Find three numbers in AP whose sum is 15 and whose product is 105.

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 15.

a − d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5

Their product is 105.

$(a-d) \times a \times(a+d)=105$

$\Rightarrow\left(a^{2}-d^{2}\right) \times a=105$

$\Rightarrow\left((5)^{2}-d^{2}\right) \times 5=105$

$\Rightarrow(5)^{2}-d^{2}=21$

$\Rightarrow 25-d^{2}=21$

$\Rightarrow d^{2}=25-21$

$\Rightarrow d^{2}=4$

$\Rightarrow d=2,-2$

For $d=2$

The numbers are $5-2,5,5+2$

i.e. $3,5,7$

For $d=-2$,

The numbers are $5+2,5,5-2$

i. e. $7,5,3$

Hence, the three numbers are 3, 5, 7 or 7, 5, 3.