three numbers whose sum is 21 are in AP . if the product of the first and the third numbers exceeds the second number by 6 , find the numbers
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Let the three numbers be
(a-d) , a , (a+d) , then;
a-d+a+a+d = 21
3a = 21
a = 7
(a-d)(a+d) = a+6
a^2 - d^2 = a+6
49 - d^2 = 7+6
d^2 = 49-13
d^2 = 36
d = 6
Hence the numbers are ;
(a-d) = (7-6) = 1
a = 7
(a+d) = (7+6) = 13
your answer is 1,7,&13
(a-d) , a , (a+d) , then;
a-d+a+a+d = 21
3a = 21
a = 7
(a-d)(a+d) = a+6
a^2 - d^2 = a+6
49 - d^2 = 7+6
d^2 = 49-13
d^2 = 36
d = 6
Hence the numbers are ;
(a-d) = (7-6) = 1
a = 7
(a+d) = (7+6) = 13
your answer is 1,7,&13
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