(c) Given that first term of AP=p ⇒a=p Where, a denotes first term. and a3=q,a5=3 ⇒a+2d=q a+4d=3 pq=a(a+2d) =(3−4d)(3−4d+2d) =(3−4d)(3−2d) =9−18d+8d2 f=9−18d+8d2 f′=0−18+16d =−18+16d For maxima and minima f′=0 ⇒−18+16d=0 ⇒d=
18
16
=
9
8
Now, f′′=16 (Positive) So, f will be maximum at d=