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Insert 3 geometric means between 16 and 256.

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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.

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