A) 0
B) 1
C) 2
D) 3
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 getYou need to login to perform this action.
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