Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is

  • [JEE MAIN 2013]
  • A

    $n + 4n^2$

  • B

    $6n^2 - n$

  • C

    $n^2 + 4n$

  • D

    $3n + 2n^2$

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