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The sum of first m terms of an AP is (4m^2-m). If its nth term is 107, find the value of n. Also, find the 21st term of this AP.

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This is the basic and conceptual question from arithmetic progression chapter in which we are to calculate the value of n and also we need to find 21st term of this AP, if the sum of first m terms of an AP is (4m^2-m) and its nth term is 107.

RS Aggarwal, Class 10, Chapter 5C, question no 37

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  1. From the question it is given that,
    an​=107
    The sum of first m terms of an A.P.Sm​=4m^2−m
    nth  term Sn​ is =4n2−n
     Then, 
    Sn−1​=4(n−1)^2−(n−1)
            =4(n^2−2n+1)−n+1
            =4n^2−8n+4−n+1
            =4n^2−9n+5
    Therefore an=Sn​−Sn−1​107=4n^2−n−4n2+9n−5
    107=8n−5
    107+5=8n
    112=8n
    n=112/8
    n=14​
    an​=8n−5
    a21​=(8×14)−5
    a21​=168−5
    a21​=163​

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