Question

if B minus C whole square, C minus a whole square A minus b whole square are in AP then prove that one by b minus C, 1 by C minus A One by a minus b are also in AP

Answer 1

Solution :

Given (b-c)², (c-a)²,(a-b)² are in A.P

(c-a)² - (b-c)² = (a-b)² - (c-a)²

=> [(c-a)+(b-c)][(c-a)-(b-c)]

= [(a-b)+(c-a)][(a-b)-(c-a)]

=> [-(a-b)[(c-a)-(b-c)] = [-(b-c)][(a-b)-(c-a)]

=> - (a-b)(c-a)+(a-b)(b-c)

= -(b-c)(a-b) + (b-c)(c-a)

=> 2(a-b)(b-c) = (b-c)(c-a) + (b-c)(a-b)

Divide both sides with (a-b)(b-c)(c-a),

we get

=> 2/(c-a) = 1/(a-b) + 1/(c-a)

Therefore ,

1/(b-c) , 1/(c-a) , 1/(a-b) are in A.P

••••

=>

Answer 2

Answer:

${(b-c)}^2, {(c-a)}^2, {(a-b)}^2 are\: in\: AP$

Then, ${t_2}-{}t_1}={t_3}-{t_2}$

${(c-a)}^2-{(b-c)}^2={(a-b)}^2-{(c-a)}^2$

(c-a+b-c)(c-a-b+c)=(a-b+c-a)(a-b-c+a)

-(a-b)[(c-a)-(b-c)]=-(b-c)[(a-b)-(c-a)]

(a-b)(b-c)-(a-b)(c-a)=(b-c)(c-a)-(a-b)(b-c)

divide both sides by (a-b)(b-c)(c-a)

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

This implies

$\frac{1}{(b-c)}, \frac{1}{(c-a)}, \frac{1}{(a-b)} \:are\: in\: A.P$