Question

If $a,b,c,$ are in A.P., then prove that:
$\left(ii\right)$ $a^2+c^2+4ac=2(ab+bc+ca)$

Answer 1

Given : $a,b,c,$ are in A.P.
By the defination of A.P. difference of two consecative terms is same.
$b-a=c-b$
$\Rightarrow 2b=a+c$
$\Rightarrow b=\frac{a+c}{2}\dots \left(i\right)$
Now, we have to prove: $a^2+c^2+4ac=2(ab+bc+ca)$
R.H.S=$2(ab+bc+ca)=2[a\left(\frac{a+c}{2}\right)+c\left(\frac{a+c}{2}\right)+ac]=2\left[\frac{a^2+ac+ac+c^2+2ac}{2}\right]$
$=a^2+4ac+c^2$
$=$L.H.S
Hence, proved