the sum of 3 terms of an ap is 36 and product is 7620 find the terms of ap
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S3= 36
let the terms be a-d,a and a+d
A/Q: a-d+a+a+d=36
3a=36
a=12
Again,(a-d)a(a+d)=1620
(12-d)12(12+d)=1620
144-d^=135
d=3
let the terms be a-d,a and a+d
A/Q: a-d+a+a+d=36
3a=36
a=12
Again,(a-d)a(a+d)=1620
(12-d)12(12+d)=1620
144-d^=135
d=3
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