If a, b, c are in AP, show that

Solution:

To prove $\frac{a(b+c)}{b c}, \frac{b(c+a)}{c a}, \frac{c(a+b)}{a b}$ are in A.P.

Given: a, b, c are in A.P.

Proof: a, b, c are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with (ab + bc + ac)

$\Rightarrow(a)(a b+b c+a c)$, (b) $(a b+b c+a c)$, (c) $(a b+b c+a c)$, are in A.P.

Multiplying the A.P. with $\left(\frac{1}{a b c}\right)$

$\Rightarrow\left[\frac{(a)(a b+b c+a c)}{a b c}\right],\left[\frac{(b)(a b+b c+a c)}{a b c}\right],\left[\frac{(c)(a b+b c+a c)}{a b c}\right]$, are in A.P.

$\Rightarrow\left[\frac{(a b+b c+a c)}{b c}\right],\left[\frac{(a b+b c+a c)}{a c}\right],\left[\frac{(a b+b c+a c)}{a b}\right]$, are in A.P.

If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

Substracting the A.P. with 1

$\Rightarrow\left[\frac{(a b+b c+a c)}{b c}-1\right],\left[\frac{(a b+b c+a c)}{a c}-1\right],\left[\frac{(a b+b c+a c)}{a b}-1\right]$, are in A.P.

$\Rightarrow\left[\frac{(a b+a c)}{b c}\right],\left[\frac{(a b+b c)}{a c}\right],\left[\frac{(b c+a c)}{a b}\right]$, are in A.P.

On rearranging

$\Rightarrow\left[\frac{a(b+c)}{b c}\right],\left[\frac{b(c+a)}{a c}\right],\left[\frac{c(a+b)}{a b}\right]$, are in A.P.

Hence Proved