If sum of n terms of an ap is 3n^2+n find the nth term
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Given:
Sn = 3n^2 + n
S(n-1) = 3(n-1)^2 + (n-1)
S(n-1) = 3(n^2 - 2n + 1) + (n - 1)
S(n-1) = 3n^2 - 6n + 3 + n - 1
S(n-1) = 3n^2 - 5n + 2
an = Sn - S(n-1)
an = (3n^2 + n) - (3n^2 - 5n + 2)
an = 6n - 2 ——> Answer
Sn = 3n^2 + n
S(n-1) = 3(n-1)^2 + (n-1)
S(n-1) = 3(n^2 - 2n + 1) + (n - 1)
S(n-1) = 3n^2 - 6n + 3 + n - 1
S(n-1) = 3n^2 - 5n + 2
an = Sn - S(n-1)
an = (3n^2 + n) - (3n^2 - 5n + 2)
an = 6n - 2 ——> Answer
Answered by
0
Answer:
an = 6n - 2
Step-by-step explanation:
Sn = 3n² + n
Substituting n = 1
S1 = 3(1)² + (1) = 4
a1 = S1 = 4
Substituting n = 2
S2 = 3(2)² + (2) = 14
a2 = S2 - S1 = 14 - 4 = 10
Substituting n = 3
S3 = 3(3)² + (3) = 30
a3 = S3 - S2 = 30 - 14 = 16
d1 = a2 - a1 = 10 - 4 = 6
d2 = a3 - a2 = 16 - 10 = 6
Since, d1 = d2
The terms 4, 10, 16,………… form an AP.
an = a1 + d(n - 1)
an = 4 + 6(n - 1)
an = 4 + 6n - 6
an = 6n - 2
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