This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we have been given an arithmetic progression. And we have to find its sum upto n terms.
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Class:- 10th
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Question 2
Given A.P. has first term(a) = 5,
Common difference(d) = 2 – 5 = -3
Sum of n terms of A.P. = Sn = n[2a + (n – 1)d] / 2
= n[2(5) + (n – 1)(-3)] / 2
= n{10 – 3n + 3)} / 2
= n / 2(13 – 3n)
Hence, the sum of the n terms of A.P. is n/2(13 – 3n).