A) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}=G\]
B) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}={{G}^{1/n}}\]
C) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}={{G}^{n}}\]
D) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}={{G}^{2/n}}\]
Correct Answer: C
Solution :
Here \[G={{(ab)}^{1/2}}\]and \[{{G}_{1}}=a{{r}^{1}},\ {{G}_{2}}=a{{r}^{2}},........{{G}_{n}}=a{{r}^{n}}\] Therefore \[{{G}_{1}}.\ {{G}_{2}}.\ {{G}_{3}}.....{{G}_{n}}={{a}^{n}}{{r}^{1+2+...+n}}={{a}^{n}}{{r}^{n(n+1)/2}}\] But \[a{{r}^{n+1}}=b\Rightarrow r={{\left( \frac{b}{a} \right)}^{1/(n+1)}}\] Therefore, the required product is \[{{a}^{n}}{{\left( \frac{b}{a} \right)}^{1/(n+1)\ .\ n(n+1)/2}}={{(ab)}^{n/2}}={{\left\{ {{(ab)}^{1/2}} \right\}}^{n}}={{G}^{n}}\]. Note: It is a well-known fact.
You need to login to perform this action.
You will be redirected in
3 sec
