Question

If $a,b,c$ are in A.P., then show that:
(i) $a^2(b+c),b^2(c+a),c^2(a+b)$ are also in A.P.
(ii) $b+c-a,c+a-b,a+b-c$ are in A.P.

Answer 1

If $a,b,c$ are in $A.P$ $b=\frac{a+c}{2}$

$\left(a\right)$

$a^2(b+c),b^2(c+a),c^2(a+b)$

$\Rightarrow [\frac{a+c}{2}+c],{\left(\frac{B+C}{2}\right)}^2(a+c),c^2[2+\frac{a+c}{2}]$

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

$\frac{(a+c{)}^3}{2}=\frac{a^2\left(\frac{a+3c}{2}\right)+c^2\left(\frac{3a+c}{2}\right)}{2}$ Hence $a^2(b+c),b^2(c+a),c^2(a+b)$ are in $A.P$

$\left(b\right)$

$b+c-a,c+a-b,a+b-c$

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

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

$\Rightarrow \frac{a+c+2c-2a}{2},\frac{2c+2a-a-c}{2},\frac{2a+a+c-2c}{2}$

$\Rightarrow \frac{3c-a}{2},\frac{a+c}{2},\frac{3a-c}{2}$

$I$ $II$ $III$

$II=\frac{I+III}{2}\Rightarrow \frac{a+c}{2}=\frac{\frac{3c-a}{2}+\frac{3a-c}{2}}{2}=\frac{3c-a+3a-c}{4}$

$\Rightarrow \frac{a+c}{2}=\frac{2a+2c}{4}=\frac{a+c}{2}$

Hence, $b+c-a,c+a-b,a+b-c$ are in $A.P$