Question

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

Answer 1

hello Frnd
here is your answer

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



hope it helps
jerri

Answer 2

Here is your solution


let,

Three numbers be a,a-d,a+d

so a+a-d+a+d = 21

a = 7

A/q
(a)(a-d)(a+d) = 231

(7)(7-d)(7+d) = 231

49- d2 = 231/7

d2 = 16

therefore d = +4 or - 4

so no(s) r : 3, 7 and 10 or 10, 7 and 3

hope it helps you