Question

three numbers ate in A.P. if their sum is 27 and their product is 648, find the numbers.

Answer 1

Let three numbers are (a - r), a , (a + r)

A/C to question,
(a - r) + a + (a + r) = 27
3a = 27
a = 9

also,
(a - r)a(a + r ) = 648
a³ - ar² = 648
(9)³ - 9r² = 648
729 - 9r² = 648
9r² = 729 - 648 = 81
r² = 9 => r = ±3

hence,
three numbers are : 6, 9, 12 or, 12 , 9, 6

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.