If the first and the nth terms of a GP are a and b respectively and if P is the product of the first n terms, then `P^(2)` is equal to

Explanation

Let `a` be the first term & `r` be the common ratio of G.P.

Given,
First term of G.P.=`a`

We know that

`n^(th)` term of G.P. `= ar^(n-1)`

`b=ar^(n-1) " " " " .....(1)`

Now P is the product of `n` terms

`P= a_1xxa_2xxa_3xx........a_n`

` " " " " = axxarxxar^2xxar^3xx.......xxar^(n-1)`

` " " " " = (axxaxx......a)xx(rxxr^2xx......r^(n-1))`

` " " " " = a^nr^(1+2+....+(n-1))`

` " " " " = a^nr^((n(n-1))/2)`

Thus, `P=a^nr^((n(n-1))/2)`


We need to prove `P^2=(ab)^n`

Taking L.H.S.
`P^2`

` " " " " = (a^nr^((n(n-1))/2))^2`

` " " " " = (a^(nxx2)r^((n(n-1))/2xx2))`

` " " " " = (a^(2n)r^(n(n-1)))`

` " " " " = (a^(2)r^((n-1)))^n`

` " " " " = (axxaxxr^((n-1)))^n`

` " " " " = (axx(ar^((n-1))))^n`

` " " " " = (axxb)^n`

` " " " " = (ab)^n`

` " " " " =` R.H.S.

Thus `P^2=(ab)^n`

Hence Proved