Let a,b,c,d be four numbers of the sequence. Now, according to the question b2=ac and c−b=6 and a −c=6 Also, given a=d ∴b2=ac⇒b2=a[
a+b
2
](∵2c=a+b) ⇒a2−2b2+ab=0 Now, c−b=6 and a−c=6 gives a−b=12 ⇒b=a−12 ∴a2−2b2+ab=0 ⇒a2−2(a−12)2+a(a−12)=0 ⇒a2−2a2−288+48a+a2−12a=0 ⇒36a=288⇒a=8 Hence, last term is d=a=8.