Question

Prove that if `a,b` and `c` are in A.P., `b^2-ac>0`, if they are in G.P. then `b^2-ac=0` and if they are in H.P. then `b^2-ac<0`?

Answer 1

Please see below.

Explanation:

We already know that if `a,b`and `c` are in G.P., we have `b/a=c/b` or `b^2=ac`

Now let us work out for other two relations

When `a,b` and `c` are in A.P., we have `b-a=c-b=d` where `d` is common difference.

Then we can write `b^2-ac` as

= `(a+c)^2/4-ac`

= `(a^2+c^2+2ac-4ac)/4`

= `(a^2+c^2-2ac)/4`

= `(a-c)^2/4`

As `(a-c)^2/4` is always positive `b^2-ac>0` i.e. `b^2>ac`

When `a,b` and `c` are in H.P., we have `1/a,1/b` and `1/c` are in A.P. and

`2/b=1/a+1/c=(a+c)/(ac)` or `b=(2ac)/(a+c)`

and `b^2-ac=(4a^2c^2)/(a+c)^2-ac`

= `(4a^2c^2-ac(a+c)^2)/(a+c)^2`

= `(ac(4ac-(a+c)^2))/(a+c)^2`

= `(-ac((a+c)^2-4ac))/(a+c)^2`

= `(-ac(a-c)^2)/(a+c)^2`

Hence `b^2-ac<0`