Question

If $a,b,c$ are in A.P.,prove that:$(a+2b-c)(2b+c-a)(c+a$
$-b)=4abc$
$[Hint:putb=\frac{a+c}{2}onL.H.S.andR.H.S.]$

Answer 1

Given:$a,b,c$ are in A.P
$\therefore a+c=2b...\left(i\right)$
$\Rightarrow b=\frac{a+c}{2}...\left(ii\right)$
Now,taking LHS i.e. $(a+2b-c)(2b+c-a)(c+a-b)$
$\Rightarrow (a+2\times \frac{a+c}{2}-c)(2\times \frac{a+c}{2}+c-a)(c+a-\frac{a+c}{2})$
[from$\left(ii\right)$]
$\Rightarrow (a+a+c-c)(a+c+c-a)\left(\frac{2c+2a-a-c}{2}\right)$
$\Rightarrow \left(2a\right)\left(2c\right)\left(\frac{a+c}{2}\right)$
$\Rightarrow 4abc$
[from$\left(ii\right)$]
$=RHS$
Hence Proved