Explanation

(d)
Let the four numbers be $\frac{a}{r},\ a,ar,\ 2ar - a$ …(i)
Where first three numbers are in GP and last three in AP.
Given that the common difference of AP is 6, so …(ii)
$ar - a = 6$
And also given $\frac{a}{r} = 2ar - a$
$\Rightarrow \frac{a}{r} = 2(ar - a) + a$
$\Rightarrow \ \frac{a}{r} = 2(6) + a$ [from Eq.(ii)]
$\Rightarrow \ \ \left( \frac{a}{r} \right) - a = 12$
$\Rightarrow \ \ a(1 - r) = 12r$
$\Rightarrow \ - 6 = 12r$ [from Eq.(ii)]
$\Rightarrow \ \ r = - \frac{1}{2}$
From Eq. (ii), we get
$a\left\lbrack \left( - \frac{1}{2} \right) - 1 \right\rbrack = 6$
$\Rightarrow \ \ a = - 4$
On putting the value of $a\ and\ r$ in Eq. (i), the required numbers are $8, - 4,\ 2,\ 8.$