A) \[\frac{m}{n}\]
B) \[\sqrt{\frac{m}{n}}\]
C) \[\sqrt{mn}\]
D) \[\sqrt{\left( \frac{n}{m} \right)}\]
Correct Answer: C
Solution :
In GP,\[(p+q)th\]term\[=m\] \[\Rightarrow \] \[a{{r}^{p+q-1}}=m\] ...(i) and \[(p-q)th\,term=n\] \[\Rightarrow \] \[a{{r}^{p-q-1}}=n\] ...(ii) Dividing Eq. (i) by Eq. (ii), \[\frac{a{{r}^{p+q-1}}}{a{{r}^{p-q-1}}}=\frac{m}{n}\] \[\Rightarrow \] \[{{r}^{p+q-1-p+q+1}}=\frac{m}{n}\] \[\Rightarrow \] \[{{r}^{2q}}=\frac{m}{n}\] \[\Rightarrow \] \[r={{\left( \frac{m}{n} \right)}^{1/2q}}\] ...(iii) Now, pth term\[=a{{r}^{P-1}}\] \[=a{{r}^{P+q-1}}.{{r}^{-q}}\] \[=m{{\left[ {{\left( \frac{m}{n} \right)}^{1/2q}} \right]}^{-q}}\] [using Eqs. (i) and (iii)] \[=m{{\left( \frac{m}{n} \right)}^{-1/2}}\] \[=m{{\left( \frac{m}{n} \right)}^{1/2}}=\sqrt{nm}\]
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