In the AP (25,33,41,\ldots), find the sum from the (40)th term to the (70)th term.

Correct answer: (14167)

The sequence \(25,33,41,\ldots\) is an arithmetic progression because the difference between consecutive terms is always \(8\). Its first term is \(25\), so the \(n\)th term is \(a_n=25+(n-1)8\). The required range contains terms 40 through 70, giving \(70-40+1=31\) terms.

The 40th term is \(25+39(8)=337\), and the 70th term is \(25+69(8)=577\). The sum of a finite AP is \(\frac{\text{number of terms}}{2}(\text{first term} + \text{last term})\). Hence the required sum is \(\frac{31}{2}(337+577)=\frac{31}{2}(914)=31(457)=14167\). Therefore option B is correct; using 30 instead of 31 terms would incorrectly exclude one endpoint.