In an AP,if S_{n} = n(4n + 1),find the AP. | vedclass

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(A) We know that the $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = n(4n + 1) = 4n^{2} + n$.Now,$S_{n-1} = (n-1)(4(n-1)

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$5, 13, 21, \ldots$

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(A) We know that the $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = n(4n + 1) = 4n^{2} + n$.Now,$S_{n-1} = (n-1)(4(n-1) + 1) = (n-1)(4n - 4 + 1) = (n-1)(4n - 3) = 4n^{2} - 3n - 4n + 3 = 4n^{2} - 7n + 3$.Therefore,$a_{n} = (4n^{2} + n) - (4n^{2} - 7n + 3) = 4n^{2} + n - 4n^{2} + 7n - 3 = 8n - 3$.To find the $AP$,we substitute values for $n$:For $n = 1$,$a_{1} = 8(1) - 3 = 5$.For $n = 2$,$a_{2} = 8(2) - 3 = 16 - 3 = 13$.For $n = 3$,$a_{3} = 8(3) - 3 = 24 - 3 = 21$.Thus,the required $AP$ is $5, 13, 21, \ldots$

In an $AP$,if $S_{n} = n(4n + 1)$,find the $AP$.

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