Question

# If $$3+\dfrac{1}{4} (3+d)+\dfrac{1}{4^{2}}(3+2d)+..+$$ upto $$\infty =8$$, then the value of $$d$$ is :

A
9
B
5
C
1
D
none of these

Solution

## The correct option is B $$9$$$$S=( 3+\dfrac{3}{4}+\dfrac{3}{4^{2}}+....)+ (\dfrac{d}{4}+\dfrac{2d}{4^{2}}+\dfrac{3d}{4^{3}}+..)$$$$S= 3(1+\dfrac{1}{4}+\dfrac{1}{4^{2}}+..)+ (s)$$ (lets's say)$$s= \dfrac{d}{4}+\dfrac{2d}{4^{2}} +\dfrac{3d}{4^{3}} +.. (1)$$$$\dfrac{s}{4}= \dfrac{d}{4^{2}} +\dfrac{2d}{4^{3}}+...(2)$$subtrating  (1)-(2)$$\dfrac{3s}{4}=\dfrac{d}{4}+\dfrac{d}{4^{2}}+\dfrac{d}{4^{3}} +....$$$$s=\dfrac{d}{3}(1+\dfrac{1}{4}+\dfrac{1}{4^{2}}+..)$$$$s=\dfrac{4d}{9}$$$$S=3(\dfrac{4}{3})+\dfrac{4d}{9}$$$$S=4+\dfrac{4d}{9}=8$$$$d=9$$Maths

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