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For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 .

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Your problem `->` For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 .


We know that, `f(n) = a + (n - 1)d`

`f(5) = 25`

`=> a + (5 - 1)d = 25`

`=> a + 4d = 25 ->(1)`


`f(11) = 49`

`=> a + (11 - 1)d = 49`

`=> a + 10d = 49 ->(2)`

Solving `(1)` and `(2)`, we get `a = 9` and `d = 4`


Let `n` be the term such that `f(n) = 105`

We know that, `f(n) = a + (n - 1)d`

`9 + (n - 1)(4) = 105`

`(n - 1)(4) = 96`

`n - 1 = 24`

`n = 25`






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