If a, b, c are in A.P., then show that:

Since $a, b, c$ are in $A$. P., we have:

$2 b=a+c$

(i) We have to prove the following:

$2 b^{2}(a+c)=a^{2}(b+c)+c^{2}(a+b)$

LHS : $2 b^{2} \times 2 b \quad$ (Given)

$=4 b^{3}$

RHS : $a^{2} b+a^{2} c+a c^{2}+c^{2} b$

$=a c(a+c)+b\left(a^{2}+c^{2}\right)$

$=a c(a+c)+b\left[(a+c)^{2}-2 a c\right]$

$=a c(2 b)+b\left[(2 b)^{2}-2 a c\right]$

$=2 a b c+4 b^{3}-2 a b c$

$=4 b^{3}$

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

hence proved.

(ii) We have to prove the following:

$2(c+a-b)=(b+c-a)+(a+b-c)$

$\operatorname{LHS}: 2(c+a-b)$

$=2(2 b-b) \quad(\because 2 b=a+c)$

$=2 b$

RHS : $(b+c-a)+(a+b-c)$

$=2 b$

LHS = RHS

Hence, proved.

(iii) We have to prove the following:

$2\left(c a-b^{2}\right)=\left(b c-a^{2}+a b-c^{2}\right)$

RHS : $b c-a^{2}+a b-c^{2}$

$=c(b-c)+a(b-a)$

$=c\left(\frac{a+c}{2}-c\right)+a\left(\frac{a+c}{2}-a\right) \quad(\because 2 b=a+c)$

$=c\left(\frac{a+c-2 c}{2}\right)+a\left(\frac{a+c-2 a}{2}\right)$

$=\frac{c(a-c)}{2}+a\left(\frac{c-a}{2}\right)$

$=\frac{c a}{2}-\frac{c^{2}}{2}+\frac{a c}{2}-\frac{a^{2}}{2}$

$=a c-\frac{1}{2}\left(c^{2}+a^{2}\right)$

$=a c-\frac{1}{2}\left(4 b^{2}-2 a c\right) \quad\left(\because a^{2}+c^{2}+2 a c=4 b^{2} \Rightarrow a^{2}+c^{2}=4 b^{2}-2 a c\right)$

$=a c-2 b^{2}+a c$

$=2 a c-2 b^{2}$

$=2\left(a c-b^{2}\right)$

$=\mathrm{LHS}$

Hence proved.