Question : If 1/b+c, 1/a+c, 1/a+b are in AP, prove that a², b², c² are also in AP.
Doubt by Divya
Solution :
1/b+c, 1/a+c, 1/a+b
d1=d2 (Given)
a2-a1=a3-a2
1/a+c-1/b+c = 1/a+b-1/a+c
(b+c)-(a+c)/(a+c)(b+c)=(a+c)-(a+b)/(a+c)(a+b)
(b+c-a-c)/(b+c)= (a+c-a-b)/(a+b)
(b-a)/(b+c) = (c-b)/(a+b)
(b-a)(a+b)=(c-b)(b+c)
ab+b²-a²-ab=bc+c²-b²-bc
b²-a²=c²-b²
d1'=d2'
Hence, a², b², c² are also in AP
d1=d2 (Given)
a2-a1=a3-a2
1/a+c-1/b+c = 1/a+b-1/a+c
(b+c)-(a+c)/(a+c)(b+c)=(a+c)-(a+b)/(a+c)(a+b)
(b+c-a-c)/(b+c)= (a+c-a-b)/(a+b)
(b-a)/(b+c) = (c-b)/(a+b)
(b-a)(a+b)=(c-b)(b+c)
ab+b²-a²-ab=bc+c²-b²-bc
b²-a²=c²-b²
d1'=d2'
Hence, a², b², c² are also in AP