Q15 of 110 Page 53

The sum of first n terms of a certain series is given as 3n2– 2n. Show that the series is an arithmetic series.

For n = 1,


Sum = 3(1)2–2(1) = 1


Therefore, first term = 1


For n = 2,


Sum = 3(2)2 – 2(2) = 12 – 4 = 8


Second term = 8–1 = 7


For n = 3,


Sum = 3(3)2 – 2(3) = 21


Third term = 21– 8 = 13


Series : 1, 7, 13…..


This is an arithmetic progression as the difference between two terms is constant.


Common difference = 7–1 = 13–7 = 6


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