The sum of first n terms of an ap is given by sn=3n²-n . Find ap and 25th term
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Sum n terms of an AP(Sn) = (n/2)[2a + (n-1)d]
Sn = an + n²d/2 - nd/2
Sn = 3n² - n
an + n²d/2 - nd/2 = 3n² - n
n(a - d/2) + n²d/2 = -n + 3n²
Equating the coefficients :
d/2 = 3
d = 6
a-d/2 = -1
a - (6/2) = -1
a - 3 = -1
a = 2
Therefore the AP is 2, 8, 14, 20, ....
Tn = a + (n-1)d
T25 = 2 + (25-1)6
T25 = 2 + 24 * 6
T25 = 2 + 144 = 146
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