Question

If Sn=n^2+1 find the a12 ​

Answer 1

.Given Sum of n terms is Sn=3n^2−4n

So Sum of n−1 terms is S_n−1=3(n−1)^2−4(n−1)

=3n^2−6n+3−4n+4

=3n2−10n+7

Sum of n terms is equal to sum of n−1 terms plus nth term

⟹nth term =S_n−S_n−1=3n^2−4n−(3n^2−10n+7)

nth term=6n−7 

Answer 2

Answer:

25

Step-by-step explanation:

Remember this formula:

                      $a_{_{n}}= S_{n+1}- S_{n}$

So,

      $a_{_{12}} = S_{12+1}- S_{12} =S_{13} -S_{12}$

Here,

      $S_{13}=13^2+1=169+1=170$

      $S_{12}=12^2+1 = 144+1=145$

∴ $a_{_{12}}=S_1_3-S_1_2=170-145=25$

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