In an Ap the 8th term and 4th term is 20 and the 8th term times the fourth term is 1½.what is the common difference

Let the first term of the AP be $a$ and the common difference be $d$.
The $n$-th term of an AP is given by $a_n = a + (n-1)d$.
The 8th term is $a_8 = a + 7d$.
The 4th term is $a_4 = a + 3d$.

From the problem statement:
The sum of the 8th term and 4th term is 20:
$a_8 + a_4 = 20$
$(a + 7d) + (a + 3d) = 20$
$2a + 10d = 20$
Dividing by 2, we get:
$a + 5d = 10$ (Equation 1)

The product of the 8th term and 4th term is $1\frac{1}{2}$ (which is 1.5):
$a_8 \times a_4 = 1.5$
$(a + 7d)(a + 3d) = 1.5$ (Equation 2)

From Equation 1, we know $a+5d = 10$.
We can rewrite $a_8$ and $a_4$ using this:
$a_8 = a + 7d = (a + 5d) + 2d = 10 + 2d$
$a_4 = a + 3d = (a + 5d) - 2d = 10 - 2d$

$(10 + 2d)(10 - 2d) = 1.5$

This is a difference of squares formula $(X+Y)(X-Y) = X^2 - Y^2$, where $X=10$ and $Y=2d$.
$10^2 - (2d)^2 = 1.5$
$100 - 4d^2 = 1.5$
Subtract 1.5 from both sides and add $4d^2$ to both sides:
$100 - 1.5 = 4d^2$
$98.5 = 4d^2$
Divide by 4:
$d^2 = \frac{98.5}{4}$
$d^2 = 24.625$

$d = \pm \sqrt{24.625}$
$d \approx \pm 4.962358$

Rounding to two decimal places:
$d \approx \pm 4.96$