Q1. How many numbers of a GP series of 100 terms ends with 6 which first term and common ratio is 2.
Q2. What is the smallest integer n such that the sum of the first n positive integers is divisible by 13?
The series is
2, 22, 23, 24 , 25 . . .
and the terms with
2, 4 , 8 , 6, 2 , 4 . . .
So for every 4 terms 1 term ends with 6
as the series is consist of 100 terms, 25 terms 'll end with 6
Let the term be n
then sum of first n terms = n * (n+1)/2
As 13 is a prime number the minimum value of n = 12
Q2
Minimum N is '13'.
Sorry, yes it is 12
n/a
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The series is
2, 22, 23, 24 , 25 . . .
and the terms with
2, 4 , 8 , 6, 2 , 4 . . .
So for every 4 terms 1 term ends with 6
as the series is consist of 100 terms, 25 terms 'll end with 6
Let the term be n
then sum of first n terms = n * (n+1)/2
As 13 is a prime number the minimum value of n = 12
Q2
Minimum N is '13'.
Sorry, yes it is 12
n/a
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