Question

answer please 12 the zeroes are in ap

Answer 1

Make the roots as a-d, a, a+d
Then by property of sum and product of roots simplify the equations
For the answer the easiest way is to put the values you get..

Answer 2

solution:
given: the zeroes of the polynomial are in AP,
f(x) = x³-3px²+qx-r,

let the zeroes be, a-d,a,a+d

sum of the zeores= -b/a
     (a-d)+(a)+(a+d)= -(-3p)/1
                          3a  =3p
                              a=p....(i)
 
(a-d)(a)+(a+d)(a)+(a-d)(a+d)= c/a
a²-ad+a²+ad+a²-d²=q/1
                      3a²-d²=q.....(ii)

product of the zeores = -d/a
                (a-d)(a)(a+d)=-(-r)/1
                      (a²-d²)(a)=r
                       a³-ad²=r........(iii)

LHS=>2p³=2(a)³
RHS=>pq-r
         =>a(3a²-d²)- (a³-ad²)
         =>3a³-ad²-a³+ad²
         ⇒2a³=LHS hence, option (a) is correct.

                    Hope it Helps