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Question

Let bi>1 for i = 1, 2, .... ,101. Suppose loge b1,loge b2,loge b101 are in AP with the common difference loge2. Suppose a1,a2,a101 are in AP, 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<b101
If log b1, log b2,log b101 are in AP, with common difference loge 2, then b1,b2,b101 are in GP, with common ratio 2.
b1=20b1,b2=21b1, b3=22b1,b101=2100b1 (i)
Also, a1,a2a101 are in AP
Given a1=b1 and a51=b51
a1+50D=250b1
a1+50D=250a1 [a1=b1](ii)
Now, t=b1+b2++b51
t=b1(2511)21 (iii)
and s=a1+a2++a51
=512(2a1+50D) (iv)
t=a1(2511) [a1=b1]
or t=251a1a1<251a1 (v)
and s=512[a1+(a1+50D)] [fromEq.(ii)]
=512[a1+250a1]=512a1+512250a1s>251a1(vi)
From Eqs. (v) and (vi), we get s > t
Also, a101=a1+100D and b101=2100b1
a101=a1+100(250a1a150) and b101=2100a1
a101=a1+251a12a1=251a1a1
a101<251a1 and b101>251a1
b101>a101

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