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If ${A_1},\,{A_2}$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{{24}}$ , then their values are
$\frac{7}{{72}},\,\frac{5}{{36}}$
$\frac{{17}}{{72}},\,\frac{5}{{36}}$
$\frac{7}{{36}},\,\frac{5}{{72}}$
$\frac{5}{{72}},\,\frac{{17}}{{72}}$
Solution
(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 $n = 2,\;a = \frac{1}{3},\;b = \frac{1}{{24}}$
$\therefore $ ${A_1} = \frac{1}{3} + \frac{{ – 7/24}}{3} = \frac{{17}}{{72}}$
${A_2} = \frac{1}{3} + \frac{{ – 14/24}}{3} = \frac{{10}}{{72}} = \frac{5}{{36}}$
