Question

The sum of 4 integers in AP is 24 and their product is 945.find the numbers.

Answer 1

Let the integers are a-3d,a-d,a+d and a+3d where d is the common ratio of the given A.P.
∴, a-3d+a-d+a+d+a+3d=24
or, 4a=24
or, a=6
Also, (a-3d)(a-d)(a+d)(a+3d)=945
or, (a²-9d²)(a²-d²)=945
or, (6²-9d²)(6²-d²)=945
or, (36-9d²)(36-d²)=945
or, 9(4-d²)(36-d²)=945
or,144-36d²-4d²+d⁴=945/9
or, d⁴-40d²+144=105
or, d⁴-40d²+144-105=0
or, d⁴-40d²+39=0
or, d⁴-39d²-d²+39=0
or, d²(d²-39)-1(d²-39)=0
or, (d²-39)(d²-1)=0
Either, d²-39=0
or, d²=39
or, d=+-√39
Or, d²-1=0
or, d²=1
or, d=+-1
∵, d is a integer therefore,  d=+-1
When d=1 the numbers are: 6-3,6-1,6+1,6+3
i.e., 3,5,7,9.
When d=-1 the numbers are: 6+3,6+1,6-1,6-3
i.e., 9,7,5,3
∴, The required numbers are 3,5,7,9.