If a, b, c are in G.P., then prove that:
Question:

If $a, b, c$ are in G.P., then prove that: $\frac{a^{2}+a b+b^{2}}{b c+c a+a b}=\frac{b+a}{c+b}$

Solution:

$a, b$ and $c$ are in G. P.

$\therefore b^{2}=a c \quad \ldots \ldots \ldots(\mathrm{i})$

Now, LHS $=\frac{a^{2}+a b+b^{2}}{b c+c a+a b}$

$=\frac{\mathrm{a}^{2}+\mathrm{ab}+\mathrm{ac}}{\mathrm{bc}+\mathrm{b}^{2}+\mathrm{ab}}$      [Using (i)]

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

$=\frac{a}{b}$

$=\frac{1}{r}$

Here, $r=$ common ratio

$\mathrm{RHS}=\frac{b+a}{c+b}$

$=\frac{a r+a}{a r^{2}+a r}$

$=\frac{a(r+1)}{a r(r+1)}$

$=\frac{1}{r}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

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