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

S₂₂ represents the sum of the first 22 terms, while S₁₁ represents the sum of the first 11 terms. Removing the first 11 terms from the first 22 terms leaves exactly the 12th through 22nd terms, inclusive. Therefore the required sum is S₂₂ − S₁₁ = 1474 − 407 = 1067. There are 11 terms in this range, but it is not necessary to find the first term or common difference because the two given partial sums already provide the required subtraction directly. Hence option C, 1067, is correct. The other options result from addition, an incorrect subtraction, or a small arithmetic error.

To find the nth term of an AP from its sum, use \(a_n=S_n-S_{n-1}\). Here, \(a_n=(4n^2+9n)-[4(n-1)^2+9(n-1)]=8n+5\). Therefore, \(a_{27}=8\times27+5=221\). Hence, 221 is the correct option. The value 223 may result from an error of 2 during simplification. Exam tip: When \(S_n\) is given, obtain a term by subtracting two consecutive partial sums.

This is an AP with first term 30 and common difference 6. The 25th row has \(30+24\times6=174\) seats, and the 50th row has \(30+49\times6=324\) seats. There are \(50-25+1=26\) rows from the 25th through the 50th row. Therefore, the sum is \(\frac{26}{2}(174+324)=6474\). Both the 25th and 50th rows must be included, so the number of terms is 26, not 25. Exam tip: for a sum over an inclusive range, use last position minus first position plus 1 for the number of terms.

Solving (\frac{n}{2}[32+9(n-1)]=2030) gives (n=20). The number of terms must be a positive integer.

For an arithmetic progression with n terms, the sum is Sₙ = n/2(a + l), where a is the first term and l is the last term. The problem directly gives a + l = 340 and Sₙ = 5780. Substituting these values, 5780 = n/2 × 340 = 170n. Dividing both sides by 170 gives n = 5780/170 = 34. Therefore option B is correct. There is no need to determine the common difference, the individual end terms, or any intermediate term. Options 32, 36, and 38 arise from using an incorrect factor of 2 or making an arithmetic error while dividing.

The numbers are (13,22,\ldots,94), and the sum of (10) terms is (535). In remainder questions, the common difference equals the divisor.

The required sum is (S_{50}-S_{17}=7722). When starting from the (18)th term, subtract the sum up to (17) terms.

Sₙ denotes the sum of the first n terms. To obtain the sum from the 41st through the 60th term, subtract the sum of the first 40 terms from the sum of the first 60 terms: required sum = S₆₀ − S₄₀. Using the given expression, S₆₀ = 7(60²) − 2(60) = 7(3600) − 120 = 25,080, and S₄₀ = 7(40²) − 2(40) = 7(1600) − 80 = 11,120. Hence the required sum is 25,080 − 11,120 = 13,960. Therefore option A is correct. Subtracting S₄₁ or S₆₀ − S₄₁ would omit the 41st term.

The given sums give (a=5) and (d=4), so (S_{45}=4185). First determine the AP from the two partial sums.

The required sum is (S_{24}-S_{12}=1104). The sum of consecutive terms is quickly found by subtracting partial sums.

From the two terms, (d=5) and (a=9), so (S_{35}=3290). First find (a,d), then apply the sum formula.

The condition gives (5+11d=4(5+2d)), so (d=5) and (S_{20}=1050). Convert the term condition into an equation first.

(S_{33}=4983) and (S_{34}=5287), so the sum first exceeds (5000) at (34) terms. In such questions, also check the previous sum.

The first term is (3), and the (70)th term is (348), so the sum is (12285). Finding the first and last terms from (a_n) is an easy method.

The sum of the first \(n\) AP terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B gives the \(n\)th term \(a_n\), not the sum. In exams, carefully check the \(n-1\) factor.

From (S_{18}=9(a+99)), (a=14), then (d=5) and (S_{36}=3654). Treat the given (n)th term as the last term.

Here \(S_n=n(6n-1)\) represents the sum of the first \(n\) terms. A range beginning at the 31st term is found by taking the total through the 50th term and removing the total through the 30th term. This works because \(S_{30}\) contains precisely the terms that come before the required range.

Evaluate \(S_{50}=50(6\cdot50-1)=50(299)=14950\). Next, \(S_{30}=30(6\cdot30-1)=30(179)=5370\). Thus the sum from the 31st through the 50th term is \(S_{50}-S_{30}=14950-5370=9580\). Therefore option D is correct. The number of included terms is 20, from 31 to 50 inclusive, which is consistent with this subtraction method.

The selected terms are (a_5,a_{10},\ldots,a_{50}), and their sum is (1875). In position-based questions, form the new AP of selected terms.

The sum of all three-digit numbers is (494550), and the sum of multiples of (11) is (44550), so the answer is (450000). For not divisible, the complement method is fast.

Subtracting the sum of multiples of (50) from the sum of multiples of (10) gives (39600). Remove the common multiples of both conditions.