if the sum of first n terms of an A.P is given by Sn=4n2-3n,than find the nth term of an AP
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sn=4n^2-3n
s1=4*1-3*1=1
s2=4*2^2-3*2=16-6=10
s3=4*3^2-3*3=36-9=27
a1=s1=1
a2=s2-s1=10-1=9
a3=s3-s2=27-10=17
first term=a=a1=1
common difference=d=a2-a1=9-1=8
n th term=an= a+(n-1)d
=1+(n-1)8
=1+8n-8
=8n-7
s1=4*1-3*1=1
s2=4*2^2-3*2=16-6=10
s3=4*3^2-3*3=36-9=27
a1=s1=1
a2=s2-s1=10-1=9
a3=s3-s2=27-10=17
first term=a=a1=1
common difference=d=a2-a1=9-1=8
n th term=an= a+(n-1)d
=1+(n-1)8
=1+8n-8
=8n-7
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