What is the smallest integer that can possibly be the sum of an infinite geometric series whose first term is 13?
The sum of an infinite geometric series is given by the formula:
S = a/(1-r)
where 'a' is the first term of the series and 'r' is the common ratio.
For the given series, the first term is 13. To find the smallest possible sum, we need to find the smallest possible value of 'r' such that the series converges.
For a series to converge, the absolute value of the common ratio must be less than 1. That is:
|r| < 1
Let's consider two cases:
Case 1: r is positive In this case, the smallest possible value of 'r' that satisfies the condition |r|<1 is r=1/2. Therefore, the sum of the infinite geometric series is:
S = a/(1-r) = 13/(1-1/2) = 26
Case 2: r is negative In this case, the smallest possible value of 'r' that satisfies the condition |r|<1 is r=-1/2. Therefore, the sum of the infinite geometric series is:
S = a/(1-r) = 13/(1+1/2) = 8.6666...
So the smallest possible integer that can be the sum of an infinite geometric series with a first term of 13 is 26.
Thanks guest answerer,
a=13
You are certainly right that |r|<1
S∞=a1−r S∞=131−r
Now for this question S∞ must be an integer so I will let it be N
r must be between -1 and 1 which means that 1-r is between 0 and 2
so N is between (not including) 13/2 and infinity
So the smallest is infinitesimally bigger than 6.5
Which means ths smallest integer value is 7
S∞=131−r N=131−r 7=131−r 7(1−r)=13 −7r=6 r=−67