If (a – b), (b – c), (c – a) are in GP then prove that

To prove: $(a+b+c)^{2}=3(a b+b c+c a)$.

Given: $(a-b),(b-c),(c-a)$ are in GP

Formula used: When $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}, \mathrm{b}^{2}=\mathrm{ac}$

As, $(a-b),(b-c),(c-a)$ are in GP

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

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

$\Rightarrow a^{2}+b^{2}+c^{2}-b c-a c-a b=0$

Adding 3(ab + bc + ac) both side

$\Rightarrow a^{2}+b^{2}+c^{2}-b c-a c-a b+3(a b+b c+a c)=3(a b+b c+a c)$

$\Rightarrow a^{2}+b^{2}+c^{2}+2 b c+2 a c+2 a b=3(a b+b c+a c)$

$\Rightarrow(a+b+c)^{2}=3(a b+b c+a c)$

Hence Proved