Asked by warnar please chek my answer
the sum of n terms of an A.P. is 3n^2+2n, then find its n th term.
answer-we know that
tn=s(n)-s(n-1)
=(3n^2+2n)-[3(n-1)^2+2(n-1)]
=(3n^2+2n)-[3n^2-6n+3+2n-2]
=3n^2+2n-[3n^2-6n+2n+3-2]
=3n^2+2n-(3n^2-4n+1]
=3n^2+2n-3n^2+4n-1
=6n-1 ans
answer-we know that
tn=s(n)-s(n-1)
=(3n^2+2n)-[3(n-1)^2+2(n-1)]
=(3n^2+2n)-[3n^2-6n+3+2n-2]
=3n^2+2n-[3n^2-6n+2n+3-2]
=3n^2+2n-(3n^2-4n+1]
=3n^2+2n-3n^2+4n-1
=6n-1 ans
Answers
Answered by
Steve
or, we know that
Sn = n/2 (2a + (n-1)d) = 3n^2+2n
an + n(n-1)/2 d = 3n^2+2n
an + d/2 n^2 - d/2 n = 3n^2+2n
d/2 n^2 + (a - d/2)n = 3n^2+2n
d/2 = 3 so d=6
(a-3)=2 so a=5
Tn = 5 + 6(n-1) = 6n-1
Sn = n/2 (2a + (n-1)d) = 3n^2+2n
an + n(n-1)/2 d = 3n^2+2n
an + d/2 n^2 - d/2 n = 3n^2+2n
d/2 n^2 + (a - d/2)n = 3n^2+2n
d/2 = 3 so d=6
(a-3)=2 so a=5
Tn = 5 + 6(n-1) = 6n-1
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.