If a, b, c are in AP, x is the GM between a and b; y is the GM between
Question:

If a, b, c are in AP, x is the GM between a and b; y is the GM between b and c; then show that $b^{2}$ is the AM between $x^{2}$ and $y^{2}$.

 

Solution:

To prove: $b^{2}$ is the AM between $x^{2}$ and $y^{2}$.

Given: (i) a, b, c are in AP

(ii) $x$ is the GM between $a$ and $b$

(iii) $y$ is the GM between $b$ and $c$

Formula used: (i) Arithmetic mean between a and $b=\frac{a+b}{2}$

(ii) Geometric mean between $a$ and $b=\sqrt{a b}$

As a, b, c are in A.P.

⇒ 2b = a + c … (i)

As x is the GM between a and b

$\Rightarrow x=(\sqrt{a b})$

$\Rightarrow x^{2}=a b$

As y is the GM between b and c

$\Rightarrow y=(\sqrt{b c})$

$\Rightarrow \mathrm{y}^{2}=\mathrm{bc} \ldots$

Arithmetic mean of $x^{2}$ and $y^{2}$ is $\left(\frac{x^{2}+y^{2}}{2}\right)$

Substituting the value from (ii) and (iii)

$\left(\frac{x^{2}+y^{2}}{2}\right)=\left(\frac{a b+b c}{2}\right)$

$=\frac{b(a+c)}{2}$

Substituting the value from eqn. (i)

$=\frac{b(2 b)}{2}$

$=b^{2}$

Hence Proved

 

 

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