Determinants Matrices Question 116
Question: If $ a_1,a_2…a _{n}…. $ form a G.P. and $ a _{i} $ >0, for all $ i\ge 1 $ , then $ \begin{vmatrix} \log a _{n} & \log {a _{n+1}} & \log {a _{n+2}} \\ \log {a _{n+3}} & \log {a _{n+4}} & \log {a _{n+5}} \\ \log {a _{n+6}} & \log {a _{n+7}} & \log {a _{n+8}} \\ \end{vmatrix} $ is equal to
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] we have, $ {a _{n+1}}^{2}=a _{n}{a _{n+2}} $
$ \Rightarrow 2\log {a _{n+1}}=\log a _{n}+\log {a _{n+2}} $ Similarly, $ 2\log {a _{n+4}}=\log {a _{n+3}}+\log {a _{n+5}} $
$ 2\log {a _{n+7}}=\log {a _{n+6}}+\log {a _{n+8}} $ Substituting these values in second column of determinant, we get
BETA
