insert three nos. between 1/3 and 432 so that the resulting sequnce is a G.P
rnANS DIS PLZZZ
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Let G1, G2, G3 be three geometric means inserted between and 432 so that , G1, G2, G3, G32 forms a G.P
Let r be the common ratio.
Then
G2 = Gr = 2 × 6 = 12
G3 = G2r = 12 × 6 = 72
Hence, ,2, 12, 72, 432 is the resulting G.P.