The average of a non-decreasing sequence of N numbers a1,a2,…,aN is 300 . If a1 is replaced by 6a1, the new average becomes 400. Then, the number of possible values of a1 is
Started 3 months ago by Shashank in
Explanatory Answer
a1 + a2 + a3 + … + an = 300(n)
6a1 + a2 + a3 + … + an = 400(n)
5a1 = 100(n)
a1 = 20(n)
The constraints of the problem are:
● Sum of n terms = 300(n)
● The first term is 20(n)
● All other terms (terms other than the first term) should be greater than or equal to the first term. (Because the sequence is a non-decreasing sequence.)
Image a scenario where all the other terms are equal to the first term. That is a case where all the terms are equal. Since the average is 300. All of them should be 300. The first term is 300.
20(n) = 300
n = 15
Now imagine that the sequence has 16 terms the first term will be 320, and all the other terms will be greater than or equal to 320. So the average can’t be 300.
Or basically, the maximum number of terms in the sequence is 15.
What is the minimum number of terms in the sequence?
Can the sequence have one term?
n = 1
a1 = 20
But the sum of terms is not 300.
The sequence should have more than one term.
n = 2
a1 = 2(20) = 40
The sum of terms can be 300(2). This happens by having the second term as 560.
The minimum number of terms in the sequence is 2.
The maximum number of terms in the sequence is 15.
The number of terms in the sequence can have 14 different values.
And in each case the value of a1 is distinct.
Hence, a1 can have 14 different values.
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