Question

# Let bi>1 for i=1,2,...,101. Suppose logeb1,logeb2,...,logeb101 are in Arithmetic Progression (A.P.) with the common difference loge2. Suppose a1,a2,...,a101 are in A.P. such that a1=b1 and a51=b51. If t=b1+b2+⋯+b51 and s=a1+a2+⋯+a51, then

A
s>t and a101>b101
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B
s>t and a101<b101
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C
s<t and a101>b101
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D
s<t and a101<b101
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Solution

## The correct option is B s>t and a101<b101As logeb1,logeb2,.....,logeb101 are in A.P. with common difference loge2, So logebi+1−logebi=loge2, i=1,2,...,100⇒bi+1=2bi ∴b1,b2,...,b101 are in Geometric Progression(G.P.) with common ratio 2. Let a1=b1=y and a51=b51=x=b1.251−1 then x=y.250 s=512(a1+a51) =51(x+y)2=51(y.250+y)2 t=b1(251−1)=y(251−1) Now s−t=51.249y+532y−251.y =47.249y+532y>0 i.e., s−t>0⇒s>t Also, a1,a51, and a101 are in A.P. ∴2a51=a101+a1⇒a101=2a51−a1 ⇒a101=2x−y and b1,b51, and b101 are in G.P. ∴b101=(b51)2b1=x2y b101−a101=x2y−2x+y=(x−y)2y>0 ∴b101>a101

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