We have, f(x)=x3+3x−9 f′(x)=3x2+3>0‌∀x∈R ∴‌‌‌f(−2)=(−2)3+3(−2)−9 =−8−6−9=−23 f(3)=(3)3+3(3)−9 =27+9−9=27 ∴‌‌‌‌fmax=f(3)=27 f′(0)=3 Let the GP is a,ar,ar2,..,|r|<1 S∞=
a
1−r
27=
a
1−r
  [∵‌S∞=fmax=27]
and a−ar=3  [∵‌f′(0)=3] a(1−r)=3⇒a=