Question

The sum of three consecutive numbers in A.P. is 27,and their product is 504,find them.

Answer 1

Hey

Here is your answer,

Let the three numbers be a-d , a ,a+d

a-d+a+a+d=27
3a=27
a=27/3
a=9

(a-d)a(a+d)=504
(a^2-d^2)a=504
(9^2-d^2)9=504
81-d^2 = 504/9
81-d^2= 56
-d^2 = 56-81
-d^2 = -25
d=+/-5

If d=+5
Terms = 4,9,14

If d=-5
Terms=14,9,4

Hope it helps you!

Answer 2

Answer:

Given: Sum of first three terms is 27

Let us assume the first three terms as

a – d, a, a + d [where a is the first term and d is the common difference]

So,

sum of first three terms is a – d + a + a + d = 27

3a = 27

a = 9

It is given that the product of three terms is 648

So,

a³ – ad²= 648

Substituting the value of a = 9,

we get 9³– 9d²= 648

729 – 9d² = 648

81 = 9d²

d = 3 or d = – 3

Hence, the given terms are a – d, a, a + d which is 6, 9, 12.