If A_1, A_2 are two arithmetic means between | vedclass

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

Question

(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$. This is an arithmetic progression $(A.P.)$ with $n=4$ terms.The common difference $d$ i

Vedclass · Accepted Answer

$\frac{17}{72}, \frac{5}{36}$

Vedclass · Answer

(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$. This is an arithmetic progression $(A.P.)$ with $n=4$ terms.The common difference $d$ is given by $d = \frac{b-a}{n+1}$,where $a = \frac{1}{3}$,$b = \frac{1}{24}$,and $n=2$ (number of means).$d = \frac{\frac{1}{24} - \frac{1}{3}}{2+1} = \frac{\frac{1-8}{24}}{3} = \frac{-7/24}{3} = -\frac{7}{72}$.Now,$A_1 = a + d = \frac{1}{3} - \frac{7}{72} = \frac{24-7}{72} = \frac{17}{72}$.$A_2 = A_1 + d = \frac{17}{72} - \frac{7}{72} = \frac{10}{72} = \frac{5}{36}$.Thus,the values are $\frac{17}{72}$ and $\frac{5}{36}$.

If $A_1, A_2$ are two arithmetic means between $\frac{1}{3}$ and $\frac{1}{24}$,then their values are

Similar Questions

The sum of the series $(1 + 0.6 + 0.06 + 0.006 + 0.0006 + \dots)$ is

The product $(32) (32)^{1/6} (32)^{1/36} \ldots \infty$ is equal to

The sum $\sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}}$ is equal to

The sum of $24$ terms of the following series $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \dots$ is

If the sum of the series $1 + \frac{2}{x} + \frac{4}{x^2} + \frac{8}{x^3} + \dots \infty$ is a finite number,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module