A) \[m/n\]
B) \[\sqrt{mn}\]
C) \[mn\]
D) 0
Correct Answer: B
Solution :
Given that \[m=a{{r}^{p+q-1}}\] and \[n=a{{r}^{p-q-1}}\] \[{{r}^{p+q-1-p+q+1}}=\frac{m}{n}\Rightarrow r={{\left( \frac{m}{n} \right)}^{1/(2q)}}\] and \[a=\frac{m}{{{\left( \frac{m}{n} \right)}^{(p+q-1)/2q}}}\] Now \[{{p}^{th}}\]term \[=a{{r}^{p-1}}=\frac{m}{{{\left( \frac{m}{n} \right)}^{(p+q-1)/2q}}}{{\left( \frac{m}{n} \right)}^{(p-1)/2q}}\] \[=m{{\left( \frac{m}{n} \right)}^{(p-1)/2q-(p+q-1)/(2q)}}=m{{\left( \frac{m}{n} \right)}^{-1/2}}={{m}^{1-1/2}}{{n}^{1/2}}\] \[={{m}^{1/2}}{{n}^{1/2}}=\sqrt{mn}\]. 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 \[{{(p-q)}^{th}}\]\[i.e.\]\[\sqrt{mn}\].
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