Question

The sum of the first n terms of an AP is given by Sn = ( 3n^2 - n ). Find its

i ) nth term,

ii ) first term and

iii) Common difference.

Answer 1

♠ ANSWER :

Sn = ( 3 n² - n )

Putting value of n = 1,

S1 = ( 3 - 1 ) = 2.

Putting value of n = 2 ,

S2 = ( 3 * 2² - 2 ) =. 12 - 2 = 10

Putting value of n = 3,

S3 = ( 3 * 3² - 3 ) = 27 - 3 = 24.

S1 = a

⏺️a = 2.

⏺️S2 - S1 = a2 = 10 - 2 =8.

⏺️Common difference, d = a2 - a =. 8 - 2 = 6.

( i ) nth Term.

An = a+ ( n - 1 ) d = 2 + ( n - 1 ) 6

An = 2 + 6n - 6

✔️An = 6n - 4.

( ii ) First term, a = 2.

( iii ) Common difference, d = 6.

Answer 2

HEY THERE!!

Question;-

The sum of the first n terms of an AP is given by Sn = ( 3n^2 - n ). Find its

i ) nth term,

ii ) first term and

iii) Common difference.

Method of Solution;-

Let to be first term 'a' and it's CommOn difference 'd' and it's nth terms 'nth'.

Given;- Sn = ( 3n^2 - n ).

Let to be n replace (n-1)

°•° Sn=3n²-n

Sn-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;-

Tn=Sn-Sn-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 = T2-T1

= 8-2

= 6

Conclusion;-

i ) nth term= 6n-4

ii ) first term = 2

iii) Common difference. = 6