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The sum of first nth terms of an A.P is given by Sn = 3n2 - 4n . Determine the A.P and the 12th term.

Let Sn denote the sum of first n terms of the A.P.

Given, Sn = 3n2 – 4n    ...(1)

Replacing n by (n –1) in (1), we get

Sn – 1 = 3(n – 1)2 – 4(n – 1)

nth term of the A.P. an = Sn – Sn – 1

∴ an = (3n2 – 4n) – [3(n – 1)2 – 4(n – 1)]

⇒ an = 3 [ n2 – (n – 1)2] – 4 [n – (n – 1)]

⇒ an = 3 (n2 – n2 + 2n – 1) – 4 (n – n + 1)

⇒ an = 3(2n –1) – 4

⇒ an = 6n – 3 – 4 = 6n – 7

Thus, the nth term of the A.P is 6n – 7.

Putting n = 1, we get  a1 = 6 × 1 – 7 = 6 – 7 =  –1

Putting n = 2, we get  a2 = 6 × 2 – 7 = 12 – 7 = 5

Putting n = 3, we get  a3 = 6 × 3 – 7 = 18 – 7 = 11  and so on.

5 – ( – 1) = 11 – 5 = ......= 6, which is a constant.

∴ –1, 5, 11, .... are in A.P.

Putting n = 12, we get 

 a12 = 6 × 12 – 7 = 72 – 7 = 65

Thus, the 12th term of A.P. is 65.

  • 11
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Sn=3n2-4n
S1=3-4=-1
S2=12-8=4
S3=27-12=15

A.P:-1,5,11,..........

a=-1 , d=6
a12=-1+(12-1)6
  = -1+66
  =65

  • 24

let  the value of n be 1-------

now s1  =  3 x 1 2 - 4 x 1

  s = a =  ( - 1 )--------------------  (x)

taking 2 as a value for n,

S2 = 3 x 22 - 4 x 2

( s )  a + a2  =  12 - 8 = 4

now, putting the value of  ( x):

-1 + a2 = 4

a2 =  5------------ ( y )

now a2 - a = d

5 + 1 = 6 = d --------- (z)

AP :  -1, 5, 11, 17....

and, a + 11d =  -1 + 11 ( 6)

 -1 + 66

65 .....

  • 6

sn = 3n2 - 4n

s1 = 3(1)2 - 4(1) = -1

s2 = 3(2)2 - 4(2) = 4

now, a1 + a2 = 4

a2 = 4 + 1 = 5

a = -1, d = 5 - (-1) = 6

a12 = -1 +(12 - 1)6 = -1 + 66 = 65

  • 2

Can this method be considered coz i have mostly seen in books Sn in the 1st Sn-1 in the second step . I can also solve this way but i got confused when i reffered to other books coz nowhere this kind of sum is solved in this way it is solved in diff. manner . 

  • -3

ya....this method can be considered.There's another mehod also but that's a little bit confusing so u can refer 2 dis method

  • -1
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