Little Different

The second through the sixth terms form a geometric progression according to:

$2^{0}d, 2^{1}d, 2^{2}d, 2^{3}d, 2^{4}d$ .

If $2^{4}d = 160$ , then $d=10.$ The first term can then be computed as $r = 2^{-1}d = \frac{10}{2} = \boxed{5}.$

This question is little different.

Here r is radius and is diameter.

So. by evaluating d is 10 and therefore r is 5