Question

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

Answer 1

Hi


Let ( a - d ) , a , ( a + d ) are

three consecutive

numbers in A.P

whose common difference = d

sum of the numbers = 21

a - d + a + a + d = 21

3a = 21

a = 21/3

a = 7

according to the problem given

( a - d ) a ( a + d ) = 231

( a² - d² ) a = 231

( 7² - d² ) 7 = 231

49 - d² = 231/7

49 - d² = 33

- d² = 33 - 49

- d² = - 16

d² = 16

d = ± 4

Therefore ,

required three numbers are ,

1 ) if a = 7 , d = 4

a - d = 7 - 4 = 3

a = 7 ,

a + d = 7 + 4= 11

( 3 , 7 , 11 )

2 )if a = 7 , d = -4

required numbers are

( 11 , 7 , 3 )

I hope this helps you.

:)

Answer 2

HELLO DEAR,


 LET THREE NUMBERS BE :-


(a - d) , a , (a+d)

SO a+a-d+a+d = 21


a = 7


(a)(a-d)(a+d) = 231

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

49- d² = 231/7

d² = 16

THEREFORE

d = +4 or - 4

THEREFORE

A.P IS  3, 7 , 11


I HOPE ITS HELP YOU DEAR,
THANKS