If 8 = 3 + frac{1}{4}(3 + p) + frac{1}{4^2}(3 | vedclass

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

Question

(A) The given series is an Arithmetico-Geometric Progression $(AGP)$ of the form $\sum_{n=0}^{\infty} (a + np)r^n$,where $a = 3$,$d = p$,and $r = \fra

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The given series is an Arithmetico-Geometric Progression $(AGP)$ of the form $\sum_{n=0}^{\infty} (a + np)r^n$,where $a = 3$,$d = p$,and $r = \frac{1}{4}$.The sum of an infinite $AGP$ is given by $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.Substituting the values into the formula:$8 = \frac{3}{1 - \frac{1}{4}} + \frac{p \cdot \frac{1}{4}}{(1 - \frac{1}{4})^2}$$8 = \frac{3}{\frac{3}{4}} + \frac{\frac{p}{4}}{(\frac{3}{4})^2}$$8 = 4 + \frac{\frac{p}{4}}{\frac{9}{16}}$$8 - 4 = \frac{p}{4} \cdot \frac{16}{9}$$4 = \frac{4p}{9}$$p = 9$.

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

Similar Questions

The sum of infinite terms of the following series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

For a certain function $u_{x}$,given that $u_{0}=3, u_{1}=12, u_{2}=81, u_{3}=200, u_{4}=100, u_{5}=8$,then $\Delta^{5} u_{x}$ is equal to

If $u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51$,then the value of $\Delta^{3} u_{0}$ is

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$ is equal to

If $7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \frac{1}{7^3}(5 + 3\alpha) + \dots \infty$,then the value of $\alpha$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module