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If the sum of first n terms of an AP is n2, then find its 10th term.

 

 

Answers (1)

Given    \rightarrow    S_n = n^2

        \Rightarrow S_n = \frac{n}{2}\left[2a_1 + (n-1)d \right ] = n^2

                     \Rightarrow \frac{n}{2}\left[2a_1 + (n-1)d \right ] = n^2

                     = 2a_1 + (n-1)d \right = 2n\quad -(i)

Also    S_1 = a_1 = 1^2

                 \Rightarrow a_1 = 1\quad-(ii)

Now,    a_{10} = a_1 = 9d \quad -(iii)

            rearranging equation (i)

            a_1 + a_1 + (n-1)d = 2n

     \Rightarrow a_1 + (n-1)d = 2n-a_1

     Put n = 10

     \Rightarrow a_1 + (10-1)d = 2\times 10 -a_1

     \Rightarrow a_1 + 9d = 20 -a_1

     \Rightarrow a_{10} = 20 -a_1    (given (ii))

     \Rightarrow a_{10} = 20-1    (given (iii))

     \Rightarrow a_{10} = 19

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Safeer PP

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