CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
5
loader
C
1
loader
D
none of these
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image