natural numbers are 2, 4, 6, …, 42, so they form an AP with a = 2, d = 2, n = 21, and last term l = 42. Using Sₙ = n(a + l) ÷ 2 gives S₂₁ = 21(2 + 42) ÷ 2 = 21 × 44 ÷ 2 = 21 × 22 = 462. Equivalently, the sum of the first n even numbers is n(n + 1), which gives 21 × 22 = 462. Therefore, option B is correct; the nearby alternatives reflect simple arithmetic errors.

The governing concept is the sum of the first n terms of an arithmetic progression. The first term is a = 17 and the common difference is d = 20 − 17 = 3. The twelfth term is a₁₂ = a + 11d = 17 + 11 × 3 = 50. Applying Sₙ = n(a + l) ÷ 2 gives S₁₂ = 12(17 + 50) ÷ 2 = 6 × 67 = 402. Therefore, option B is correct. The multiplier 11 appears because reaching the twelfth term requires 12 − 1 differences. The other options can result from using an incorrect final term or from an arithmetic mistake in the formula.

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.