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


If abc are in A.P., prove that:

(i) (a − c)2 = 4 (a − b) (b − c)

(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)

(iii) a3 + c3 + 6abc = 8b3.


Since a, b, c are in A.P., we have:

2b = a+c

$\Rightarrow b=\frac{a+c}{2}$

(i) Consider RHS:

4 (a − b) (b − c)

Substituting $b=\frac{a+c}{2}:$

$\Rightarrow 4\left\{a-\frac{a+c}{2}\right\}\left\{\frac{a+c}{2}-c\right\}$

$\Rightarrow 4\left\{\frac{2 a-a-c}{2}\right\}\left\{\frac{a+c-2 c}{2}\right\}$



Hence, proved.

(ii) Consider RHS:

2 (ab + bc + ca)

Substituting $b=\frac{a+c}{2}$ :

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

$\Rightarrow 2\left\{\frac{a^{2}+a c+a c+c^{2}+2 a c}{2}\right\}$

$\Rightarrow a^{2}+4 a c+c^{2}$

Hence, proved.

(iii) Consider RHS:

$8 b^{3}$

Substituting $b=\frac{a+c}{2}$ 

$\Rightarrow 8\left(\frac{a+c}{2}\right)^{3}$

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

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

$\Rightarrow a^{3}+c^{3}+6 a b c$

Hence, proved.

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