Question

if abc are in A.P show a(b+c)/bc,b(c+a)/ca,c(a+b)/ab are in a.p

Answer 1

hope this helps you☺️

Answer 2

Answer:

Given :

a, b, c are in AP,

To prove :

$\frac{a(b+c)}{bc}, \frac{b(a+c)}{ac}, \frac{c(b+a)}{ab}\text{ are in AP}$

Proof :

Since, if we operate each term of the AP by the same number then the resultant series would be also an AP,

⇒ $\frac{a}{abc}, \frac{b}{abc}, \frac{c}{abc}\text{ are in AP}$

⇒ $\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}\text{ are in AP}$

⇒ $\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ac}, \frac{ab+bc+ca}{ab}\text{ are in AP}$

⇒ $\frac{ab+bc+ca}{bc}-1, \frac{ab+bc+ca}{ac}-1, \frac{ab+bc+ca}{ab}-1\text{ are in AP}$

⇒ $\frac{ab+bc+ca-bc}{bc}, \frac{ab+bc+ca-ac}{ac}, \frac{ab+bc+ca-ab}{ab}\text{ are in AP}$

⇒ $\frac{ab+ca}{bc}, \frac{ab+bc}{ac}, \frac{bc+ca}{ab}\text{ are in AP}$

⇒ $\frac{a(b+c)}{bc}, \frac{b(a+c)}{ac}, \frac{c(b+a)}{ab}\text{ are in AP}$

Hence proved.....