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 the given series be $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  --- $(1)$Multiply by $\frac{1}{4}$:$\frac{S}{4} = \frac{3}{4} +

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the given series be $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  --- $(1)$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 + \left( \frac{3+d}{4} - \frac{3}{4} \right) + \left( \frac{3+2d}{4^2} - \frac{3+d}{4^2} \right) + \dots \infty$$\frac{3S}{4} = 3 + \frac{d}{4} + \frac{d}{4^2} + \frac{d}{4^3} + \dots \infty$The series $3 + \frac{d}{4} + \frac{d}{4^2} + \dots$ is an Arithmetico-Geometric series where the sum of the infinite geometric part $\frac{d}{4} + \frac{d}{4^2} + \dots$ is $\frac{a}{1-r} = \frac{d/4}{1 - 1/4} = \frac{d/4}{3/4} = \frac{d}{3}$.So,$\frac{3S}{4} = 3 + \frac{d}{3}$.Given $S = 8$,substitute $S = 8$ into the equation:$\frac{3(8)}{4} = 3 + \frac{d}{3}$$6 = 3 + \frac{d}{3}$$3 = \frac{d}{3}$$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

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms is

The ratio of the sum of $m$ and $n$ terms of an $A.P.$ is $m^2 : n^2$. Then the ratio of the $m^{th}$ and $n^{th}$ term will be:

The first term of an $A.P.$ is $2$ and the common difference is $4$. The sum of its $40$ terms will be:

Which number should be added to the numbers $13, 15, 19$ so that the resulting numbers are the consecutive terms of a $H.P.$?

The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive,then its $7^{th}$ term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module