if pth term of a gp is p and qth term is q prove that the nth term is (p - Maths - Sequences and Series

Question

if pth term of a gp is p and qth term is q prove that the nth
term is (p (to the power n-q) /q (to the power n-p) ) to the power
1/p-q

Aakash Sharma · Accepted Answer

Let the first term of the G P be a The common ratio be r Then the G P will be a, ar, ar2 arnâ€“1 Given: pth term (ap) = p â‡’arp-1 = p $\Rightarrow a = \frac{p}{r^{p - 1}} (1)$ and qth term (aq) = q â‡’arq-1 = q $\Rightarrow a = \frac{q}{r^{q - 1}} (2)$ from (1) and (2), we get $\Rightarrow \frac{p}{r^{p - 1}} = \frac{q}{r^{q - 1}} \Rightarrow \frac{r^{p - 1}}{r^{q - 1}} = \frac{p}{q} \Rightarrow r^{p - 1 - (q - 1)} = r^{p - 1 - q + 1} = r^{p - q} = \frac{p}{q} \Rightarrow r = (\frac{p}{q})^{\frac{1}{p - q}}$ Putting the value of r in (1), we get $a = \frac{p}{((\frac{p}{q})^{\frac{1}{p - q}})^{p - 1}} = \frac{p}{(\frac{p}{q})^{\frac{p - 1}{p - q}}} = p (\frac{q}{p}) \frac{p - 1}{p - q} = \frac{(q)^{\frac{p - 1}{p - q}}}{(p)^{\frac{p - 1}{p - q} - 1}} = \frac{(q)^{\frac{p - 1}{p - q}}}{(p)^{\frac{p - 1 - (p - q)}{p - q}}} = \frac{(q)^{\frac{p - 1}{p - q}}}{(p)^{\frac{p - 1}{p - q}}} = (\frac{(q)^{p - 1}}{(p)^{q - 1}})^{\frac{1}{p - q}} = (\frac{p - (q - 1)}{q - (p - 1)})^{\frac{1}{p - q}} = (\frac{p^{1 - q}}{q^{1 - p}})^{\frac{1}{p - q}}$ Hence nth term (an) = arnâ€“1 $= (\frac{p^{1 - q}}{q^{1 - P}})^{\frac{1}{p - q}} \times [(\frac{p}{q})^{\frac{1}{p - q}}]^{n - 1} = (\frac{p^{1 - q}}{q^{1 - P}})^{\frac{1}{p - q}} [(\frac{p}{q})^{n - 1}]^{\frac{1}{p - q}} = [\frac{p^{1 - q + n - 1}}{q^{1 - p + n - 1}}]^{\frac{1}{p - q}} = (\frac{p^{n - q}}{q^{n - p}})^{\frac{1}{p - q}}$

if pth term of a gp is p and qth term is q.prove that the nth term is (p (to the power n-q) /q (to the power n-p) ) to the power 1/p-q