Q3 of 202 Page 9

Find the sum of n terms of an A.P. whose nth term is given by

Put n = 1

a1 = 5 – 6(1) = 5 – 6 = -1


Put n = 2


a2= 5 – 6(2) = 5 – 12 = -7


a = -1, d = -7 + 1 = -6


Sn = [2a + (n – 1) d]


= [2(-1) + (n -1) (-6)]


= [-1 + 3n + 3]


= n [2 -3n]


More from this chapter

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1

Find the sum of the following arithmetic progressions:

(i)to 10 terms


(ii)to 12 terms


(iii)to 25 terms


(iv)to 12 terms


(v)to 22 terms


(vi)to n terms


(vii)to n terms


(viii)to 36 terms

 2

Find the sum on n term of the A.P.

4

If the n sum of a certain number of terms starting from first term of an A.P. is is 116. Find the last term.

 5

How many terms of the sequence should be taken so that their sum is zero?