Question

if a squarev,b square, c squeare are in ap then show thatc1/b+c, 1/c+A, 1/a+b are also in ap​

Answer 1

Answer:

Step-by-step explanation:

Question :

a² ,b², c²  are in AP

To Show

$\sf\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}$ are also in AP.

Solution:

We have  a² ,b², c²  are in AP

Thus , 2b²=a²+c²  ....(1)

Now if $\sf\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}$ are  in AP.

$\sf\implies\dfrac{2}{c+a}=\dfrac{1}{a+b}+\dfrac{1}{b+c}$

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

$\sf\implies\:2(ab+ac+b^2+bc)=(c+a)(a+2b+c)$

$\sf\implies\:2ab+2ac+2b^2+2bc=ac+2bc+c^2+a^2+2ab+ac$

From equation (1) [ 2b²=a²+c²]

$\sf\implies\:2ab+2ac+2b^2+2bc=2ac+2ab+2b^2+2bc$

Hence , Proved

$\sf\implies\dfrac{2}{c+a}=\dfrac{1}{a+b}+\dfrac{1}{b+c}$ are also in AP

Answer 2

Given :

a² ,b², c² are in AP

To Prove :

$\sf\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}$ are also in AP.

Solution:

$\mathfrak\red{We\:Have}\begin{cases}\sf\pink{\underline{a^2,b^2,c^2\:are\:in\:AP}}\end{cases}$

a² ,b², c² are in AP

⇒ Common difference , d

⇒c² -b² =b²- a²

⇒(c+b)(c-b)=(b+a)(b-a)......(1)

Now if;

$\sf\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}$

are in AP

Then , Common difference

$\sf\implies\dfrac{1}{a+b}=\dfrac{1}{c+a}=\dfrac{1}{c+a}-\dfrac{1}{b+c}$

$\sf\implies\dfrac{c+a-(a+b)}{(a+b)(a+c)}=\dfrac{b+c-(c+a)}{(a+c)(b+c)}$

$\sf\implies\:(c-b)(c+b)(a+c)=(b-a)(b+a)(a+c)$

$\sf\implies\:(c-b)(c+b)=(b-a)(b+a)$

This is true [ From equation (1)]

Hence,

$\sf\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}$ are also in AP.