Question

Let $a_1, a_2, a_3$, ... be an arithmetic progression with nonzero common difference. It is given that $\sum^{12}_{i = 4} a_i = 63$ and $a_k = 7 $ for some k . Then the value of k is

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The Correct Option is C

Given, \($a_1, a_2, a_3, ... $\) be an arithmetic progression \($ a_{1} +a_{2} +a_{3} + ...a_{n} = \frac{n}{2}\left(a_{1}+a_{n}\right)\)

 \(\Sigma_{i=4}^{12} a_{i} = 63\)

\(\Rightarrow a_{4} +a_{5} +a_{6} + ... +a_{12} = 63\) 

\(\Rightarrow \frac{9}{2}\left(a_{4} +a_{12}\right) = 63\)

 \(\Rightarrow a_{4} +a_{12} =14\)

 \(\Rightarrow a +3d +a +11d = 14\)

 \(\Rightarrow 2a +14 d = 14\)

 \(\Rightarrow a+7d = 7\) 

\(\Rightarrow a_{8} = 7\)

 \(\Rightarrow k = 8\)
Therefore the Correct Answer is (C) 8