Find the following are in AP or not?
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Since a 2, b2, c2 r in ap
b2-a2=c2-a2
B^2=c2------------(1)
In the first subdivision
1/c+a-1/b+c=1/a+b-1/c+a
On solving we get
b-a=cab [from(1)]
a+c=2b---------equation of ap
So it is an ap
In the second subdivision
b/a+c-a/b+c=c/b+a-b/a+c
On cross multiply
We get
b-a=c-b
So it is an ap
Hope it helps you
b2-a2=c2-a2
B^2=c2------------(1)
In the first subdivision
1/c+a-1/b+c=1/a+b-1/c+a
On solving we get
b-a=cab [from(1)]
a+c=2b---------equation of ap
So it is an ap
In the second subdivision
b/a+c-a/b+c=c/b+a-b/a+c
On cross multiply
We get
b-a=c-b
So it is an ap
Hope it helps you
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