If s_{n}, the sum of first n terms of an AP is given by s_{n}= (3n^{2}-4n), then find its nth term.

#### Answers

HEre sum of first n terms is given by Sn=S(3n^{2} - 4n)

We have to find the an^{th} term

We know that,

a_{n} = S_{n} - S_{n-1}

_{ = }(3n^{2} - 4n) - {3(n-1)^{2} - 4(n-1)}

=(3n^{2} - 4n) - {3(n^{2}+1-2n) - 4n+4}

=(3n^{2}-4n) - (3n^{2}+3-6n-4n+4)

= 3n^{2}-4n-3n^{2}-7+10n

=6n-7

Hope I am right but you must wait for expert answer

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

Let the sum of first * n * terms of the A.P.=S * _{ n } *

Given: S * _{ n } * = 3

*n*

^{ 2 }– 4

*n*...(i)

Now,

Replacing * n * by ( * n * –1) in (i), we get,

S * _{ n } *

_{ – 1 }= 3(

*n*– 1)

^{ 2 }– 4(

*n*– 1)

* n * ^{ th } term of the A.P. * a _{ n } * = S

*– S*

_{ n }

_{ n }_{ – 1 }

∴ * a _{ n } * = (3

*n*

^{ 2 }– 4

*n*) – [3(

*n*– 1)

^{ 2 }– 4(

*n*– 1)]

⇒ * a _{ n } * = 3 [

*n*

^{ 2 }– (

*n*– 1)

^{ 2 }] – 4 [

*n*– (

*n*– 1)]

⇒ * a _{ n } * = 3 (

*n*

^{ 2 }–

*n*

^{ 2 }+ 2

*n*– 1) – 4 (

*n*–

*n*+ 1)

⇒ * a _{ n } * = 3(2

*n*–1) – 4

⇒ * a _{ n } * = 6

*n*– 3 – 4

⇒ * a _{ n } * = 6

*n*– 7

Thus, **the** ** ** * n *

^{ th }

**term of the A.P = 6**

**n**

**– 7.**

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