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If a(1),a(2),a(3)"....." are in GP with ...

If `a_(1),a_(2),a_(3)"....."` are in GP with first term a and common rario r, then `(a_(1)a_(2))/(a_(1)^(2)-a_(2)^(2))+(a_(2)a_(3))/(a_(2)^(2)-a_(3)^(2))+(a_(3)a_(4))/(a_(3)^(2)-a_(4)^(2))+"....."+(a_(n-1)a_(n))/(a_(n-1)^(2)-a_(n)^(2))` is equal to

A

`(nr)/(1-r^(2))`

B

`((n-1)r)/(1-r^(2))`

C

`(nr)/(1-r)`

D

`((n-1)r)/(1-r)`

Text Solution

Verified by Experts

The correct Answer is:
B
`a_(1),a_(2),"….",a_(n)` are in GP with first term a and common rario r.
`S_(n)=ubrace((a_(1)a_(2))/(a_(1)^(2)-a_(2)^(2))+(a_(2)a_(3))/(a_(2)^(2)-a_(3)^(2))+(a_(3)a_(4))/(a_(3)^(2)-a_(4)^(2))+"....."+(a_(n-1)a_(n))/(a_(n-1)^(2)-a_(n)^(2)))_((n-1)" times ")" " "......(i)"`
`T_(n)=(a_(n-1)a_(n))/(a_(n-1)^(2)-a_(n)^(2))=(a_(n-1)a_(n))/((a_(n-1)-a_(n))(a_(n-1)-a_(n)))`
`=(1)/((1-(a_(n))/(a_(n-1)))(1+(a_(n-1))/a_(n)))`
`=(1)/((1-r)(1+(1)/(r )))=(r )/((r+1)(1-r)) " " [by GP]`
`:.S_(n)=sum_(n=2)^(n)T_(n)=sum_(n=2)^(n)(r )/((1-r^(2)))=(r )/((r+1)(1-r))`.
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