Question

If the sum of two extreme numbers of an A.P. with four terms is 8 and product of remaining two middle term is 15, then greatest number of the series will be

• A. 7
• B. 5
• C. 9
• D. 11

Answer 1

The correct option is A 7

Let four numbers are $a-3d,a-d,a+d,a+3d.$
Now $(a-3d)+(a+3d)=8\Rightarrow a=4$ and $(a-d)(a+d)=15\Rightarrow a^2-d^2=15\Rightarrow d=1$ Thus required numbers are 1,3,5,7. Hence greatest number is 7.