The sum of three no in ap os 21 and their product is 231 find tha no
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let the numbers be
a-d,a,a+d
since sum = 21
a-d+a+a+d =21
3a=21
a= 7
and product = 231
(7-d)7(7+d) = 231
49-d² = 33
16 = d²
d = ±4
so a-d = 7-4 = 3 or a-d = 7-(-4)= 11
a = 7
a+d = 7+4 = 11 or a+d = 7-4 = 3
thus ap can be 3,7,11 or 11,7,3
a-d,a,a+d
since sum = 21
a-d+a+a+d =21
3a=21
a= 7
and product = 231
(7-d)7(7+d) = 231
49-d² = 33
16 = d²
d = ±4
so a-d = 7-4 = 3 or a-d = 7-(-4)= 11
a = 7
a+d = 7+4 = 11 or a+d = 7-4 = 3
thus ap can be 3,7,11 or 11,7,3
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