Question : If sn, the sum of first n terms of an AP is given by Sn=3n2-4n, then find its nth term.
Doubt by Vanshika
Solution :
Method I
Sn=3n2-4n
an=?
We know,
an=Sn-Sn-1
an=[3n2-4n]-[3(n-1)2-4(n-1)]
an=[3n2-4n]-[3(n-1)2-4(n-1)]
an=[3n2-4n]-[3(n2+1-2n)-4n+4]
an=[3n2-4n]-[3n2+3-6n-4n+4]
an=[3n2-4n]-[3n2-10n+7]
an=3n2-4n-3n2+10n-7
an=6n-7
an=[3n2-4n]-[3n2+3-6n-4n+4]
an=[3n2-4n]-[3n2-10n+7]
an=3n2-4n-3n2+10n-7
an=6n-7
Method II
Sn=3n2-4n
an=?
Sn=3n2-4n
a1=S1=3(1)2-4(1)
a1=3-4
a1=-1— (1)
a1+a2=S2=3(2)2-4(2)
a1+a2=3(4)-8
a1+a2=12-8
a1+a2=4
(-1)+a2=4 [∵ Using equation (1)]
a2=4+1
a2=5
a1=3-4
a1=-1— (1)
a1+a2=S2=3(2)2-4(2)
a1+a2=3(4)-8
a1+a2=12-8
a1+a2=4
(-1)+a2=4 [∵ Using equation (1)]
a2=4+1
a2=5
Now,
a1=(-1)
a2=5
d=a2-a1
d=5-(-1)
a2=5
d=a2-a1
d=5-(-1)
d=6
an=a+(n-1)d
an=(-1)+(n-1)6
an=-1+6n-6
an=6n-7
an=(-1)+(n-1)6
an=-1+6n-6
an=6n-7