Discussion
The sum of the first 16 terms of an A.P. whose first term and third term are 5 and 15 respectively is:
Answer: A
Given \(a=5\) and \(a_3 = a+2d = 15\). So, \(5+2d=15 \Rightarrow 2d=10 \Rightarrow d=5\). We need the sum of the first 16 terms. \(S_{16} = \frac{16}{2}[2(5) + (16-1)5] = 8[10 + 15 \times 5] = 8[10+75] = 8(85) = 680\).
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