Question

3.If $a^2,b^2,c^2$ are in AP, then prove that

(i) $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in AP.

Answer 1

Step 1. Note the given data:

Given that $a^2,b^2,c^2$ are in AP.

Here first term $t_1=a^2$

Second term $t_2=b^2$

Third term $t_3=c^2$

Step 2. Applying the condition of AP:

The general formula of common differences of an AP is $d=n^{th}term-{\left(n-1\right)}^{th}term$.

So,$d=t_2-t_1=t_3-t_2$

Thus $b^2-a^2=c^2-b^2$….(i)

Step 3. Rewrite the above equation:

From (i) we get

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

$\left(b-a\right)\left(b+a\right)=\left(c-b\right)\left(c+b\right)$

$\frac{b-a}{c+b}=\frac{c-b}{b+a}$

Multiply $\left(c+a\right)$ in the denominator

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

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

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

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

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

Thus $\frac{1}{\left(b+c\right)},\frac{1}{\left(c+a\right)},\frac{1}{\left(b+a\right)}$ are in AP.

Step 4. Checking whether $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$is in AP using the above result:

We get,

$\frac{1}{\left(b+c\right)},\frac{1}{\left(c+a\right)},\frac{1}{\left(b+a\right)}$ are in AP.

Multiply $\left(a+b+c\right)$with the numerator of each term

$\frac{\left(a+b+c\right)}{\left(b+c\right)},\frac{\left(a+b+c\right)}{\left(c+a\right)},\frac{\left(a+b+c\right)}{\left(b+a\right)}$ are in AP.

$\frac{a+\left(b+c\right)}{\left(b+c\right)},\frac{b+\left(a+c\right)}{\left(c+a\right)},\frac{c+\left(a+b\right)}{\left(b+a\right)}$ are in AP.

$\frac{a}{\left(b+c\right)}+\frac{\left(b+c\right)}{\left(b+c\right)},\frac{b}{\left(c+a\right)}+\frac{\left(a+c\right)}{\left(c+a\right)},\frac{c}{\left(b+a\right)}+\frac{\left(a+b\right)}{\left(b+a\right)}$ are in AP.

$\frac{a}{\left(b+c\right)}+1,\frac{b}{\left(c+a\right)}+1,\frac{c}{\left(b+a\right)}+1$ are in AP.

$\frac{a}{\left(b+c\right)},\frac{b}{\left(c+a\right)},\frac{c}{\left(b+a\right)}$ are in AP.

Hence, it proved that $\frac{a}{\left(b+c\right)},\frac{b}{\left(c+a\right)},\frac{c}{\left(b+a\right)}$ are in AP.