Question

If a, b, c 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.

Answer 1

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)
$$\begin{gathered} \mathrm{Substituting}b=\frac{a+c}{2}: \\ \Rightarrow 4\left\{a-\frac{a+c}{2}\right\}\left\{\frac{a+c}{2}-c\right\} \\ \Rightarrow 4\left\{\frac{2a-a-c}{2}\right\}\left\{\frac{a+c-2c}{2}\right\} \\ \Rightarrow \left(a-c\right)\left(a-c\right) \\ \Rightarrow {\left(a-c\right)}^2 \end{gathered}$$
Hence, proved.


(ii) Consider RHS:
2 (ab + bc + ca)
$$\begin{gathered} \mathrm{Substituting}b=\frac{a+c}{2}: \\ \Rightarrow 2\left\{a\left(\frac{a+c}{2}\right)+c\left(\frac{a+c}{2}\right)+ac\right\} \\ \Rightarrow 2\left\{\frac{a^2+ac+ac+c^2+2ac}{2}\right\} \\ \Rightarrow a^2+4ac+c^2 \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$


(iii) Consider RHS:
8$b^3$
$$\begin{gathered} \mathrm{Substituting}b=\frac{a+c}{2}: \\ \Rightarrow 8{\left(\frac{a+c}{2}\right)}^3 \\ \Rightarrow a^3+c^3+3ac\left(a+c\right) \\ \Rightarrow a^3+c^3+3ac\left(2b\right) \\ \Rightarrow a^3+c^3+6abc \end{gathered}$$

Hence, proved.