Let the quadratic polynomial be $f(x)=px^2+qx+r.$

Use the condition $f(1)=f(-1).$

Compute: $f(1)=p+q+r,$ $f(-1)=p-q+r.$ Since these are equal, $p+q+r=p-q+r \,\Rightarrow\, 2q=0 \Rightarrow q=0.$

Hence the polynomial becomes $f(x)=px^2+r.$

Differentiate: $f'(x)=2px.$

Since $a,b,c$ are in A.P., we have $2b=a+c.$

Now evaluate the derivatives: $f'(a)=2pa, \quad f'(b)=2pb, \quad f'(c)=2pc.$

Check whether these are in A.P.: $f'(a)+f'(c)=2pa+2pc=2p(a+c)=2p(2b)=4pb=2f'(b).$ Therefore, $f'(a), f'(b), f'(c)$ are in arithmetic progression.

Hence the correct option is $\boxed{\text{B: A.P.}}$