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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 ${A_1},\;{A_2}$ are the two $A.M.'s$ between two numbers $a$ and $b$ and ${G_1},\;{G_2}$ be two $G.M.'s$ between same two numbers, then $\frac{{{A_1} + {A_2}}}{{{G_1}.{G_2}}} = $
  • ✓
    $\frac{{a + b}}{{ab}}$
  • B
    $\frac{{a + b}}{{2ab}}$
  • C
    $\frac{{2ab}}{{a + b}}$
  • D
    $\frac{{ab}}{{a + b}}$
✓

Answer

Correct option: A.
$\frac{{a + b}}{{ab}}$
a
(a) Given that $a,\;{A_1},\;{A_2},\;b$ are in $A.P.$

Therefore ${A_1} = \frac{{a + {A_2}}}{2},\;{A_2} = \frac{{{A_1} + b}}{2}$

$ \Rightarrow $ ${A_1}+{A_2} = \frac{{a +b+{A_1}+{A_2}}}{2}$

$ \Rightarrow $ $ \frac{{{A_1}+{A_2}}}{2}$ $=\frac{{{a+b}}}{2}$ or ${A_1} + {A_2} = a + b$ …..$(i)$

and $a,\;{G_1},\;{G_2},\;b$ are in $G.P.$

Therefore $G_1^2 = a{G_2},\;G_2^2 = b{G_1}$ …..$(ii)$

$ \Rightarrow G_1^2G_2^2 = ab{G_1}{G_2}$

$\Rightarrow {G_1}{G_2} = ab$

Hence $\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}} = \frac{{a + b}}{{ab}}$

Trick : Let $a = 1,\;b = 2$, then ${A_1} + {A_2} = 1 + 2 = 3$

and ${G_1}\;.\;{G_2} = 2 \times 1 = 2$

$\therefore \;$$\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}} = \frac{3}{2}$,

which is given by $(a).$

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