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If S be the sum P the product and R be t...

If `S` be the sum `P` the product and `R` be the sum of the reciprocals of `n` terms of a GP then `p^2` is equal to

A

`(S/R)^(n)`

B

`S/R`

C

`(R/S)^(n)`

D

`R/S`

Text Solution

Verified by Experts

The correct Answer is:
a
Let `S = a+ar +ar^(2) +...+ar^(n-1) = (a(1-r^(n)))/(1-r)" " [ :' r lt 1] `
P ` = a* ar* ar^(2) ....ar6(n-1) = a^(n) r^(1+2+...+(n-1))=a^(n)r (n(n-1))/2`
and `R = 1/a +1/ar +1/(ar^(2)) +....+ 1/(ar^(n-1))`
` = (1/2 (1/(r^(n))-1))/(1/r-1)=(1/a(1-r^(n)))/(r^(n-1(1-r)))" " [ :' 1/r gt1]`
Now, `(S/R)^(n) = {((a(1-r^(2)))/((1-r)))/(((1)/(a)(1-r^(n)))/(r^(n-1)(1-r)))}^(n)a^(2n)r^(n(n-1))`
`={a^(n)r^((n(n-1))/(2))}^(2)=P^(2)`
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