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Question
If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)th term is `(q^p/p^q)^(1/(p - q))`
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Solution
Let a be the first term and r be the common ratio of a G.P.
Given that ap = q
⇒ arp–1 = q ....(i)
And aq = p
⇒ arq–1 = p ....(ii)
Dividing equation (i) by equation (ii) we get,
`(ar^(p - 1))/(ar^(q - 1)) = q/p`
⇒ `(r^(p - 1))/(r^(q - 1)) = q/p`
⇒ `r^(p - q) = q/p`
⇒ r = `(q/p)^(1/(p - q))`
Putting the value of r in equation (i), we get
`a[q/p]^(1/(p- q) xx p - 1)` = q
`a[q/p]^((p - 1)/(p - q))` = q
∴ a = `q * [p/q]^((p - 1)/(p - q))`
Now Tp+q = `ar^(p + q - 1)`
= `q[p/q]^((p - 1)/(p - q)) [q/p]^(1/(p - q)(p + q - 1)`
= `q(p/q)^((p - 1)/(p - q)) * (q/p)^((p + q - 1)/(p - q))`
= `q(p/q)^((p - 1)/(q - q)) * (p/q)^((-(p + q - 1))/(p - q))`
= `q(p/q)^((p - 1)/(p - q) - (p + q - 1)/(p - q))`
= `q(p/q)^((p - 1 - p - q + 1)/(p - q))`
= `q(p/q)^((-q)/(p - q))`
= `q(p/q)^(q/(p - q))`
= `(q^(q/(p - q) + 1))/(p^(q/(p - q))`
= `(q^(p/(p - q)))/(p^(q/(p - q))`
= `[q^p/p^q]^(1/(p - q))`
Hence, the required term = `[q^p/p^q]^(1/(p - q))`.
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