Three numbers are in Arithmetic progression (AP) whose sum is 30 and the product is 910. Then the greatest number in the AP is:

Solution (By Examveda Team)

Let the three number is a - d, a, a + d
a is first term, d is common difference
a + d + a + a - d = 30     (Given)
3a = 30
a = 10
(a + d)(a)(a - d) = 910
(10 + d)(10)(10 - d) = 91 × 10
(10 + d)(10 - d) = 91
Put d = 3
$$\boxed{\left( {10 + 3} \right)\left( {10 - 3} \right) = 91}$$
So, d = 3
So, greater number is = a + b = 10 + 3 = 13