Question

The sum of the first n terms of an ap is given by sn=3n2-n find the ap and the 12th term

Answer 1

so the 12th term is 68

Answer 2

AP is 2 , 8 ,14 ,..... and  the 12th term is 68

Step-by-step explanation:

We are given that  sum of the first n terms of an ap is given by $S_n=3n^2-n$

$a_n=S_n-S_{n-1}\\A_n=3n^2-n-(3(n-1)^2-(n-1))\\A_n=3n^2-n-(3(n^2+1-2n)-(n-1))\\A_n=3n^2-n-(3n^2+3-6n-(n-1))\\A_n=3n^2-n-(3n^2+3-6n-n+1)\\A_n=3n^2-n-3n^2-4+7n\\A_n=6n-4$

Substitute n = 1

$a_1=6(1)-4=2\\a_1=6(2)-4=8\\a_3=6(3)-4=14$

hence AP is 2 , 8 ,14 ,.....

Substitute n = 12

a_{12}=6(12)-4=68

Hence AP is 2 , 8 ,14 ,..... and  the 12th term is 68

#Learn more:

If Sn,the sum of first n terms of an AP is given by An=3n2-4n,find the n th term

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