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Question

Question: What is the \({{n}^{th}}\) term of the geometric sequence\[360,\text{ }180,\text{ }90,\text{ }45\] ?...

What is the nth{{n}^{th}} term of the geometric sequence360, 180, 90, 45360,\text{ }180,\text{ }90,\text{ }45 ?

Explanation

Solution

For the type of question we need to find out the common ratio, as in geometric progression each term differs by a common ratio. As in general form geometric progression occurs in a way a,ar,ar2...........arn1a,ar,a{{r}^{2}}...........a{{r}^{n-1}} where a is the 1st term and arn1a{{r}^{n-1}}is the nth{{n}^{th}} term. And to find the common ratio of a geometric sequence is just to divide the 2nd{{2}^{nd}} term by 1st{{1}^{st}} term.

Complete step by step solution:
In order to know the nth{{n}^{th}} term of our question we just need to know the 1st term and common ratio and put their value in the general nth{{n}^{th}}terms of geometric expression.
Since the expression given to us 360, 180, 90, 45360,\text{ }180,\text{ }90,\text{ }45in which the first term is 360 and 2nd term is 180. So to find the common ratio divide the 2nd{{2}^{nd}} term by 1st{{1}^{st}} term as a way to find common ratio in geometric progression.
So the common ratio of the given sequence is180360=12\dfrac{180}{360}=\dfrac{1}{2} .
So 12\dfrac{1}{2}is the common ratio, whose 1st term i.e. a is 360.
So as we know the general nth{{n}^{th}}term of the sequence isarn1a{{r}^{n-1}}.
The general nth{{n}^{th}}term of our sequence is 360×(12)n1360\times {{\left( \dfrac{1}{2} \right)}^{n-1}}
Hence 360×(12)n1360\times {{\left( \dfrac{1}{2} \right)}^{n-1}}is the nth{{n}^{th}} term of the sequence.

Note: By just finding the common ratio and 1st term in the geometric sequence we can easily find any of the terms in the sequence. The common ratio can be greater than 1 or less than 1. As in our case, it is12\dfrac{1}{2}, less than 1.