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

Answer: (C)

8

Answer 2

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