The sum of three numbers in AP is 18.

Solution:

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 18.

a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6

The product of the first and third term is 5 times the common difference.

$(a-d)(a+d)=5 d$

$\Rightarrow a^{2}-d^{2}=5 d$

$\Rightarrow(6)^{2}-d^{2}=5 d$

$\Rightarrow 36-d^{2}=5 d$

$\Rightarrow d^{2}+5 d-36=0$

$\Rightarrow d^{2}+9 d-4 d-36=0$

$\Rightarrow d(d+9)-4(d+9)=0$

$\Rightarrow(d+9)(d-4)=0$

$\Rightarrow d=4,-9$

For $d=4$

The numbers are $6-4,6,6+4$

i. e. $2,6,10$

For $d=-9$

The numbers are $6+9,6,6-9$

i.e. $15,6,-3$

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.