Question

Q.6.. The sum of three numbers in an
Arithmetic Progression is 45 and their
product is 3000. What are the three
numbers?
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Answer 1

let the three consecutive numbers are (a-d), a, (a+d).........(1)

their sum

(a-d) + a + (a+d) =45

a-d + a+a+d = 45

3a = 45

a= 15

their product

a(a-d)(a+d)= 3000

a(a²-d²)=3000

15(15²-d²) = 3000

15²-d² = 3000/15 = 200

225 -d² = 200

d² = 225-200

d² = 25

d = ±5

Hence the number are

when d = -5,

put in (1)

15-(-5),15,(15-5)

20,15,10 are numbers

when d=5

15-5 , 15, 15+5

10,15,20 are three numbers