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

Here, the first term is \(a=6\), the common difference is \(d=9-6=3\), and the number of terms is \(n=14\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{14}=\frac{14}{2}[2(6)+13(3)]=7(51)=357\). Hence, 357 is correct. An option such as 367 usually results from an error while calculating \((n-1)d\) or adding the terms. Exam tip: write down \(a\), \(d\), and \(n\) separately before applying the sum formula.

Using (S_n=\frac{n}{2}[2a+(n-1)d]), we get (S_{13}=442). Write ((n-1)d) carefully in the formula.

The tenth term is (51), so (S_{10}=\frac{10}{2}(15+51)=330). Finding the last term first is an easy method.

The sum of the first (n) natural numbers is (\frac{n(n+1)}{2}), so the answer is (253). Put the value of (n) directly in such questions.

The sum of the first (n) odd numbers is (n^2), so (15^2=225). This short formula saves time in exams.

The sum of the first (n) even numbers is (n(n+1)), so (19\times20=380). Even numbers start from (2,4,6,\ldots).

The governing concept is the sum of a finite arithmetic progression. When the first term a, last term l, and number of terms n are known, the efficient formula is Sₙ = n(a + l) ÷ 2. Here a = 8, l = 62, and n = 10. Substitution gives S₁₀ = 10(8 + 62) ÷ 2 = 10 × 70 ÷ 2 = 350. Thus option C is correct. There is no need to find the common difference because the first and last terms are already supplied. Options A, B, and D result from an incorrect average, an arithmetic slip, or using the wrong number of terms.

The governing concept is the sum of the first n terms of an arithmetic progression. Here the first term is a = 40 and the common difference is d = 36 − 40 = −4. The eighth term is a₈ = a + 7d = 40 + 7(−4) = 12. Now use S₈ = n(a + l) ÷ 2: S₈ = 8(40 + 12) ÷ 2 = 4 × 52 = 208. Therefore, option A is correct. The negative common difference must be retained because the progression is decreasing. The other options reflect an incorrect final term or an arithmetic mistake.

The ninth term is (47), so (S_9=\frac{9}{2}(7+47)=243). With an odd number of terms, the middle term can also check the sum.

Here (a=9), (d=3), and (n=12), so there will be (306) seats. In word problems, convert the pattern into an arithmetic progression.

The sum of the (7)th to (12)th terms is (S_{12}-S_6=180). Subtract partial sums to find the sum of middle terms.

The eleventh term is (73), so (S_{11}=\frac{11}{2}(3+73)=418). Finding the last term correctly is the key step.

Sum equals average (\times) number of terms, so (27\times8=216). If the average is given, the long formula is not needed.

The fifteenth term is (74), so (S_{15}=\frac{15}{2}(18+74)=690). The sum can also be found using the average of the first and last terms.

This is the sum of the first (10) multiples of (4), so the total is (220) questions. Treat the number of days as (n).

For an arithmetic progression, the first term is \\(a=20\\), the common difference is \\(d=-3\\), and the number of terms is \\(n=7\\). The seventh term is \\(a_7=a+(7-1)d=20+6(-3)=2\\). The sum formula is \\(S_n=\frac{n}{2}[2a+(n-1)d]\\), so \\(S_7=\frac{7}{2}[40+6(-3)]=\frac{7}{2}(22)=77\\).

The same result comes from pairing the first and last terms: the average is \\(\frac{20+2}{2}=11\\), and seven terms give \\(7\times11=77\\). Thus option A is correct. The negative difference decreases each term, but it must be included with its sign; replacing \\(d=-3\\) by 3 would produce an incorrect sum.

The twelfth term is (80), so (S_{12}=\frac{12}{2}(25+80)=630). Keep both the last term and (n) correct.

The sum of the first (n) odd numbers is (n^2), so (13^2=169). Remember this pattern for odd numbers.

The governing idea is summing the specified initial terms of an arithmetic progression. The formula aₙ = 3n + 2 gives a₁ = 3(1) + 2 = 5, a₂ = 8, a₃ = 11, a₄ = 14, and a₅ = 17. These terms form an AP with first term 5 and common difference 3. Their sum is 5 + 8 + 11 + 14 + 17 = 55. Equivalently, S₅ = 5[2(5) + (5−1)3] ÷ 2 = 55. Hence option B is correct. The distractors commonly come from using n = 0 initially, omitting a term, or making an arithmetic error.

Here (a=200), (d=50), and (n=8), so the total is (3000) rupees. In money questions, keep the unit in mind.