A) 1
B) \[\frac{1}{2}\]
C) \[\frac{2}{3}\]
D) \[\frac{3}{2}\]
Correct Answer: B
Solution :
Let the first series be \[a+ar+a{{r}^{2}}+.........\]then the second series is \[{{a}^{2}}+{{a}^{2}}{{r}^{2}}+{{a}^{2}}{{r}^{4}}+..........\]their sums are given as 3. So, we have \[\frac{a}{1-r}=3\] or \[a=3(1-r)\] and \[\frac{q}{2}=\frac{pr}{p+r}=K\] or \[{{a}^{2}}=3(1-{{r}^{2}})\] Eliminating \[a,\ {{\left\{ 3\,(1-r) \right\}}^{2}}=3\,(1-{{r}^{2}})\] \[\Rightarrow \]\[3\,(1-r)=(1+r)\], \[\left\{ \because \ r\ne 1 \right\}\] \[\Rightarrow \]\[4r=2\] or \[r=\frac{1}{2}\].
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