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Let b(i)gt1" for "i=1,2,"......",101. Su...

Let `b_(i)gt1" for "i=1,2,"......",101`. Suppose `log_(e)b_(1),log_(e)b_(2),log_(e)b_(3),"........"log_(e)b_(101)` are in Arithmetic Progression (AP) with the common difference `log_(e)2` . Suppose `a_(1),a_(2),a_(3),"........"a_(101)` are in AP. 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

(a)`sgtt " and "a_(101)gtb_(101)`

B

(b)`sgtt " and "a_(101)ltb_(101)`

C

(c)`sltt " and "a_(101)gtb_(101)`

D

(d)`sltt " and "a_(101)ltb_(101)`

Text Solution

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The correct Answer is:
B
`:.log_(e)b_(1),log_(e)b_(2),log_(e)b_(3),"........"log_(e)b_(101)` are in AP.
`impliesb_(1),b_(2),b_(3),"........"b_(101)` are in GP with common ratio 2.
`(therefore " common difference " =log_(e)2)`
Also,`a_(1),a_(2),a_(3),"........"a_(101)` are in AP.
where, `a_(1)=b_(1)" and "a_(51)=b_(51)`
`:.b_(2),b_(3),"......",b_(50)` are GM's and `a_(2),a_(3),"......",a_(50)` are AM's between `b_(1)" and "b_(51)`.
`therefore GMltAM`
`impliesb_(2)lta_(2),b_(3)lta_(3),"......",b_(50)lta_(50)`
`:.b_(1)+b_(2)+b_(3)+"......"+b_(51)lta_(1)+a_(2)+a_(3)+"........."+a_(51)`
`implies tlts`
Also,`a_(1),a_(2),a_(3),"......",a_(101)" are in AP and "b_(1),b_(2),b_(3),".........",b_(101)` are in GP.
`:.a_(1)=b_(1)" and "a_(51)=b_(51)`
`:.b_(101)gta_(101)`.
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