Let 8 = 3 + 3+p/4 + 3+2p/4^2 + .... ∞, then p is

Question

Let 8 = 3 + \(\frac {3+P}{4} + \frac {3 + 2P}{4^2} + ... \infty,\) then p is

(1) 9 

(2) 5/4 

(3) 3 

(4) 1

Answer 1

(1) 9 

\(8 = 3+\frac {3+p}{4} +\frac {3+2p}{4^ 2}+.....(i)\)

multiply both sides by 1/4 , we get

\(2=\frac{3}{4}+\frac {3+P}{4^2}+...(ii)\)

Equation (i) – equation (ii)