Q18 of 74 Page 9

How many terms of the A.P. –6, -11/2, -5, ….. are needed to give the sum –25?


a= -6, d = (-11/2) + 6 = (-11 + 12)/2 = ½, Sn = -25

-25 = (n/2)[2(-6) + (n -1)(1/2)]

-25 4 = n(-24 + n –1)

- 100 = n(-25 + n)

n2 -25n + 100 = 0

n2 -20n – 5n + 100 = 0

n(n – 20) – 5(n – 20) = 0

(n – 5)(n – 20) = 0

n = 5, n = 20

Therefore the required number of terms is 5 or 20.

More from this chapter

All 74 →
16
Find the sum of all natural numbers between 100 and 1000 which are multiples of 5.

17
In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the sum of the next five terms. Show that the 20th term is –112.

19
If the pth term of an A.P is and the qth term is , prove that the sum of the first pq terms must be (pq + 1).

20
If the sum of a certain number of terms of the A.P. 25, 22, 19,….. is 116, find the last term.