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

From (\frac{n}{2}[50-4(n-1)]=105), (n=7). Even in a decreasing AP, (n) must be a positive integer.

The natural numbers from 1 through 32 form an arithmetic progression with first term 1, last term 32, and 32 terms. The standard sum formula is S_n = n(n + 1)/2. Substituting n = 32 gives S_32 = 32 × 33 / 2 = 16 × 33 = 528. Therefore, option B is correct. The same result can be obtained by pairing the terms: 1 + 32 = 33, 2 + 31 = 33, and so on; there are 16 such pairs, giving 16 × 33 = 528. The other options reflect arithmetic errors in multiplication, division, or in counting the terms.

The sequence is an AP because consecutive terms differ by the constant amount d = 16 − 9 = 7. Thus a = 9 and n = 13. Use S_n = n/2[2a + (n − 1)d]: S_13 = 13/2[2(9) + 12(7)] = 13/2[18 + 84] = 13/2 × 102 = 13 × 51 = 663. Hence option C is correct. An equivalent check uses the last term: a_13 = 9 + 12 × 7 = 93, so S_13 = 13/2(9 + 93) = 663. The other options arise from an incorrect difference or an arithmetic error.

The sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+l)\). Using \(l=a+(n-1)d\) gives option A. Option B represents only the \(n\)th term. Exam tip: distinguish the formula for a sum from that for an \(n\)th term.

When the first term, last term, and number of terms are known, the most efficient AP formula is S_n = n/2(a + l), where a is the first term and l is the last term. Substituting n = 19, a = 14, and l = 104 gives S_19 = 19/2(14 + 104) = 19/2 × 118 = 19 × 59 = 1121. Therefore, option C is correct. There is no need to find the common difference or list all terms. The nearby options result from small arithmetic errors, such as mishandling 118/2 or multiplying 59 by an incorrect number.

Using (S_n=\frac{n}{2}[2a+(n-1)d]), the sum is (3038). In exams, calculate ((n-1)d) separately.

The required sum is (S_{30}-S_{12}=1566). To find a middle block sum, subtract the previous partial sum from the larger sum.

First, (192=12+(n-1)6) gives (n=31), and the sum is (3162). When the last term is given, find the number of terms first.

The numbers are (110,121,\ldots,495), and their sum is (10890). When between is written, check carefully whether endpoints are included.

This is an AP with (a=1500), (d=250), (n=18), and the total is (65250). In word problems, treat each amount as a term.

The first number is (104), the last is (988), and there are (69) terms, so the sum is (37674). In divisibility questions, choose the first and last values carefully.

The governing concept is the sum of a consecutive block of terms of an arithmetic progression. Here the first term is a = 50 and the common difference is d = -3. The 6th term is a6 = 50 + 5(-3) = 35, and the 20th term is a20 = 50 + 19(-3) = -7. There are 20 - 6 + 1 = 15 terms in the required block. Using the AP sum formula, block sum = 15/2 × (35 + (-7)) = 15/2 × 28 = 210. Thus option C is correct. A and B are smaller incorrect totals, while D does not follow from the two endpoint terms and the correct number of terms.

The governing concept is that the sum from the rth term to the sth term of a sequence is Sₛ - Sᵣ₋₁. Since the required terms begin with the 12th term and end with the 20th term, calculate S20 - S11. From the given expression, S20 = 4(20)² - 3(20) = 1600 - 60 = 1540. Also, S11 = 4(11)² - 3(11) = 484 - 33 = 451. Hence the required sum is 1540 - 451 = 1089. Therefore option D is correct. Subtracting S12 or S10 would exclude or include the wrong boundary term, which explains the likely distractors.

The two sums give (a=7) and (d=4), so (S_{30}=1950). When two partial sums are given, find (a,d) first.

Here (a=80), (d=-5), (n=18), and the sum is (675). Do not forget the negative sign of (d) in a decreasing AP.

Putting (a=11), (d=6), (n=19) in the formula gives (S_{19}=1235). Do not worry about (\frac{n}{2}) when (n) is odd.

In \(S_n=\frac{n}{2}[2a+(n-1)d]\), the coefficient of \(n^2\) is \(\frac d2\). Since \(d\ne0\), \(S_n\) is quadratic, not linear. Exam tip: check the \(n^2\) term first.

The governing concept is the sum formula for the first n terms of an AP: Sₙ = n/2 [2a + (n - 1)d]. Substitute n = 16, d = 7 and S₁₆ = 1176: 1176 = 16/2 [2a + 15(7)] = 8(2a + 105). Dividing by 8 gives 147 = 2a + 105, so 2a = 42 and a = 21. Therefore option A is correct. The other numerical choices result from an arithmetic error, using the wrong factor for n - 1, or failing to divide the total sum by the correct multiplier. The value of d is positive, so the progression increases from its first term.

The numbers are (117,135,\ldots,999), and the sum of (50) terms is (27900). For odd multiples, the common difference is (18).

The AP is (56,63,\ldots,245) with (28) terms, and the sum is (4214). Choose the first and last multiples within the limits correctly.