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Arithmetic Progression 



Problem 3 of 19 


3. For arithemetic progression f( ) = , f( ) = , then find n such that f(n) = .










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Arithmetic Progression 
3. 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`







