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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


Your Answer
A
s>t and a101>b101
Correct Answer
B
s>t and a101<b101
Your Answer
C
s<t and a101>b101
Your Answer
D
s<t and a101<b101

Solution

The correct option is B s>t and a101<b101
As  logeb1,logeb2,.....,logeb101 are in A.P. with common difference loge2,
So logebi+1logebi=loge2,  i=1,2,...,100bi+1=2bi 
b1,b2,...,b101 are in Geometric Progression(G.P.) with common ratio 2.

Let a1=b1=y and a51=b51=x=b1.2511
then x=y.250 
s=512(a1+a51)  =51(x+y)2=51(y.250+y)2

t=b1(2511)=y(2511)

Now st=51.249y+532y251.y                 =47.249y+532y>0
i.e., st>0s>t

Also, a1,a51, and a101 are in A.P.
2a51=a101+a1a101=2a51a1
a101=2xy
and  b1,b51, and b101 are in G.P.
b101=(b51)2b1=x2y
b101a101=x2y2x+y=(xy)2y>0
b101>a101

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