If {A_1},,{A_2} be two arithmetic means betwe | vedclass

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

Question

(b) Here $\frac{1}{3},\;{A_1},\;{A_2},\;\frac{1}{{24}}$ will be in $A.P.,$

then ${A_1} - \frac{1}{3} = \frac{1}{{24}} - {A_2}$

$ \Rightarrow $${

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Here $\frac{1}{3},\;{A_1},\;{A_2},\;\frac{1}{{24}}$ will be in $A.P.,$

then ${A_1} - \frac{1}{3} = \frac{1}{{24}} - {A_2}$

$ \Rightarrow $${A_1} + {A_2} = \frac{3}{8}$......$(i)$

Now, ${A_1}$ is a arithmetic mean of $\frac{1}{3}$ and ${A_2}$, we have

$2{A_1} = \frac{1}{3} + {A_2} \Rightarrow 2{A_1} - {A_2} = \frac{1}{3}$ ......$(ii)$

From $(i)$ and $(ii),$ we get, ${A_1} = \frac{{17}}{{72}}$ and ${A_2} = \frac{5}{{36}}$.

Aliter : As we have formula ${A_m} = a + \frac{{m(b - a)}}{{n + 1}}$

where $n = 2,\;a = \frac{1}{3},\;b = \frac{1}{{24}}$

$	herefore $ ${A_1} = \frac{1}{3} + \frac{{ - 7/24}}{3} = \frac{{17}}{{72}}$

${A_2} = \frac{1}{3} + \frac{{ - 14/24}}{3} = \frac{{10}}{{72}} = \frac{5}{{36}}$

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

Similar Questions

The number of terms common between the two series $2 + 5 + 8 +.....$ upto $50$ terms and the series $3 + 5 + 7 + 9.....$ upto $60$ terms, is

If the sum of the $10$ terms of an $A.P.$ is $4$ times to the sum of its $5$ terms, then the ratio of first term and common difference is

Which of the following sequence is an arithmetic sequence

Given that $n$ A.M.'s are inserted between two sets of numbers $a,\;2b$and $2a,\;b$, where $a,\;b \in R$. Suppose further that ${m^{th}}$ mean between these sets of numbers is same, then the ratio $a:b$ equals

The arithmetic mean of the nine numbers in the given set $\{9,99,999,...., 999999999\}$ is a $9$ digit number $N$, all whose digits are distinct. The number $N$ does not contain the digit