Question

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

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{p^{th}\,\,\,\,and\,\,\,\,q^{th}\,\,\,term\,\,\,a\,\,\,GP\,\,\,are\,\,\,respectively\,\,\,q\,\,\,and\,\,\,p}$

Ler the first term and the common ratio be 'A' and 'R' respectively

So,

$\sf{A\,R^{p-1}=q\,\,\,,\,\,\,A\,R^{q-1}=p}$

Divide

$\sf{\dfrac{A\,R^{p-1}}{A\,R^{q-1}}=\dfrac{q}{p}}$

$\sf{\implies\,R^{p-1-q+1}=\dfrac{q}{p}}$

$\sf{\implies\,R^{p-q}=\dfrac{q}{p}}$

$\sf{\implies\,R=\left(\dfrac{q}{p}\right)^{\dfrac{1}{p-q}}}$

Now, put the value of R in 1st equation

$\sf{A\,\left(\dfrac{q}{p}\right)^{\dfrac{p-1}{p-q}}=q}$

$\sf{\implies\,A=\left(\dfrac{p}{q}\right)^{\dfrac{p-1}{p-q}}\cdot\,q}$

So,

$\sf{(p+q)^{th} term=\,A\,R^{p+q-1}}$

$\sf{=q\cdot\left(\dfrac{p}{q}\right)^{\dfrac{p-1}{p-q}}\cdot\,\left(\dfrac{q}{p}\right)^{\dfrac{p+q-1}{p-q}}}$

$\sf{=q\cdot\left(p\right)^{\dfrac{p-1}{p-q}-\dfrac{p+q-1}{p-q}}\cdot\,\left(q\right)^{\dfrac{p+q-1}{p-q}-\dfrac{p-1}{p-q}}}$

$\sf{=q\cdot\left(p\right)^{\dfrac{p-1-p-q+1}{p-q}}\cdot\,\left(q\right)^{\dfrac{p+q-1-p+1}{p-q}}}$

$\sf{=q\cdot\left(p\right)^{\dfrac{-q}{p-q}}\cdot\,\left(q\right)^{\dfrac{q}{p-q}}}$

$\sf{=\left(p\right)^{\dfrac{-q}{p-q}}\cdot\,\left(q\right)^{\dfrac{q}{p-q}+1}}$

$\sf{=\left(p\right)^{\dfrac{-q}{p-q}}\cdot\,\left(q\right)^{\dfrac{q+p-q}{p-q}}}$

$\sf{=\left(p\right)^{\dfrac{-q}{p-q}}\cdot\,\left(q\right)^{\dfrac{p}{p-q}}}$

$\sf{=\left(p^{-q}\right)^{\dfrac{1}{p-q}}\cdot\,\left(q^{p}\right)^{\dfrac{1}{p-q}}}$

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