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If `a_(1) , a_(2) , a_(3),"………."a_(n)` are in H.P. , then `a_(1), a_(2) + a_(2) a_(3) + "………" + a_(n-1)a_(n)` will be equal to `:`

A

`2a_(1)a_(5)`

B

`3a_(1)a_(5)`

C

`4a_(1)a_(5)`

D

`6a_(1)a_(5)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given that ` a_(1),a_(2),a_(3),a_(4),a_(5)` are in HP.
`:. (1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)),(1)/(a_(4)),(1)/(a_(5))` are in AP.
`implies (1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a_(2))=(1)/(a_(4))-(1)/(a_(3))=(1)/(a_(5))-(1)/(a_(4))=d" " ["say "]`
`:. a_(1)-a_(2)=a_(1)a_(2)d " " implies a_(2)-a_(3)=a_(2)a_(3)d`
` a_(3)-a_(4)=a_(3)a_(4)d " " implies a_(4)-a_(5)=a_(4)a_(5)d`
On adding all, we get
`a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+a_(4)a_(5)=(a_(1)-a_(5))/(d)=a_(1)a_(5)(((1)/(a_(5))-(1)/(a_(1)))/(d))=4a_(1)a_(5)`.
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