Let G1, G2, G3 be the required means.
Then 16, G1, G2, G3, 256 form a G.P.
Let r be the common ratio.
⇒ 256 = 5th term = ar4 = 16 × r4
⇒ 16r4 = 256 ⇒ r4 = 16 ⇒ r = 2
∴ G1 = ar = 16 x 2 = 32
G2 = ar2 = 16 x 4 = 64
G3 = ar3 = 16 x 8 = 128
Hence 32, 64 and 128 are the required G.M’s between 16 and 256.