In the AP (18,25,32,\ldots), find the sum from the (30)th term to the (55)th term.

Correct answer: (8021)

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.