If a1,a2,a3⋯,an,⋯ are in G.P., then the determinant Δ=∣∣
∣∣loganlogan+1logan+2logan+3logan+4logan+5logan+6logan+7logan+8∣∣
∣∣ is equal to
A
1
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B
\N
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C
4
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D
2
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Solution
The correct option is B \N Since, a1,a2,a3⋯,an,⋯ are in G.P. an=a1rn−1 ⇒logan=loga1+(n−1)logr an+1=a1rn ⇒logan+1=loga1+nlogr an+2=a1rn+1 ⇒logan+2=loga1+(n+1)logr
and so on
Now, Δ=∣∣
∣∣loganlogan+1logan+2logan+3logan+4logan+5logan+6logan+7logan+8∣∣
∣∣ =∣∣
∣
∣∣loga1+(n−1)logrloga1+nlogrloga1+(n+1)logrloga1+(n+2)logrloga1+(n+3)logrloga1+(n+4)logrloga1+(n+5)logrloga1+(n+6)logrloga1+(n+7)logr∣∣
∣
∣∣
Applying R2→R2−R1 and R3→R3−R1, Δ=2∣∣
∣
∣∣loga1+(n−1)logrloga1+nlogrloga1+(n+1)logr3logr3logr3logr3logr3logr3logr∣∣
∣
∣∣
=0 (since two rows are identical)