If 3 + frac{1}{4} (3 + d) + frac{1}{4^2} (3 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  $(1)$Multiply by $\frac{1}{4}$:$\frac{S}{4} = \frac{3}{4} + \frac{3+d}{4^2} + \frac

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  $(1)$Multiply by $\frac{1}{4}$:$\frac{S}{4} = \frac{3}{4} + \frac{3+d}{4^2} + \frac{3+2d}{4^3} + \dots \infty$  $(2)$Subtract $(2)$ from $(1)$:$S - \frac{S}{4} = 3 + \frac{d}{4} + \frac{d}{4^2} + \frac{d}{4^3} + \dots \infty$$\frac{3S}{4} = 3 + \frac{d/4}{1 - 1/4} = 3 + \frac{d/4}{3/4} = 3 + \frac{d}{3}$Given $S = 8$,so $\frac{3(8)}{4} = 3 + \frac{d}{3}$$6 = 3 + \frac{d}{3}$ $\Rightarrow 3 = \frac{d}{3}$ $\Rightarrow d = 9$

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

Similar Questions

$2.\overline{357} = $

Let $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right]n$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then $\sum_{n=1}^{56} f(n)$ is equal to

The sum of the series $3 + 33 + 333 + \dots$ to $n$ terms is

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is

The first term of the ${11^{th}}$ group in the following sequence of groups $(1), (2, 3, 4), (5, 6, 7, 8, 9), \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module