The sequence converges, so you know that |r|<1
But that's not really an issue here.
a/(1-r) = 27
(6/r)/(1-r) = 27
6 = 27r(1-r)
27r^2 - 27r + 6 = 0
3(3r-1)(3r-2) = 0
r = 1/3 or 2/3
"The sum of an infinite geometric sequence is 27. The second term of the sequence is 6. Find the possible values of r."
I made a formula 'u1=6/r' using the info. given above, and I was going to insert it into the formula for the sum of an infinite geometric series 'sum infinity = u1/r-1, [r]<1'. But how do you know if r is less than 1 or greater than 1? The markscheme used u1/1-r, how did they know?
Thanks in advance.
5 answers
oobleck, when does a sequence converge or diverge?
think about it. The sequence is
a, ar, ar^2, ...
if r>1 the terms keep getting bigger and bigger.
If r=1, the sequence is constant, so the sum will diverge (unless a=0)
So for a GP to converge, |r| < 1
a, ar, ar^2, ...
if r>1 the terms keep getting bigger and bigger.
If r=1, the sequence is constant, so the sum will diverge (unless a=0)
So for a GP to converge, |r| < 1
That makes sense. But which part in the question gives away that it is a convergent geometric sequence because they haven't mentioned r values? Sorry to keep asking.
The sum of an infinite geometric sequence is 27.
The fact that the sum of an infinite number of terms is a finite value is the definition of convergent. Maybe you should google the topic and do some further reading. You will find much more information than I can provide here.
The fact that the sum of an infinite number of terms is a finite value is the definition of convergent. Maybe you should google the topic and do some further reading. You will find much more information than I can provide here.