Question

if the pth and qth terms of a GP are q and P respectively. show that (p+q)th term is (q^p/p^q)^1/p-q​

Answer 1

Given,

$ar^{p - 1} = q\quad\longrightarrow\quad(1)\\\\ar^{q - 1} = p\quad\longrightarrow\quad(2)$

Dividing (1) by (2),

$\dfrac {q}{p}= r^{p - q}\\\\r = \left(\dfrac {q}{p}\right)^{\frac {1}{p - q}}$

So from (1),

$a = \dfrac {q}{r^{p - 1}}\\\\a = \dfrac {q}{\left(\dfrac {q}{p}\right)^{\frac {p - 1}{p - q}}}\\\\a=\dfrac {q}{\left (\dfrac {q^{\frac {p-1}{p-q}}}{p^{\frac {p-1}{p-q}}}\right)}\\\\a=\dfrac {p^{\frac {p-1}{p-q}}\cdot q}{q^{\frac {p-1}{p-q}}}\\\\a=p^{\frac {p-1}{p-q}}\cdot q^{\frac{1-q}{p-q}}\\\\a=\left (\dfrac {p^{p-1}}{q^{q-1}}\right)^{\frac {1}{p-q}}$

Now,

$a_{p+q}=ar^{p+q-1}\\\\a_{p+q}=\left (\dfrac {p^{p-1}}{q^{q-1}}\right)^{\frac {1}{p-q}}\left (\dfrac {q}{p}\right)^{\frac {p+q-1}{p-q}}\\\\a_{p+q}=\left (\dfrac {p^{p-1}}{q^{q-1}}\right)^{\frac {1}{p-q}}\left (\dfrac {p}{q}\right)^{\frac {1-p-q}{p-q}}\\\\a_{p+q}=\left (\dfrac {p^{p-1}}{q^{q-1}}\cdot\dfrac {p^{1-p-q}}{q^{1-p-q}}\right)^{\frac {1}{p-q}}$

$a_{p+q}=\left (\dfrac {p^{p-1+1-p-q}}{q^{q-1+1-p-q}}\right)^{\frac {1}{p-q}}\\\\a_{p+q}=\left (\dfrac {p^{-q}}{q^{-p}}\right)^{\frac {1}{p-q}}\\\\a_{p+q}=\left (\dfrac {p^{q}}{q^{p}}\right)^{-\frac {1}{p-q}}\\\\\large\boxed{\mathbf{a_{p+q}=\left (\dfrac {q^{p}}{p^{q}}\right)^{\frac {1}{p-q}}}}$

Done!