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For geometric progression f( 1 ) = 2 , f( 4 ) = 54 , then find n such that f(n) = 18 .

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Your problem `->` For geometric progression f( 1 ) = 2 , f( 4 ) = 54 , then find n such that f(n) = 18 .


We know that, `a_n = a × r^(n-1)`

Here `a_1 = 2`

`=> a × r^(1 - 1) = 2`

`=> a × r^0 = 2`

`=> a = 2 ->(1)`



`a_4 = 54`

`=> a × r^(4 - 1) = 54`

`=> a × r^3 = 54 ->(2)`

Solving `(1)` and `(2)`, we get `a = 2` and `r = 3`


Let n be the term such that `f(n) = 18`

We know that, `a_n = a × r^(n-1)`

`=> 2 × 3^(n-1) = 18`

`=> 3^(n-1) = 9`

`=> 3^(n-1) = 3^2`

`=> n - 1 = 2`

`=> n = 2 + 1`

`=> n = 3`






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