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)*n*th term

(ii)first term

(iii)common difference

This is the Question Number 25 Of Exercise 11 C of RS Aggarwal Solution.

# 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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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