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

Here the last term is (79), and (S_{20}=\frac{20}{2}(3+79)=820). Finding the last term can often be easier.

This is an arithmetic progression with (a=8), (d=3), and (n=10), so there will be (215) chairs. In word problems, convert the pattern into a progression.

The sum of the first 10 terms is \(S_{10}=175\), while the sum of the first 5 terms is \(S_5=50\). Hence, the sum from the 6th to the 10th term is \(S_{10}-S_5=175-50=125\). Subtracting \(S_5\) removes exactly the first five terms; 120 does not follow from the correct partial-sum relation. Exam tip: To find the sum of consecutive terms, subtract the smaller partial sum from the larger partial sum.

The first (6) terms go from (20) to (10), and the average is (15), so the sum is (90). The average of equally spaced terms is useful.

The governing concept is direct substitution in the AP sum formula S_n = n/2[2a + (n - 1)d]. Substituting a = 9, d = 1, and n = 14 gives S_14 = 14/2[2(9) + (14 - 1)(1)] = 7[18 + 13] = 7 × 31 = 217. Therefore option A is correct. A common error is to use n instead of n - 1 in the difference term, which would give an incorrect result. The other choices do not equal the correctly evaluated expression and are distractors caused by arithmetic or substitution mistakes.

Here (a=1), (d=3), and (n=11), so (S_{11}=176). For an odd number of terms, you can also check using the middle term.

Sum equals average (\times) number of terms, so (25\times9=225). When the average is given, the long formula is not needed.

The sum of n terms of an arithmetic progression can be found from the first term, last term, and number of terms using \(S_n=\frac{n}{2}(a+l)\). This works because the terms can be paired from the two ends, and every pair has the same total. Equivalently, the sum equals the number of terms multiplied by the average of the first and last terms.

Here \(n=13\), the first term is \(a=15\), and the last term is \(l=51\). Therefore, \(S_{13}=\frac{13}{2}(15+51)=\frac{13}{2}\times66=13\times33=429\). The average term is \((15+51)/2=33\), and 13 times 33 also gives 429. Hence option B, 429, is correct. The number of terms must be 13, not the number of gaps between them.

The first (5) terms are (30,25,20,15,10), so the sum is (100). In easy questions, writing the terms can also verify the answer.

Here (a=5), (d=2), and (n=12), so (192) questions are solved. Treat days as the number of terms.

The first (6) terms are (3,5,7,9,11,13), whose sum is (48). When (a_n) is given, find the first and last terms.

Here, the first term is \(a=4\), the common difference is \(d=9-4=5\), and \(n=16\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{16}=\frac{16}{2}[2(4)+15(5)]=8(83)=664\). Therefore, 664 is the correct answer. A value such as 672 usually results from an addition or multiplication error. Exam tip: calculate \((n-1)d\) first, then substitute carefully in the formula.

The sum formula for the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Here, \(S_7=140\), \(n=7\), and \(a=5\). Thus, \(140=\frac{7}{2}(5+l)\), which gives \(40=5+l\) and hence \(l=35\). Option \(40\) is not correct because it is the value of \(a+l\), not the last term. Exam tip: in questions involving the last term, first calculate \(a+l=\frac{2S_n}{n}\).

These are the first (10) multiples of (6), whose sum is (330). Treat multiples as an arithmetic progression.

(\frac{30\times31}{2}=465), so the correct sum is (465). Remember the direct formula for natural numbers.

For an arithmetic progression, S_n = n/2(a + l), where l is the nth or last term. Here a = 9, d = 13 - 9 = 4, and n = 15. First calculate the fifteenth term: l = a + (n - 1)d = 9 + 14(4) = 65. Thus S_15 = 15/2(9 + 65) = 15/2 × 74 = 15 × 37 = 555. Hence option A is correct. The other options are close numerical distractors, usually produced by finding an incorrect last term or making an addition or multiplication error.

This is the sum of the first (9) multiples of (4), so there are (180) plants. Even in real situations, write (a), (d), and (n).

For an arithmetic progression, the sum of n terms can be found using Sₙ = n/2 × (first term + last term). Substituting n = 13, the first term 2, and the last term 50 gives S₁₃ = 13/2 × (2 + 50) = 13/2 × 52 = 13 × 26 = 338. Therefore, option B is correct; the other values result from an arithmetic or formula error.

The first term is (100), and the eighth term is (30), so (S_8=\frac{8}{2}(130)=520). The same formula applies to decreasing progressions.

The sum of the first (n) odd numbers is (n^2), so (12^2=144). This pattern gives a quick answer.