The sum of first nth terms of an A.P is given by Sn = 3n2 - 4n . Determine the A.P and the 12th term.
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sn=3n^2-4n
s1=3*1*1-4*1
=-1
s2=3*2*2-4*2
=4
a1=s1= -1
a2=s2-s1
=5
d=a2-a1=6
AP is -1,5,11..........
a12=a+11d
=-1+11*6
=65
hope it helps u
s1=3*1*1-4*1
=-1
s2=3*2*2-4*2
=4
a1=s1= -1
a2=s2-s1
=5
d=a2-a1=6
AP is -1,5,11..........
a12=a+11d
=-1+11*6
=65
hope it helps u
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