1.

The `(m+n)t ha n d(m-n)t h`terms fa G.P. ae `pa n dq`respectively. Show that the mth and nth terms are `sqrt(p q)a n dp(q/p)^(m//2n)`respectively.

Answer» Let `a` is the first term of the G.P. is `a` and the common ratio is `r`.
Then, it is given that ,
`ar^(m+n-1) = p->(1)`
`ar^(m-n-1) = q->(2)`
Multiplying (1) and (2),
`a^2r^(m+n-1+m-n-1) = pq`
`=>a^2r^(2(m-1)) = pq`
`=>ar^(m-1) = sqrt(pq)`
`:. m` th term of the given G.P. is `sqrt(pq).`
Now, dividing (1) by (2),
`=>(ar^(m+n-1) )/(ar^(m-n-1) ) = p/q`
`=>r^(2n) = p/q`
`=>r = (p/q)^(1/(2n))`
Now, putting value of `r` in (1),
`=>a((p/q)^(1/(2n)))^(m+n-1) = p`
`=>a = p/((p/q)^((m+n-1)/(2n)))`
`=>ar^(n-1) = p/((p/q)^((m+n-1)/(2n)))**((p/q)^(1/(2n)))^(n-1)`
`=>ar^(n-1) = p(p/q)^(((n-1)/(2n) - (m+n-1)/(2n)))`
`=>ar^(n-1) = p(p/q)^((-m)/(2n))`
`=>ar^(n-1) = p(q/p)^((m)/(2n))`
`:. n` th term of the given G.P. is ` p(q/p)^((m)/(2n)).`


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