Question

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

Answer 1

HELLO DEAR,

let the three no. be (a-d),a,(a+d)

given that:-

a-d+a+a+d=21

3a=21

a=21/3

a=7----------(1)

and

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

(7²-d²)7=231-------using(1)

49-d²=231/7

d²=49-33

d²= 16

d= +-4

when d= 4

no. be 3,7,11

when d=-4

ni. be 11,7,3

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hi Aiswarya

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

a - d + a + a + d = 21

a = 7......(1)

a × ( a + d ) × ( a - d ) = 231

a × [ a² - d² ] = 231

49 - d² = 33

d² = 16

d = ± 4

So numbers are

when d = +4

3 , 7 , and 11

when d = -4

11 , 7 and 3

hope it helps