Question

Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3/4 then

• A. a = 4/7, r = 3/7
• B. a = 2, r = 3/8
• C. a = 3/2, r = 1/2
• D. a = 3, r = 1/4

Answer 1

Answer: (D)

a = 3, r = 1/4

Answer 2

Since, sum = 4 and second term = $\frac{3}{4}$
It is given first term a and common rayio r .
\Rightarrow \hspace25mm \frac{a}{1-r} = 4, ar = \frac{3}{4}
\Rightarrow \hspace35mm r = \frac{3}{4a}
\Rightarrow \hspace25mm \frac{a}{1-\frac{3}{4a}} = 4
\Rightarrow \hspace25mm \frac{4a^2}{4a - 3} = 4
$\Rightarrow \, \, \, \, \, (a-1)\, (a - 3) = 0$
\Rightarrow \hspace40mm a = 1\, \, \, or\, \, \, \, 3
When a = 1, r = 3/4 and when a = 3, r = 1/4