If sum of three consecutive no. In an ap is 21 and their product is 231 find the no.
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Let x be the middle number and d be the common difference. Then the numbers are (x-d), x, and (x+d)
Sum=21
3x=21
X=7
Product=231
(x-d)x(x+d)=231
x(x^2-d^2)=231
7(49-d^2)=231
49-d^2=33
d^2=16
d=4
So numbers are 3, 7, 11
Sum=21
3x=21
X=7
Product=231
(x-d)x(x+d)=231
x(x^2-d^2)=231
7(49-d^2)=231
49-d^2=33
d^2=16
d=4
So numbers are 3, 7, 11
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