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If a1, a2, a3,.....an are in H.P. and ...

If `a_1, a_2, a_3,.....a_n` are in H.P. and `a_1 a_2+a_2 a_3+a_3 a_4+.......a_(n-1) a_n=ka_1 a_n`, then k is equal to

A

`(bn-1)(a_(1)-a_(n))`

B

`na_(1)a_(n)`

C

`(n-1)a_(1)a_(n)`

D

`n(a_(1)-a_(n))`

Text Solution

Verified by Experts

The correct Answer is:
c
Since , `a_(1)a_(2),a_(3),…a_(n)` are in HP .
` :. 1/(a_(1)),1/(a_(2)),1/(a_(3)) , ..., 1/(a_(n))` are in AP.
Let d be common difference of AP. Then `1/(a_(2)) -1/(a_(1))=d`
` rArr a_(1) -a_(2) =a_(1) a_(2) d`
Similarly , `underset(" "a_(n-1)-a_(n)=a_(n-1)a_(n)d)(underset(.)underset(.)underset(.)(a_(2)-a_(3))=underset(.)underset(.)underset(.)(a_(2)a_(3)d))`
On adding all above equations , we get
`a_(1)-a_(n) = d(a_(1)a_(2)+a_(2)a_(3)+...+a_(n-1)a_(n))" "` ...(i)
Also, ` 1/(a^(n))=1/(a^(1))+(n-1)d rArr d = (a_(1)-a_(n))/(a_(1)a_(n)(n-1))`
On putting the value of d in Eq. (i) , we get
` a_(1) -a_(n) = (a_(1)-a_(n))/(a_(1)a_(n)(n-1)) (a_(1)a_(2)+a_(2)a_(3)+...+ a_(n-1)a_(n))`
` rArr a_(1)a_(2) +a_(2)a_(3)+...+ a_(n-1)a_(n) = a_(1)a_(n) (n-1)`
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Knowledge Check

  • If a_1, a_2, a_3,….,a_(2n) are in AP, then the value of a_1^2-a_2^2+a_3^2-a_4^2+....+a_(2n-1)^2 is equal to

    A
    `n/(2n-1)(a_1^2-a_(2n)^2)`
    B
    `n/(2n+1)(a_1^2-a_(2n)^2)`
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    D
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    A
    `(n-1)(a_1-a_n)`
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    C
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    D
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