The sum of n terms of an ap is 5n^2/2 + 3n/2 find the nth term and 20th term
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Sn = (5n2 + 3n) / 2
Therefore, S20 = [5 (20)2 + 3 (20) ] / 2
Thus S20 = 2060 / 2 = 1030
S19 = [5 (19)2 + 3 (19)] / 2 = 1862 / 2
Thus, S19 = 931
Tn = Sn - Sn-1
Therefore, T20 = S20 - S19
T20 = 1030 - 931 = 99
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Therefore, S20 = [5 (20)2 + 3 (20) ] / 2
Thus S20 = 2060 / 2 = 1030
S19 = [5 (19)2 + 3 (19)] / 2 = 1862 / 2
Thus, S19 = 931
Tn = Sn - Sn-1
Therefore, T20 = S20 - S19
T20 = 1030 - 931 = 99
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