Question

The sum of n terms of an
a.P. Is 3n2+n; then its pth term is

Answer 1

$P_{th}$ term is (6P - 2).

Step-by-step explanation:

1.  The sum of n terms of A.P series

    $S_{n} = 3\times n^{2} +n$

    This above equation is varied for every natural number.

 so      $S_{1} = 3\times 1^{2} +1$ = 4

          $S_{2} = 3\times 2^{2} +2$ = 14

          $S_{3} = 3\times 3^{2} +3$ = 30

         $S_{4} = 3\times 4^{2} +4$ = 52

         .

         .

         .

So      $S_{n} = 3\times n^{2} +n$        

2.   Now

  $S_{1}$ is sum of first term = 4 = $a_{1}$      ...1)

 $S_{2}$ is sum of first and second term = 14 = $a_{1} +a_{2}$      ...2)

  $S_{3}$ is sum of first, second and third term = 30 = $a_{1} +a_{2}+a_{3}$     ...3)

  $S_{4}$ is sum of first, second, third and fourth term = 52 = $a_{1} +a_{2}+a_{3}+a_{4}$    ...4)

3. On solving equation 1,2,3,and 4

We get

$a_{1}= 4$ = First term = a

$a_{2} = 14-4 =10$ = Second term

$a_{3}= 30-10-4= 16$ = Third term

$a_{4}= 52-16-10-4= 22$ = Fourth term

4.  So common difference (d) =10-4=6

5. Then $P_{th}$ term

   $P_{th} = a + (P-1)\times d$

  $P_{th} = 4 + (P-1)\times 6$

     So   $P_{th} = 6P-2$