The sum of three consecutive numbers in AP is 21 and their product is 231. Find the numbers
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3
a-d a a+d
a-d +a+a+d=21
3a=21
a=7
(7-d)7(7+d)=231
49-7d(7+d)
343-49d-7d^2+49d=231
7d^2=112
d^2=16
d=+-4
if+4
3,7,11
if-4
11,7,3
a-d +a+a+d=21
3a=21
a=7
(7-d)7(7+d)=231
49-7d(7+d)
343-49d-7d^2+49d=231
7d^2=112
d^2=16
d=+-4
if+4
3,7,11
if-4
11,7,3
Answered by
2
ans is 3 ,7 ,11 .....................
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