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SEQUENCES AND SERIES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSEQUENCES AND SERIES1 Mark
MCQ
If a, b, c are in G.P. is 2 and x, y are AM's between a, b and b, c respectively, then:
  • A
    $\frac{1}{\text{x}}+\frac{1}{\text{y}}=2$
  • B
    $\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{2}$
  • C
    $\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{a}}$
  • ✓
    $\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$
✓

Answer

Correct option: D.
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$
a, b and c are in G.P.
$\therefore\text{b}^2=\text{ac}\cdots(\text{i})$
a, x and b are in A.P.
$\therefore2\text{x}=\text{a}+\text{b}\ \cdots(\text{ii})$
Also, b, y and c are in A.P.
$\therefore2\text{y}=\text{b}+\text{c}$
$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}^2}{\text{a}}$ [Using (i)]
$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}^2}{(2\text{x}-\text{b})}$ [Using (ii)]
$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}(2\text{x}-\text{b})+\text{b}^2}{(2\text{x}-\text{b})}$
$\Rightarrow2\text{y}=\frac{2\text{bx}-\text{b}^2+\text{b}^2}{(2\text{x}-\text{b})}$
$\Rightarrow2\text{y}=\frac{2\text{b}\text{x}}{(2\text{x}-\text{b})}$
$\Rightarrow\text{y}=\frac{\text{bx}}{(2\text{x}-\text{b})}$
$\Rightarrow\text{y}(2\text{x}-\text{b})=\text{bx}$
$\Rightarrow2\text{xy}-\text{by}=\text{bx}$
$\Rightarrow\text{bx}+\text{by}=2\text{xy}$
Dividing both the sides by xy:
$\Rightarrow\frac{1}{\text{y}}+\frac{1}{\text{x}}=\frac{2}{\text{b}}$

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