If sum of n terms of an ap is 3n^2+5n,then which terms of seriess is 164?
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Answered by
6
Answer:
Sn=3n2+5n
S1=a1=3(1)2+5(1)=8
S2=3(2)2+5(2)=22
S2=22=a1+a2
a2=22-8=14
d=a2-a1=14-8=6
nth term value is 164,then what is n?
nth term=a+(n-1)d
164=8+(n-1)6
164-8 / 6=n-1
156/6=n-1
26+1=n
n=27
So 27th term is 164.
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Answered by
1
Answer:
Sn=3n2+5n
S1=a1=3(1)2+5(1)=8
S2=3(2)2+5(2)=22
S2=22=a1+a2
a2=22-8=14
d=a2-a1=14-8=6
nth term value is 164,then what is n?
nth term=a+(n-1)d
164=8+(n-1)6
164-8 / 6=n-1
156/6=n-1
26+1=n
n=27
So 27th term is 164
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