# If the sum of the First n term of an AP is given by S_n=(3n^2-n), Find its (i)nth term (ii) first term (iii) common difference.

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ICSE & CBSE Board Question Based on Arithmetic Progression of RS Aggarwal
Here the sum of the First n terms of AP is given
You have to find
(i)nth term
(ii)first term
(iii)common difference
This is the Question Number 25 Of Exercise 11 C of RS Aggarwal Solution.

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### 1 Answer

1. Let to be first term ‘a’ and it’s CommOn difference ‘d’ and it’s nth terms ‘nth’.
Given;- S_n = ( 3n^2 – n ).
Let to be n replace (n-1)
°•° S_n=3n²-n
S_n-1= 3(n-1)²-(n-1)
= 3(n²-2n+1-(n-1)
= 3n²-6n+3-n+1
= 3n²-7n+4
Now, According to the Question statement;-
T_n=S_n-S_n-1
= 3n²-n-(3n²-7n+4)
= 3n²-n-3n²+7n-4
=6n-4
Hence, It’s Nth term= 6n-4
2) Substitute 1 in Equation as replace n
=> 6n-4
=> 6(1)-4
= 6-4
=> 2
Hence, 2 is the first term of this Arithmetic Sequence or Progression Equation!-
3) Substitute 2 in Equation as replace n
=> 6(2)-4
=> 12-4
=> 8
Hence, Arithmetic Sequence:- 2, 8
Now, Considering on Question as Third follow;-
CommOn Difference = T_2-T_1
= 8-2
= 6
Conclusion;-
i ) nth term= 6n-4
ii ) first term = 2
iii) Common difference. = 6

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