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Let bi > 1 for i =1, 2,....,101. Suppose...

Let `b_i > 1` for i =1, 2,....,101. Suppose `log_e b_1, log_e b_2,....,log_e b_101` are in Arithmetic Progression (A.P.) with the common difference `log_e 2`. Suppose `a_1, a_2,...,a_101` are in A.P. such that `a_1 = b_1 and a_51 = b_51`. If `t = b_1 + b_2+.....+b_51 and s = a_1+a_2+....+a_51` then

A

`s gt t and a_(101) gt b_(101)`

B

`s gt t and a_(101) lt b_(101)`

C

`s lt t and a_(101) gt b_(101) gt b_(101)`

D

`s lt t and a_(101) lt b_(101)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have log `b_(2)-logb_(1)=log(2)`
`rArr(b_(1))/(b_(2))=2`
`rArrb_(1),b_(2)…..` are in G.P. with common ratio 2
`thereforet=b_(1)+2b_(1)+…+2^(50)b_(1)=b_(1)(2^(51)-1)`
`s=a_(1)+a_(2)+…+a_(51)`
`=51/2(a_(1)+a_(51))`
`=51/2(b_(1)+b_(51))`
`=51/2b_(1)(1+2^(50))`
`rArrs-t=b_(1)(51/2+51xx2^(49)-2^(51)+1)`
`=b_(1)(53/2+2^(49)xx47)gt0`
`rArrsgtt`
`b_(101)=2^(100)b_(1)`
`a_(101)=a_(1)+100d`
`=2(a_(1)+50d)-a_(1)`
`=2a_(51)-a_(1)`
`=2b_(51)-b_(1)`
`=(2xx2^(51)-1)b_(1)`
`=(2^(51)-1)b_(1)`
`thereforeb_(101)gta_(101)`
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