(d)
Given $a,b,c$ are in GP and $4a,5b,4c$ are in AP.
$\therefore\ b^{2} = ac$ and $5b = \frac{4a + 4c}{2}$
$\Rightarrow b^{2} = ac$ and $5b = 2a + 2c$
Now, $a + b + c = 70$ (given)
$\Rightarrow 2a + 2c + 2b = 140$
$\Rightarrow \ 5b + 2b = 140$
$\Rightarrow \ b = 20$