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

From the two terms, (d=5) and (a=2), so (S_{30}=2235). First find (a,d), then apply the sum formula.

The condition gives (6+8d=2(6+3d)), so (d=3) and (S_{25}=1050). Convert the term condition into an equation first.

(S_{25}=2800) and (S_{26}=3029), so the sum first exceeds (3000) at (26) terms. In such questions, also check the previous sum.

The first term is (11), and the (60)th term is (365), so the sum is (11280). Finding the first and last terms from (a_n) is an easy method.

For an AP with an odd number of terms, let the middle term be \(m\). The sum of the first \(n\) terms is \(S_n=n m\). Since \(n\neq0\) and \(S_n=0\), \(m=0\). The common difference need not be zero. Exam tip: for odd terms, relate the sum directly to the middle term.

From (S_{15}=\frac{15}{2}(a+68)), (a=12), then (d=4) and (S_{30}=2100). Treat the given (n)th term as the last term.

Here the sum of the first n terms is given directly by \\(S_n=n(4n+3)\\). For a range beginning with the 21st term, subtract the sum of the first 20 terms from the sum of the first 35 terms. This leaves exactly terms 21 through 35, including both endpoints.

Calculate \\(S_{35}=35[4(35)+3]=35(140+3)=35(143)=5005\\). Also, \\(S_{20}=20[4(20)+3]=20(80+3)=20(83)=1660\\). Therefore the required sum is \\(S_{35}-S_{20}=5005-1660=3345\\). Hence option D is correct. The subtraction uses 20, not 21, because the 20th term is the last term that must be excluded.

The selected terms are (a_4,a_8,\ldots,a_{40}), and their sum is (1060). 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 (9) is (55350), so the answer is (439200). For not divisible, the complement method is fast.

Subtracting the sum of multiples of (24) from the sum of multiples of (8) gives (33368). In a but-not condition, subtract the complement.

The two terms give (d=4) and (a=0), so (S_{40}=3120). Finding the common difference from distant terms is the first step.

The sequence is an arithmetic progression with first term 18 and common difference 7. Its nth term is \(a_n=18+(n-1)7\). The requested terms are the 30th through the 55th, so their number is \(55-30+1=26\). The first term of this block is \(a_{30}=221\), and the last is \(a_{55}=396\). The sum of consecutive AP terms equals the number of terms multiplied by the average of the first and last terms.

Hence the sum is \(26\times\frac{221+396}{2}=13\times617=8021\). The same result follows from \(S_{55}-S_{29}\), because subtracting through the 29th term leaves precisely terms 30 to 55. Therefore option A is correct. Subtracting \(S_{30}\) would incorrectly omit the 30th term, so the endpoint handling is important.

Use the AP sum formula S_n = n/2[2a + (n−1)d]. For n = 31, d = −9, and S₃₁ = 0, we get 0 = 31/2[2a + 30(−9)]. Since 31/2 is non-zero, the bracket must be zero: 2a − 270 = 0. Therefore, 2a = 270 and a = 135. Thus option B is correct. The zero sum occurs because the 31 terms are symmetric about the middle term, and the middle term is zero when a + 15d = 0; this also gives a = 135. The other choices do not make the stated partial sum zero.

There are 30 − 16 + 1 = 15 terms from the 16th through the 30th term. The first of these is t₁₆ = a + 15d = 5 + 15d, and the last is t₃₀ = a + 29d = 5 + 29d. Using the sum formula for these 15 terms, 1395 = 15/2[(5 + 15d) + (5 + 29d)] = 15/2(10 + 44d) = 75 + 330d. Hence 330d = 1320, so d = 4. Therefore option C is correct. Subtracting S₁₅ from S₃₀ gives the same result; forgetting that both endpoints are included would incorrectly count only 14 terms.

The given sums give (a=4) and (d=3), so (S_{35}=1925). From two sums, first find the AP values.

Adding sums of multiples of (4) and (7), then subtracting multiples of (28), gives (87850). Avoiding double counting is important.

The numbers are (11,17,\ldots,95), and the sum of (15) terms is (795). A remainder-based AP has common difference equal to the divisor.

The conditions give (a=18) and (d=5), so (S_{14}=707). Convert the given term and sum into two equations.

Apply S_n = n/2[2a + (n−1)d] with n = 10, a = x, and d = 2x + 1. Then 445 = 10/2[2x + 9(2x + 1)] = 5[2x + 18x + 9] = 5(20x + 9). Dividing by 5 gives 89 = 20x + 9, so 20x = 80 and x = 4. Therefore option D is correct. Checking the result, a = 4 and d = 9, so the sum is 5(8 + 81) = 445. The other options fail when substituted into the same formula.

Subtracting the sum of multiples of (36) from the sum of multiples of (12) gives (26460). Use the least common multiple to remove overlap.