The sum of the consecutive terms of a.p. is21 and the product of first and third term the exceed the second term by 6 find the he term
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Let the numbers in A.P be a-d,a,a+d
a-d+a+a+d=21
3a=21
a=7
Also given (a-d)×(a+d)-a=6
a=7
(7-d)(7+d)-7=6
49-d^2=6+7
49-d^2=13
49-13=d^2
36=d^2
d=6
a=7,
a-d=7-6=1
a+d=7+6=13
a-d+a+a+d=21
3a=21
a=7
Also given (a-d)×(a+d)-a=6
a=7
(7-d)(7+d)-7=6
49-d^2=6+7
49-d^2=13
49-13=d^2
36=d^2
d=6
a=7,
a-d=7-6=1
a+d=7+6=13
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