Question

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

• A. $a=\frac{4}{7},r=\frac{3}{7}$
• B. $a=3,r=\frac{1}{4}$
• C. $a=2,r=\frac{3}{8}$
• D. $a=\frac{3}{2},r=\frac{1}{2}$

Answer 1

Answer: (B)

$a=3,r=\frac{1}{4}$

Sum of infinite G.P. is $=\frac{a}{1-r}=4$

$\Rightarrow a=4(1-r)=4-4r$.......(1)

and second term is $ar=\frac{3}{4}$.....(2)

Multiply (1) by $r$

$ar=4r-4r^2$

$\Rightarrow \frac{3}{4}=4r-4r^2$

$\Rightarrow 16r^2-16r+3=0$

$\Rightarrow r=\frac{16\pm \sqrt{{16}^2-16\cdot 12}}{32}=\frac{16\pm 8}{32}=\frac{1}{4},\frac{3}{4}$

Now if $r=\frac{1}{4}$, then $a=3$

and if $r=\frac{3}{4}$, then $a=1$