the sum of n terms of an AP is 3n^2 - n find which term of the AP is 50
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0
Step-by-step explanation:
Given, sum of first n terms of an AP, Sn = 3n2 – n
Sn = 3n2 – n
Replacing n by n – 1, we get
Let the nth term of AP be an.
Now,put n = 1 in an, we get a = 6×1−4 = 2put n = 2 in an, we get a2 = 6×2−4 = 8put n = 3 in an, we get a3 = 6×3−4 = 14and so on.So, required AP is, 2,8,14,.....
Putting n = 50 we get
a50 = 6 × 50– 4 = 300– 4 = 296
Thus, the 50th term of the given A.P. is 296
Answered by
2
Answer:
9
Step-by-step explanation:
an= Sn-Sn-1
50 = 3n^2-n-(3(n-1)^2-(n-1))
50=3n^2-n-(3(n^2-2n+1)-n+1)
50=3n^2-n-3n^2+6n-3+n-1
50=6n-4
6n=54
n=54/6
n=9
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