If a1,a2,a3,... from a geometric progression and ai>0 for all i≥1, then ∣∣
∣∣logamlogam+1logam+2logam+3logam+4logam+5logam+6logam+7logam+8∣∣
∣∣ is equal to
A
logam+8−logam
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B
logam
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C
2logam+1
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D
0
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Solution
The correct option is B0 Since a1,a2,a3.... are in G.P ∴loga1,loga2,loga3,.... are in G.P ⇒logam+logam+2=2logam+1 logam+3+logam+5=2logam+4 logam+6+logam+8=2logam+7 applying C2→C2−(1/2)(C1+C3) we get Δ=∣∣
∣∣logam0logam+2logam+30logam+5logam+60logam+8∣∣
∣∣=0