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Subject: Math , asked on 23/1/13

# 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.=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 n   – S 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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• 28
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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

• 15

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