Consider an infinite geometric series with first term $$a$$ and common ratio $$r$$ . If its sum is 4 and the second term is $$\frac{3}{4}$$ , then

Solution :
Sum = 4 and second term = $$\frac{3}{4}$$ , it is given that
first term is $$a$$ and common ratio $$r$$
$$\eqalign{ & \Rightarrow \,\,\frac{a}{{1 - r}} = 4{\text{ and }}ar = \frac{3}{4} \cr & \Rightarrow r = \frac{3}{{4a}} \cr & {\text{Therefore, }}\frac{a}{{1 - \frac{3}{{4a}}}} = 4 \cr & \Rightarrow \,\frac{{4{a^2}}}{{4a - 3}} = 4 \cr & {\text{or }}{a^2} - 4a + 3 = 0 \cr & \Rightarrow \,\,\left( {a - 1} \right)\left( {a - 3} \right) = 0 \cr & \Rightarrow \,\,a = 1{\text{ or 3}} \cr & {\text{When }}a = 1,r = \frac{3}{4}{\text{ and when }}a = 3,r = \frac{1}{4} \cr} $$