Question

The sum of three numbers in AP is 21 and their product is 231. Find the numbers.​

Answer 1

Let the required numbers be (a−d),a,(a+d).

Then,

a−d+a+a+d=21

3a=21

a=7

Also,

(a−d)a(a+d)=231

a(a^2 −d^2 )=231

7(49−d^2 )=231

7d^2 =112

d^2 =16

d=±4

Hence, the required numbers are (3,7,11) or (11,7,3).

Answer 2

The numbers are 3 , 7 , 11 and 11 , 7 , 3

sum of numbers = 21

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

a - d + a + a + d = 21

3a = 21

a = 21÷ 3 = 7

product = 231

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

( a^2 - d ^2 ) a = 231

(7^2- d^2 ) 7 = 231

(49 - d^2) = 231÷7

49 - d^2 = 33

d^2 = 49 - 33 = 16

d = root of 16 = +/- 4

when d = +4

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

when d = - 4

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