if a^2,b^2 ,c^2 are in ap then prove that a/b+c, b/c+a,c/a+b are also in ap
Answers
ANSWER
We have,
a
2
,b
2
,c
2
are in A.P
⇒b
2
−a
2
=c
2
−b
2
⇒(b−a)(b+a)=(c−b)(c+b)
⇒
(c+b)
(b−a)
=
(b+a)
(c−b)
⇒
(c+b)(c+a)
(b−a)
=
(b+a)(c+a)
(c−b)
⇒
(c+b)(c+a)
(b+c−c−a)
=
(b+a)(c+a)
(c+a−a−b)
⇒
c+a
1
−
b+c
1
=
a+b
1
−
c+a
1
b+c
1
,
c+a
1
,
a+b
1
are in A.P
b+c
a+b+c
,
c+a
a+b+c
,
a+b
a+b+c
are in A.P
b+c
a
+1,
c+a
b
+1,
a+b
c
+1 are in A.P
b+c
a
,
c+a
b
,
a+b
c
are in A.P
Hence, this is the answer.
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