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

The first number is (1036), the last is (9990), and there are (243) terms, so the sum is (1339659). Choose the first and last multiples carefully.

Here (d=-13), and (S_{35}=2765). Do not forget the negative sign of the common difference in a decreasing AP.

The expression \(S_n=3n^2+2n\) gives the sum of the first \(n\) terms. To find the sum from the 51st through the 80th term, subtract the sum of the first 50 terms from the sum of the first 80 terms. This leaves exactly the required terms, since all terms before the 51st are removed.

Compute \(S_{80}=3(80)^2+2(80)=3(6400)+160=19360\). Also, \(S_{50}=3(50)^2+2(50)=3(2500)+100=7600\). Therefore the required sum is \(S_{80}-S_{50}=19360-7600=11760\). Hence option B is correct. The subtraction must use \(S_{50}\), not \(S_{51}\), because the 51st term must remain in the answer.

The given sums give (a=12) and (d=6), so (S_{36}=4212). From two partial sums, determine the AP first.

The last term is (-98), and (S_{30}=2280). Once the last term is found, (S_n=\frac{n}{2}(a+l)) is faster.

Option A is correct. The sequence is an arithmetic progression because the difference between consecutive terms is constant: d = 43 − 29 = 14 and 57 − 43 = 14. Here the first term is a = 29 and the number of terms is n = 31. Use S_n = n/2 [2a + (n − 1)d]. Substitution gives S_31 = 31/2 [2(29) + (31 − 1)(14)] = 31/2 [58 + 420] = 31/2 × 478 = 31 × 239 = 7409. As a check, the 31st term is 29 + 30 × 14 = 449, and the average of the first and last terms is (29 + 449)/2 = 239; multiplying this average by 31 again gives 7409. Therefore A is unambiguous. The other values result from arithmetic or formula errors.

The formula gives (S_{45}=\frac{45}{2}[-100+660]=12600). Keep the sign of the negative first term carefully.

From (4290=13[2a+275]), (a=\frac{55}{2}). In unknown-first-term questions, simplify the bracket first.

Subtracting the sum of multiples of (48) from the sum of multiples of (24) gives (1027752). In but-not cases, subtract the stricter condition.

The first multiple is (209), the last is (1197), and there are (53) terms, so the sum is (37259). Choose the first multiple within the range correctly.

The given sums determine the AP, and (S_{30}-S_{20}=910). Before finding a later block sum, determine (a,d).

For an AP, the \(n\)th term is \(a_n=S_n-S_{n-1}\). Therefore, \(a_{35}=S_{35}-S_{34}\). Here, \(S_{35}=5(35)^2-4(35)=5985\) and \(S_{34}=5(34)^2-4(34)=5644\). Hence, \(a_{35}=5985-5644=341\). Option 336 does not give the correct difference for \(n=35\). Exam tip: When \(S_n\) is given, find an individual term using \(S_n-S_{n-1}\).

This is an AP with first term 40 and common difference 8. The 30th row has \(40+29\times8=272\) seats, and the 75th row has \(40+74\times8=632\) seats. There are \(75-30+1=46\) rows from the 30th to the 75th row. Therefore, the required total is \(\frac{46}{2}(272+632)=20792\). Taking only the sum up to the 75th row would incorrectly include the first 29 rows as well. Exam tip: for terms from one position to another, count them as \(\text{last position}-\text{first position}+1\).

Solving (\frac{n}{2}[22+9(n-1)]=3973) gives (n=29). The number of terms must be a positive integer.

The governing formula for the sum of an arithmetic progression is S_n = n/2(a + l), where a is the first term and l is the last term. The question gives the combined value a + l = 420 and the total sum S_n = 7350, so substitute these directly: 7350 = n/2 × 420 = 210n. Dividing both sides by 210 gives n = 7350/210 = 35. Therefore, option C is correct. The common difference, first term separately, and last term separately are unnecessary here because their sum is already supplied. Options A, B, and D result from an incorrect division or substitution.

The numbers are (111,124,\ldots,995), and their sum is (38157). A remainder-based AP has common difference equal to the divisor.

The required sum is (S_{60}-S_{24}=12006). When starting from the (25)th term, subtract the sum up to (24) terms.

For a sequence whose sum of the first n terms is S_n, the sum from the rth term through the sth term is S_s − S_{r−1}. Here the required range is from the 61st to the 90th term, so calculate S_90 − S_60. S_90 = 4(90²) + 3(90) = 4(8100) + 270 = 32670, while S_60 = 4(60²) + 3(60) = 4(3600) + 180 = 14580. Hence the required sum is 32670 − 14580 = 18090. Therefore, option A is correct. Subtracting S_61 or S_90 − S_61 would exclude or mishandle a boundary term, producing distractor values.

The given sums give (a=6) and (d=7), so (S_{54}=10341). From two partial sums, determine the AP first.

The required sum is (S_{28}-S_{14}=1806). The sum of consecutive terms is quickly found by subtracting partial sums.