Question

The sum of the first n terms of an AP is given by  ${\displaystyle s_n=(3n^2-n)}$ Find its

(i) nth term,
(ii) first term and
(iii) common difference.

 

Answer 1

Given : ${\displaystyle s_n=(3n^2-n)}$   .............(1)

Replacing n by (n - 1) in (i), we get:

${\displaystyle s_{n-1}=3{(n-1)}^2-(n-1)}$

= 3 (n² - 2n +1 ) - n + 1  

= 3n² -7 n + 4

(i) Now , ${\displaystyle T_n=(s_n-s_{n-1})}$

${\displaystyle =(3n^2-n)- (3n^2-7n+4)=6n-4}$

∴ n^th  term , T_n = (6n -4)  ..................(ii) 

(ii) Putting n = 1 in (ii), we get:

${\displaystyle T_1=(6\times 1)-4=2}$

(iii) Putting n = 2 in (ii), we get:

${\displaystyle T_2=(6\times 2)-4=8}$

∴ Common difference, d = T₂ - T₁ =  8 - 2 = 6