Question

The sum of the three consecutive number in A.P. is 21 and product is 231? find the nubers

Answer 1

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

According to question
a-d+a+a+d = 21
=> 3a = 21
a = 7

Now,

Product = 231
(a-d) (a) (a+d) = 231
=> (7-d)(7+d) * 7 = 231
=> 49 - d^2 = 33
=> d^2 = 49-33
=> d^2 = 16
d = 4

Required numbers are 3, 7, 11

Answer 2

Let the three terms of an AP is

a - d, a , a + d

Now, according to question

a - d + a + a + d = 21
3a = 21
a = 7

Now,

(a - d ) a (a + d) = 231
a^3 -ad^2 = 231

Putting a = 7
343 - 7d^2 = 231
d^2 = 16

d = 4 or d = -4


Now, required AP is

❇When a = 7 Nd d = 4

a - d = 7 - 4= 3
a = 7
a + d = 7 + 4 = 11

Required AP is 3, 7 and 11

❇When a = 7 Nd d = -4

a - d = 7 - ( -4 ) = 11
a = 7
a + d = 7 - 4 = 3

Required AP is 11, 7 and 3