Question

find three consecutive terms in an ap whose sum is 45 and their product is 3240

Answer 1

let the three terms be (a-d),a and a+d
A.T.Q
a-d+a+a+d=45
3a=45
a=15
now. (a-d)×a×(a+d)=3240
(3^2 - d^2)×3=3240
9-d^2=3240/3
9-d^2=1080
-d^2=1080-9
-d^2=1071
-d=3.under root 119
d=-3 u r 119
therefore the three consecutive terms are
a-d=3+3u r 119
a=3
a+d =3-3 u r 119
hope this could be correct answer