Mathematics - Sequence & Series Question with Solution | TestHub

Let the G.P. be $a,ar,a{r}^2,a{r}^3,\dots$. Given $a+ar=12\Rightarrow a(1+r)=12$. Also $a{r}^2+a{r}^3=48\Rightarrow a{r}^2(1+r)=48$. Dividing the second equation by the first, $r^2=48/12=4$. Since $r$ is positive, $r=2$. So (P)-(2). Substitute $r=2$ into $a(1+r)=12\Rightarrow a(1+2)=12\Rightarrow 3a=12\Rightarrow a=4$. So (Q)-(1). The sum of the first 4 terms $S_4=a\frac{r^4-1}{r-1}=4\frac{2^4-1}{2-1}=4(15)=60$. So (R)-(3). Since $r=2$, which is not less than 1, the infinite series does not converge. So (S)-(4).