Question

If a,b and c are in ap then prove that a(b+c)/bc, b(c+a)/ca and c(b+a)/ab

Answer 1

Answer:

Hence, $\dfrac{a(b+c)}{bc} , \dfrac{b(c+a)}{ca} \ and \ \dfrac{c(b+a)}{ab}$ are in A.P.

Step-by-step explanation:

Given that-

a, b and c are in A.P. then-

$b = \dfrac{a+c}{2}$

2b = (a + c)    --------------- 1

$\dfrac{a(b + c)}{bc} , \dfrac{b(c + a)}{ca} \ and \ \dfrac{c(b + a)}{ab}$ will be in AP if -

$\dfrac{2b(c + a)}{ca} = \dfrac{a(b + c)}{bc}+\dfrac{c(a+b)}{ab}$

L.H.S-

We have-

= $\dfrac{2b(c+a)}{ca}$

But a + c = 2b

= $\dfrac{2b \times 2b}{ca}$

L.H.S. = $\dfrac{4b^{2} }{ca}$

R.H.S = $\dfrac{a(b+c)}{bc}+\dfrac{c(a+b)}{ab}$

= $\dfrac{ab+ac}{bc}+\dfrac{ac+bc}{ab}$

= $\dfrac{a(ab+ac)+c(ac+bc)}{abc}$

= $\dfrac{ab^{2}+a^{2}c+ac^{2}+bc^{2}}{abc}$

= $\dfrac{ac(a+c) + b(a^{2}+c^{2})}{abc}$

= $\dfrac{ac(2b) + b(a^{2}+c^{2}) }{abc}$

= $\dfrac{2abc + ba^{2}+bc^{2}}{abc}$

= $\dfrac{b(2ac + a^{2}+c^{2})  }{abc}$

= $\dfrac{a^{2}+c^{2}+2ac  }{ac}$

But $a^{2}+c^{2}+2ac = (a+c)^{2}$

= $\dfrac{(a+c)^{2} }{ac}$

But  (a + c) = 2b

= $\dfrac{(2b)^{2} }{ac}$

= $\dfrac{4b^{2}}{ac}$

R.H.S = $\dfrac{4b^{2}}{ac}$

So, L.H.S. = R.H.S.

Hence, $\dfrac{a(b+c)}{bc} , \dfrac{b(c+a)}{ca} \ and \ \dfrac{c(b+a)}{ab}$ are in A.P.