Question

The average of 5 consecutive numbers a b c d and e is 48. what is the product of a and e

Answer 1

Let a be the smallest number out of a,b,c,d and e. Since all of them are consecutive. 

a=a
b= a+1
c= a+2
d= a+3
e= a+4

The average= 48

Avg= a+b+c+d+e/5
48= a+a+1+a+2+a+3+a+4/5  {Substituting values}
48*5= 5a + 10
240-10=5a
230/5= a
a= 46

So, a=46, b= 47, c= 48, d=49, e=50

Product of a and e= 46*50= 2300

Answer 2

Given: 5 consecutive numbers - a,b,c,d,e.
As they are consecutive, the numbers can be represented as-
a, a+1, a+2, a+3,a+4.
Average of these numbers= 48
Average= Sum of numbers/Total count
48= (a+a+1+a+2+a+3+a+4)/5
48= (5a+10)/5
48= a+2
a= 46
Therefore,
e=46+4=50.
Product of a and e = 46×50=2300.