Question

If a square, b square, c square are in ap then prove that a /b + c, b/c+a,c/a+b are in ap

Answer 1

It is given that a^2 , b^2 and c^2 are in an AP

So they have a common difference

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

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

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

Let;

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

Now for a/(b + c) , b/(c + a) and c/(a+ b)

b/(c+a) - a/(b+c)

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

= b^2 + bc - a^2 - ac / (b+c)(c+a)

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

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

= K(a+b+c) / (c+a)

c/(a+b) - b/(c+a)

= c^2 + ac - ab - b^2 /

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

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

= K(a+b+c) / (c+a)

So;

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

Which means that terms a/(b+c) , b/(c+a) and c/(a+b) have a a common difference

Therefore a/(b+c) , b/(c+a) and c/(a+b) are in an A.P

hope it helps you...........

Answer 2

Step-by-step explanation:

plzz mark my answer as brainliest