Question

If pth term of an GP is P and qth term is Q find .nth term is​

Answer 1

Hope it helps you ! mark as brainliest

Answer 2

Answer:

$(P^{n-q}.Q^{p-n})^\frac{1}{p-q}$

Step-by-step explanation:

Since,

If a GP has first term a and common ratio r then,

pth term = $ar^{p-1}$

qth term = $ar^{q-1}$

According to the question,

$ar^{p-1}=P----(1)$

$ar^{q-1}=Q-----(2)$

$\frac{equation (1)}{equation (2)}$

$\frac{r^{p-1}}{r^{q-1}}=\frac{P}{Q}$

$r^{p-1-q+1}=\frac{P}{Q}$

$r^{p-q}=\frac{P}{Q}$

$\implies r = (\frac{P}{Q})^\frac{1}{p-q}$

From equation (1),

$a(\frac{P}{Q})^{\frac{p-1}{p-q}}=P$

$\implies a = \frac{P}{(\frac{P}{Q})^{\frac{p-1}{p-q}}}=P(\frac{Q}{P})^\frac{p-1}{p-q}$

Hence, nth term = $ar^{n-1}$

$=P(\frac{Q}{P})^\frac{p-1}{p-q}(\frac{P}{Q})^\frac{n-1}{p-q}$

$=P^{1-\frac{p-1}{p-q}+\frac{n-1}{p-q}}\times Q^{\frac{p-1}{p-q}-\frac{n-1}{p-q}}$

$=P^{\frac{p-q-p+1+n-1}{p-q}}\times Q^{\frac{p-1-n+1}{p-q}}$

$=P^{\frac{n-q}{p-q}}\times Q^{\frac{p-n}{p-q}}$

$=(P^{n-q}.Q^{p-n})^\frac{1}{p-q}$

Note : by using the properties of exponents,

$a^m\times a^n = a^{m+n}$

$a^m=\frac{1}{a^{-m}}$

$(a^m)^n=a^{mn}$