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How many terms of the A.P. - 6, - 11/2, - 5, .... are needed to give the sum - 25?
Solution:
It is given that
- 6, - 11/2, - 5, .... are in A.P.
Let the sum of n terms of the given A.P be, Sn = - 25
Here, first term a = -6
Common difference d = - 11/2 + 6
= (- 11 + 12)/2 = 1/2
We know that, Sn = n/2 [2a + (n - 1) d]
Therefore,
⇒ - 25 = n/2 [2(- 6) + (n - 1) (1/2)]
⇒ - 50 = n [- 12 + n/2 - 1/2]
⇒ - 50 = n [n/2 - 25/2]
⇒ - 100 = n[n - 25]
⇒ n2 - 25n + 100 = 0
⇒ n2 - 5n - 20n + 100 = 0
⇒ n (n - 5) - 20 (n - 5) = 0
⇒ (n - 5)(n - 20) = 0
⇒ n = 20, 5
NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 4
How many terms of the A.P. - 6, - 11/2, - 5, .... are needed to give the sum - 25 ?
Summary:
Due to the fact that this is an increasing A.P and starts from the negative side, it is possible to have two solutions which were 20 and 5
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