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Geometric Progression 



Problem 3 of 23 


3. For given geometric progression series ,... then find n such that f(n) = .










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Geometric Progression 
3. For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 .
Here `a = 3,`
`r = 6/3 = 2`
Let n be the term such that `f(n) = 1536`
We know that, `a_n = a × r^(n1)`
`=> 3 × 2^(n1) = 1536`
`=> 2^(n1) = 512`
`=> 2^(n1) = 2^9`
`=> n  1 = 9`
`=> n = 9 + 1`
`=> n = 10`







