Question

The sum of first ‘n’ terms of an arithemetic progression is given by the formula

Sn

= 3n2 + n, then its 3rd term is​

Answer 1

Answer:

t3 = 13

refer to the attachment

Answer 2

Answer:

Step-by-step explanation:

Given Sum of n terms is S  

n

​

=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

=3n  

2

−10n+7

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

th

 term

⟹n  

th

 term =S  

n  −S  n−1  =3n  

2  −4n−(3n  

2 −10n+7)

n  

th

 term=6n−7