Question

If $(p+q{)}^{th}$ term of a G.P. be m and $(p-q{)}^{th}$ term be n, then the \(p^{th}) term will be

• A. m/n
• B.
• C. mn
• D. 0

Answer 1

The correct option is B

Given that m = $a{r}^{p-q-1}$

$r^{p+q-1-p+q+1}$ = $\frac{m}{n}\Rightarrow$ r = $(\frac{m}{n}{)}^{1\left(2q\right)}$

and a = $\frac{m}{(\frac{m}{n}{)}^{\frac{(p+q-1)}{2q}}}$

Now, $p^{th}$ term = $a{r}^{p-1}$ = $a{r}^{p-1}$ = $\frac{m}{(\frac{m}{n}{)}^{\frac{(p+q-1)}{2q}}}(\frac{m}{n}{)}^{\frac{(p-1)}{2q}}$

= $m(\frac{m}{n}{)}^{\frac{(p-1)}{\frac{2q-(p+q-1)}{\left(2q\right)}}}$ = $m(\frac{m}{n}{)}^{-\frac{1}{2}}$ = $m^{1-\frac{1}{2}}{n}^{\frac{1}{2}}$

= $m^{\frac{1}{2}}{n}^{\frac{1}{2}}$ = $\sqrt{mn}$

Aliter: As we know each term in a G.P. is geometric mean of the terms equidistant from it. Here $(p+q{)}^{th}$ and $(p-q{)}^{th}$ terms are equidistant from $p^{th}$ term i.e. at a distance of q. Therefore, $p^{th}$ term will be G.M. of $(p+q{)}^{th}$ and $(p-q{)}^{th}$ i.e., $\sqrt{mn}$