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For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 .

Solution:
Your problem `->` 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^(n-1)`

`=> 3 × 2^(n-1) = 1536`

`=> 2^(n-1) = 512`

`=> 2^(n-1) = 2^9`

`=> n - 1 = 9`

`=> n = 9 + 1`

`=> n = 10`






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