three consecutive terms of an AP have sum 21 and product 315. find the numbers
a-d,a,a+d
a-d+a+a+d=21
3a=21
a=7
product=a-d*a*a+=315
multiply them u will get value of d=2
subsitute the value in a-d,a,a+d
numbers are 5,7,9
a-d,a,a+d
a-d+a+a+d=21
3a=21
a=7
product=a-d*a*a+=315
multiply them u will get value of d=2
subsitute the value in a-d,a,a+d
numbers are 5,7,9