Solve the following
Question:

If $A_{1}, A_{2}$ be two AM’s and $G_{1}, G_{2}$ be two GM’s between a and $b$, then find the value of $\frac{A_{1}+A_{2}}{G_{1} G_{2}}$.

Solution:

It is given that $A_{1}$ and $A_{2}$ are the A.M.s between $a$ and $b$.

Thus, $a, A_{1}, A_{2}$ and $b$ are in A.P. with common difference d.

Here, $d=\frac{b-a}{3}$

$\therefore A_{1}=a+\frac{b-a}{3}=\frac{2 a+b}{3}$

and $A_{2}=a+\frac{2(b-a)}{3}=\frac{a+2 b}{3}$

It is also given that $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are the G.M.s between $a$ and $b$.

Thus, $a, G_{1}, G_{2}$ and $b$ are in G.P. with common ratio $r$.

Here, $r=\left(\frac{b}{a}\right)^{\frac{1}{3}}$

$\therefore G_{1}=a\left(\frac{b}{a}\right)^{\frac{1}{3}}=b^{\frac{1}{3}} a^{\frac{1}{3}}$

and $G_{2}=a\left[\left(\frac{b}{a}\right)^{\frac{1}{3}}\right]^{2}=b^{\frac{1}{3}} a^{\frac{1}{3}}$

$\Rightarrow \frac{A_{1}+A_{2}}{G_{1} G_{2}}=\frac{\frac{2 a+b}{3}+\frac{a+2 b}{3}}{b^{\frac{1}{3}} a^{\frac{1}{3}} \times b^{\frac{1}{3}} a^{\frac{1}{3}}}=\frac{a+b}{a b}$

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