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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. |
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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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