Three numbers are in an AP. their sum is 21 and their products is 231.find these numbers?

jock

6 years ago

let a rep. the first number
let a+d rep. the 2nd number
let a+2d rep. the 3rd number

sum
a+(a+d)+(a+2d)=21
a+a+d+a+2d=21
3a+3d=21
3(a+d)=21
divide both side by 3
[3(a+d)]/3=21/3
a+d=7 .....eq1
a=7-d ......eq2

product
a(a+d)(a+2d)=231
a+2d=(a+d)+d
a(a+d)((a+d)+d)=231
when a+d=7
a(7)(7+d)=231
7(7a+ad)=231
divide both side by 7
7a+ad=231/7
7a+ad=33
in eq2,a=7-d
7(7-d)+(7-d)d=33
49-7d+7d-d^2=33
49-d^2=33
49-33=d^2
d^2=16
square root both side
d=4
put d=4 in eq2
a=7-d
a=7-4
a=3

:,1st number=a=3
2nd number=a+d=3+4=7
3rd number=a+2d=3+2(4)=3+8=11



3,7 & 11

jock

6 years ago

let a-d rep. the first number
let a rep. the 2nd number
let a+d rep. the 3rd number

sum
(a-d)+a+(a+d)=21
a-d+a+a+d=21
3a=21
3a=21
divide both side by 3
3a/3=21/3
a=7

product
(a-d)a(a+d)=231
when a=7
(7-d)7(7+d)=231
divide both side by 7
(7-d)(7+d)=231/7
49-7d+7d-d^2=33
49-d^2=33
49-33=d^2
d^2=16
square root both side
d=4
:,1st number=a-d=7-4=3
2nd number=a=7
3rd number=a+d=7+4=11



3,7 & 11