Question

find three consecutive terms in AP whose sum is 45 and their product is 3240​

Answer 1

Step-by-step explanation:

Let the three consecutive integers be a -d , a and a+d.

a-d+a+a+d = 45

3a = 45

a = 45/3

a = 15

Now the numbers are (15-d) , 15 and (15+d)

(15-d) x 15 x (15+d) = 3240

(15-d) x (15+d) = 3240/15

(15)^2 - (d)^2 = 216

-d^2 = 216 - 225

d^2 = 9

d = +/-3

If d= +3, The terms are

a-d = 15-3=12

a=15

a+d = 15+3=18

If d=-3 , the terms are

a-d = 15+3=18

a=15

a+d=15-3=12