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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
If ${A_1},\,{A_2}$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{{24}}$ , then their values are
  • A
    $\frac{7}{{72}},\,\frac{5}{{36}}$
  • ✓
    $\frac{{17}}{{72}},\,\frac{5}{{36}}$
  • C
    $\frac{7}{{36}},\,\frac{5}{{72}}$
  • D
    $\frac{5}{{72}},\,\frac{{17}}{{72}}$
✓

Answer

Correct option: B.
$\frac{{17}}{{72}},\,\frac{5}{{36}}$
b
(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 $ ${A_1} = \frac{1}{3} + \frac{{ - 7/24}}{3} = \frac{{17}}{{72}}$

${A_2} = \frac{1}{3} + \frac{{ - 14/24}}{3} = \frac{{10}}{{72}} = \frac{5}{{36}}$

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