Question

suppose 1/a^2 - 1/b^2 =1/c, ab= underroot c,c>0 and a<b. then the average value of a and b is?​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• $\sf\dfrac{1}{a^2}-\dfrac{1}{b^2}=\dfrac{1}{c}$
• where ab = √c
• c>0 and a<b

${\sf{\green{\underline{\large{To\:find}}}}}$

• Average value of a and b

${\sf{\pink{\underline{\Large{Explanation}}}}}$

$\sf\dfrac{1}{a^2}-\dfrac{1}{b^2}=\dfrac{1}{c}$

Taking LCM ab

$\implies\sf\dfrac{b^2-a^2}{a^2b^2}=\dfrac{1}{c}$

Here

ab=√c (both side squaring)

=>a²b² = c

¶utting the value of a²b

$\implies\sf\dfrac{b^2-a^2}{a^2b^2}=\dfrac{1}{c}$

$\implies\sf\dfrac{b^2-a^2}{c}=\dfrac{1}{c}$

$\implies\sf{b^2-a^2}={1}$

$\implies\sf{b-a}{b+a}={1}$

$\implies\sf{b+a}=\dfrac{1}{b-a}$

Here average of two no is

$\tt{Average=}\frac{a+b}{2}$

$\implies\sf{b+a}=\dfrac{1}{b-a}$

dividing by 2 in both side

$\implies\sf\dfrac{b+a}{2}=\dfrac{1}{2(b-a)}$

Hence average of a and b is

$\sf\dfrac{b+a}{2}=\dfrac{1}{2(b-a)}$

Answer 2

$\bigstar$ Given :

• $\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{c}$

• $ab=\sqrt{c}$

• c > 0 & a < b.

$\bigstar$ To Find :

• Average value of a and b
• i.e.,$\frac{(a+b)}{2}$

$\bigstar$ Explanation :

$\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{c}\\\\ \implies \frac{b^2-a^2}{a^2.b^2} =\frac{1}{c}\\\\ \implies \frac{b^2-a^2}{(ab)^2} =\frac{1}{c}\\\\\implies \frac{b^2-a^2}{(\sqrt{c} )^2} =\frac{1}{c} \;\;[Since,\;ab=\sqrt{c} ]\\\\\implies b^2-a^2=1\\\\\implies (b+a)(b-a)=1\\\\\implies (a+b)=\frac{1}{b-a}\\\\ \implies \frac{a+b}{2}=\frac{1}{2(b-a)}\;\;[Dividing\;\;both\;sides\;by\;2]\\\\\bold{\implies \frac{a+b}{2}=\frac{1}{2}.\frac{1}{b-a} }$