Question

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

• A. $\frac{7}{72},\frac{5}{36}$
• B. $\frac{17}{72},\frac{5}{36}$
• C. $\frac{7}{36},\frac{5}{72}$
• D. $\frac{5}{72},\frac{17}{72}$

Answer 1

The correct option is B $\frac{17}{72},\frac{5}{36}$

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 $2A_1=\frac{1}{3}+A_2\Rightarrow 2A_1-A_2=\frac{1}{3}$ ......(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, $a=\frac{1}{3},b=\frac{1}{24}$
$\therefore A_1=\frac{1}{3}+\frac{-7/24}{3}=\frac{17}{72}$
$A_2=\frac{1}{3}+\frac{-14/24}{3}=\frac{10}{72}=\frac{5}{36}$