Question

Find three consecutive terms in A.P. such that their sum is $21$ and their products is $315$.

Answer 1

A] Let three numbers be a - d, a, a + d

given

a - d + a + a + d = 21

3a = 21

$\therefore a=7$

(a + d) (a - d) a = 315

$=(7+d)(7-d)\times 7=315$

$=(7^2-d^2)=45$

$=7^2-45=d^2$

$=49-45=d^2$

$=4=d^2$

$d=\pm 2$

numbers are

7 - 2 , 7 , 7 + 2

= 5 , 7 , 9