Below is a Geometric sequence with u1 = 5, and the common ratio r5,5√3,53,53√3, .......a) Find the value of r.b) Find the general term un.c) Find which term of the sequence will be 581,d) Show that the sum the infinite terms of the sequence in the form 5√3√3−1.
Question
Below is a Geometric sequence with u1 = 5, and the common ratio r5,5√3,53,53√3, .......a) Find the value of r.b) Find the general term un.c) Find which term of the sequence will be 581,d) Show that the sum the infinite terms of the sequence in the form 5√3√3−1.
Solution
The question seems to be a bit jumbled, but I'll try to answer it based on my understanding.
a) The common ratio (r) in a geometric sequence is found by dividing any term (un) by the preceding term (un-1). From the sequence given, it seems like the common ratio is √3.
b) The general term (un) in a geometric sequence can be found using the formula un = u1 * r^(n-1). Here, u1 is the first term and r is the common ratio. So, un = 5 * (√3)^(n-1).
c) To find which term of the sequence will be 581, you can set the general term equal to 581 and solve for n. So, 5 * (√3)^(n-1) = 581. Solving this equation will give you the value of n.
d) The sum of an infinite geometric sequence can be found using the formula S = a / (1 - r), where a is the first term and r is the common ratio. So, the sum of the infinite terms of the sequence is 5 / (1 - √3). However, this doesn't seem to simplify to 5√3√3−1. There might be a mistake in the question.
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