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

The given sums give (a=1) and (d=5), so (S_{24}=1404). First determine the AP from the smaller sums.

Subtracting the sum of multiples of (30) from the sum of multiples of (6) gives (66336). Numbers divisible by both (6) and (15) are multiples of (30).

When S_n is given, the sum from the rth term to the sth term is S_s − S_{r−1}. Therefore, the required sum is S₄₅ − S₃₀. Calculate S₄₅ = 6(45²) + 45 = 6(2025) + 45 = 12,195. Also, S₃₀ = 6(30²) + 30 = 6(900) + 30 = 5,430. Hence the range sum is 12,195 − 5,430 = 6,765. Thus option D is correct. Subtracting S₃₁ would omit the 31st term, while using S₄₅ alone would include all earlier terms as well.

From (1625=\frac{25}{2}(a+113)), (a=17). Treat the (n)th term as the last term of the first (n) terms.

(a_{11}+a_{30}=a_1+a_{40}=210), so the sum of (20) terms is (2100). Sums of symmetric terms are equal in an AP.

Using (S_n=\frac{n}{2}[2a+(n-1)d]), the sum is (3523). In exams, calculate ((n-1)d) separately.

The required sum is (S_{36}-S_{17}=3116). To find a middle block sum, subtract the previous partial sum.

First, (184=-12+(n-1)7) gives (n=29), and the sum is (2494). When the last term is given, find (n) first.

The numbers are (153,170,\ldots,748), and their sum is (16218). Choose the first and last multiples within the limits correctly.

This is an AP with (a=3200), (d=225), (n=30), and the total is (193875). In word problems, treat each amount as a term.

The first number is (114), the last is (988), and there are (47) terms, so the sum is (25897). In divisibility questions, choose the first and last values carefully.

Here (d=-9), and the formula gives (S_{22}=1441). In a decreasing AP, write the common difference as negative.

The notation \(S_n=6n^2-5n\) gives the sum of the first \(n\) terms, not the value of the \(n\)th term. To obtain the sum beginning with the 31st term and ending with the 50th term, subtract the sum through the 30th term from the sum through the 50th term. This removes terms 1 to 30 and leaves exactly terms 31 to 50.

Calculate \(S_{50}=6(50)^2-5(50)=15000-250=14750\), and \(S_{30}=6(30)^2-5(30)=5400-150=5250\). Therefore the required range sum is \(S_{50}-S_{30}=14750-5250=9500\). Hence option B is correct. Subtracting \(S_{31}\) instead would wrongly omit the 31st term.

The two partial sums give (a=\frac{17}{2}) and (d=5), so (S_{48}=6048). Determine the AP first.

The last term is (-52), and (S_{25}=1100). Even in a decreasing AP, (S_n=\frac{n}{2}(a+l)) is useful.

Here (a=19), (d=11), (n=23), and the sum is (3220). The formula applies directly even when (n) is odd.

For an arithmetic progression, the sum of the first n terms is Sₙ = n/2[2a + (n − 1)d]. Here a = −35, d = 12, and n = 28. Substitution gives S₂₈ = 28/2[2(−35) + 27(12)] = 14[−70 + 324] = 14 × 254 = 3556. The negative first term must be retained while calculating 2a; changing its sign would produce an incorrect result. Therefore option C is correct. The other values are plausible arithmetic-error distractors, usually caused by mishandling the negative sign, using 26d or 28d, or making a final multiplication error.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(2450=\frac{20}{2}[2a+19\times9]\), so \(2450=10(2a+171)\). Hence \(245=2a+171\), giving \(2a=74\) and \(a=37\). If 39 is used, the sum becomes greater, so it is a close but incorrect distractor. Exam tip: remember to use \((n-1)d\), which is \(19d\) when \(n=20\).

The numbers are (105,147,\ldots,987), and their sum is (12012). In this condition, the new AP has common difference (42).

The first multiple is (128), the last is (896), and there are (49) terms, so the sum is (25088). Choose the first term according to the range carefully.