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If sn, the sum of first n terms of an AP is given by sn= (3n2-4n), then find its nth term.

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

 

Given: Sn = 3n2 – 4n    ...(i)

 

Now,

Replacing n by (n –1) in (i), 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 

⇒ an= 6n – 7

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

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HEre sum of first n terms is given by Sn=S(3n2  - 4n)

We have to find the  anth term

We know that,

an = Sn - Sn-1

  =  (3n2 - 4n) - {3(n-1)2 - 4(n-1)}

  =(3n2 - 4n) - {3(n2+1-2n) - 4n+4}

  =(3n2-4n) - (3n2+3-6n-4n+4)

  =  3n2-4n-3n2-7+10n

  =6n-7

Hope I am right but you must wait for expert answer

  • 83

sn=3n2-4n

taking n as 1 we get the first term as the sum of first term is itself.

s1=3*1*1-4*1=-1

taking n as 2 we get the sum of first 2 terms since we have found the first term we can get the second term this  way

s2=3*2*2-4*2=4

common differnce d=4-(-1)=5

a.p.=-1,4,9,14...............

nth term is given as:

an=a+(n-1)d

=-1+(n-1)5

=-1+5n-5

therefore n=5n-6

though the above answer had different method it was kinda hard this is the easy method =D

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