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Geometric Progression 



Problem 11 of 23 


11. Find geometric mean between and .










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Geometric Progression 
11. Find 3 geometric mean between 1 and 256 .
Let ` a_1, a_2, a_3` be the `3` geometric mean between `1` and `256`.
`:. 1, a_1, a_2, a_3, 256` are in GP .
Here First term `a = 1` and Last term `b = 256`
Comman ratio `r = (b/a)^(1/(n+1)) = (256/1)^(1/4) = 4`
`:.` Required Geometic Means,
`a_1 = a * r = 1 * 4 = 4`
`a_2 = a r^2 = 1 × 4^2 = 16`
`a_3 = a r^3 = 1 × 4^3 = 64`





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