Let `b_i gt 1 " for " i= 1,2 …., 101` .Suppose loge `b_1` loge `b_2` ….., loge `b_(101)` are in arihtmetic 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). If t= b_1+b_2+….+ b_51 " and " s=a_1+a_2+...+a_(51)` then

Options

A. `sgtt and a_(101)gtb_(101)`
B. `sgtt and a_(101)ltb_(101)`
C. `sltt and a_(101)gtb_(101)`
D. `sltt and a_(101)ltb_(101)`

Answer

B

Explanation

It is given `log_(e)b_(1),log_(e)b_(2), . . . .,log_(e)b_(101)` are in A.P. with common difference `log_(e)2`. Therefore,
`b_(1),b_(2),b_(3), . . . .,b_(101)` are G.P. with common ratio 2.
Let d be the common difference of the A.P. `a_(1),a_(2),a_(2),a_(101)`.
Now,
`a_(51)=b_(51)`
`rArr" "a_(1)+50d=b_(1)2^(50)`
`rArr" "a_(1)+50d=a_(1)(a^(50))" "[becauseb_(1)=a_(1)]`
`rArr" "50d=(2^(50)-1)a_(1)` . . . .(i)
`:." "a_(101)=a_(1)+100d=a_(1)+2(2^(50)-1)a_(n)=(2^(51)-1)a_(1)and,b_(101)=b_(1)xx2^(100)=a_(1)(2^(100))`
Clearly, `b_(101)gta_(101)`.
Now,
`s=(51)/(2)(2a_(1)+50d)=51a_(1)+(50xx51d)/(2)`
`rArr" "s=51a_(1)+(51)/(2)(2^(50)-1)a_(1)`
`rArr" "s=(51)/(2)a_(1)(2^(50)+1)" [Using (i)]"`
`and,t=b_(1)+b_(2)+ . . . +b_(51)`
`rArr" "t=b_(1)(2^(51)-1)/(2-1)=a_(1)(2^(51)-1)`
`:." "s-t=(51a)/(2)(2^(50)+1)-a_(1)(2^(51)-1)=(47xx2^(49)+(53)/(2))a_(1)gt0`
`rArr" "sgt t`
Hence, `sgtt anda_(101)ltb_(101)`. So, option (b) is correct.