Question

Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is $\frac{1}{4}$. then a = $-$ and r = $-$.

• A. $3,\frac{1}{4}$
• B. $1,\frac{1}{3}$
• C. $1,\frac{3}{4}$
• D. $2,\frac{3}{4}$

Answer 1

Answer: (A), (C)

The correct options are
A $3,\frac{1}{4}$
C $1,\frac{3}{4}$

Infinite geometric series $=a+ar+a^2+a{r}^3+......$ 2nd term

So, $ar=3/4$ ...(I)

Sum of I.G. S is $S=\frac{a}{1-r}\Rightarrow \frac{a}{1-r}=4$ ,,,(II)

from (I) and (II)

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

$a+4\times \frac{3}{4a}=4$

$\Rightarrow a^2+3=4a$,

$a^2-3a-a+3=0$

$a(a-3)-(a-3)=0$

$a=3,1$

$a=3,r=1/4$

$a=1,r=3/4$

$\therefore$ Hence, both (A) and (C) are correct.