If b+c, c+a, a + b are in H.P., then prove that a2,
b2, c2 are in A.P.
Answers
Step-by-step explanation:
here a^2, b^2,and c^2 are in ap
therefore b^2-a^2=c^2-b^2
2(b^2)=a^2+c^2....................1
now a^2, b^2,and c^2 are in ap
therefore 1/a^2, 1/b^2, 1/c^2 are in hp
1/(b^2)^2=1/(a^2)^2*1/(c^2)^2
b^4=a^2*c^2 (taking square root on both sides and reciprocle)
b^2=ac...............................2
from 1&2 2(ac)=a^2+c^2
a^2-2ac+c^2=0
(a-c)^2=0
a-c=0
a=c.............................3
from 3&2 b^2=c^2
b=c.........................4
from 3&4 a=b=c
a+b=b+c=c+a
(b+c)^2=(a+b)*(c+a)
and therefore
b+c,c+a,a+b are in h.p
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