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 For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 . [ Calculator, Method and examples ]Solution: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)df(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 = 4Let n be the term such that f(n) = 105We know that, f(n) = a + (n - 1)d9 + (n - 1)(4) = 105(n - 1)(4) = 96n - 1 = 24n = 25

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