Four numbers are in arithmetic progression. The sum of first and last terms is 8 and the product of both middle terms is 15. The least number of the series is

Let the four terms in A.P are

$a-3d,a-d,a+d,a+3d$

The sum of first and last terms $=8$

$\Rightarrow a-3d+a+3d=8$

$\Rightarrow 2a=8\Rightarrow a=4$

Product of both middle terms $=15$

$(a-d)(a+d)=15$

$\Rightarrow a^2-d^2=15$

$\Rightarrow 16-d^2=15$

$\Rightarrow d=\pm 1$

If $d=1$ then

The least number is  $=4-3\left(1\right)$

$a-3d$

$=1$

If $d=-1$ then

The least number is $a+3d$

$=4+3(-1)$

$=1$

$\therefore$ the least number of the series $=1$