The sum of the three consecutive number in A.P. is 21 and product is 231? find the nubers
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Answered by
0
let the numbers be x,y,z
they r in ap
x+z= 2y
x+y+z=21
xyz=231
solving them, we get y=7,x=3,z=11
they r in ap
x+z= 2y
x+y+z=21
xyz=231
solving them, we get y=7,x=3,z=11
Answered by
12
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
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
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