Question

In a gp pth term is q and qth term is p find the (p+q)th term

Answer 1

that's it the and is
${( \frac{ {q}^{p} }{ {p}^{q} })}^{ \frac{1}{p - q}}$

Answer 2

$t_{p + q} = p \times [(\frac{q}{p} )^{\frac{1}{(p - q)} } ]^{p}$

Step-by-step explanation:

Let us assume that the first term of the G.P. is a and the common ratio is r.

So, the pth term of the G.P. is given by  $t_{p} = ar^{p - 1} = q$ .........(1)

Again, the qth term of the G.P. is given by $t_{q} = ar^{q - 1} = p$ .......... (2)

Now, dividing equation (1) with equation (2) we get,

$\frac{ar^{(p - 1)} }{ar^{(q - 1)}} = \frac{q}{p}$

⇒ $r^{(p - q)} = \frac{q}{p}$

⇒ $r = (\frac{q}{p} )^{\frac{1}{(p - q)} }$ .......... (3)

Now, the (p + q) th term of the G.P. = $t_{p + q} = ar^{(p + q - 1)} = (ar^{p - 1} \times ar^{q - 1}) \times \frac{r}{a}$

⇒ $t_{p + q} = pq \times \frac{r}{a}$ ........... (4)

Now, from equation (1) we get,

$\frac{a}{r} = \frac{q}{r^{p}}$

⇒ $\frac{r}{a} = \frac{r^{p}}{q} = \frac{[(\frac{q}{p} )^{\frac{1}{(p - q)} } ]^{p} }{q}$ {From equation (3)}  ........... (5)

Now, from equation (4) we get,

$t_{p + q} = pq \times \frac{r}{a} = pq \times \frac{[(\frac{q}{p} )^{\frac{1}{(p - q)} } ]^{p} } {q} = p \times [(\frac{q}{p} )^{\frac{1}{(p - q)} } ]^{p}$ {From equation (5) (Answer)