If sum of four numbers in A.P. is 28 and product of two middle terms is 45, then product of the first and last terms is
11
13
15
17
Let four numbers in A.P. be
a−3d, a−d, a+d, a+3d
We are given
28=(a−3d)+(a−d)+(a+d)+(a+3d)=4a⇒ a=7
Also, (a−d)(a+d)=45
⇒ 49−d2=45⇒d=±2.
Now, (a−3d)(a+3d)=a2−9d2=13