Discussion

Answer: A

Let the series be \(a, ar, ar^2, ...\). Sum \(\frac{a}{1-r} = 15\). The series of squares is \(a^2, a^2r^2, a^2r^4, ...\) which is a G.P. with first term \(a^2\) and common ratio \(r^2\). Its sum is \(\frac{a^2}{1-r^2} = 45\). We can write this as \(\frac{a^2}{(1-r)(1+r)} = 45\). Substituting \(a=15(1-r)\) is complex. Let's use \(\frac{a}{1-r} \times \frac{a}{1+r} = 45 \Rightarrow 15 \times \frac{a}{1+r} = 45 \Rightarrow \frac{a}{1+r} = 3 \Rightarrow a = 3(1+r)\). Now we have \(15(1-r) = 3(1+r) \Rightarrow 5(1-r)=1+r \Rightarrow 5-5r=1+r \Rightarrow 4=6r \Rightarrow r=2/3\). Then \(a = 15(1-2/3) = 15(1/3)=5\). The series is 5, 10/3, 20/9, ...