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

Let $a_1=8, a_2, a_3, \ldots a_n$ be an $A.P.$ If the sum of its first four terms is $50$ and the sum of its last four terms is $170$ , then the product of its middle two terms is

A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{{13}}{{12}}$, then the numbers are

If the numbers $a,\;b,\;c,\;d,\;e$ form an $A.P.$, then the value of $a - 4b + 6c - 4d + e$ is

If $f(x + y,x - y) = xy\,,$ then the arithmetic mean of $f(x,y)$ and $f(y,x)$ is

Write the first three terms in each of the following sequences defined by the following:

$a_{n}=\frac{n-3}{4}$