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 $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

If the sum of the $n$ terms of a $G.P.$ is $S$,the product is $P$,and the sum of their reciprocals is $R$,then ${P^2}$ is equal to:

If the sum of the first $n$ terms of the series $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$ is $435\sqrt{3}$,then $n$ equals

If the ${n^{th}}$ terms of two arithmetic progressions $(A.P.)$ are $3n + 8$ and $7n + 15$,then the ratio of their ${12^{th}}$ terms will be:

If every term of a $G.P.$ with positive terms is the sum of its two previous terms,then the common ratio of the series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module