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Question: How do you find the indicated term of the geometric sequence where \({{a}_{1}}=\dfrac{1}{3},r=3,n=8\...

How do you find the indicated term of the geometric sequence where a1=13,r=3,n=8{{a}_{1}}=\dfrac{1}{3},r=3,n=8?

Explanation

Solution

We are given that the sequence is in GP. We know that the general term of the GP is given by arn1a{{r}^{n-1}} . Now we know the values of a and r in the sequence hence we can easily find the 8th{{8}^{th}} term by substituting the values of a, r and n in the given formula.

Complete step-by-step solution:
Now let us first understand the concept of geometric sequence and arithmetic sequence.
Arithmetic sequence is a sequence in which the difference between two consecutive terms is constant. Let us say d is the common difference then the sequence is given as a, a+d, a+2d, …
Where a is the first term of the sequence. Hence the nth{{n}^{th}} term of the sequence is given by tn=a+(n1)d{{t}_{n}}=a+\left( n-1 \right)d .
Similarly Geometric sequence is a sequence in which the ratio of two consecutive terms is constant. Let us say that r is the common ratio between the terms and a be the first term then the sequence is given as a,ar,ar2,...a,ar,a{{r}^{2}},... . Hence the nth{{n}^{th}} term of the sequence is given by arn1a{{r}^{n-1}} .
Now consider the given sequence. We have the first terms as 13\dfrac{1}{3} and r = 3.
Hence the nth{{n}^{th}} term of the sequence is given by tn=arn1=13.3n1==3n2{{t}_{n}}=a{{r}^{n-1}}=\dfrac{1}{3}{{.3}^{n-1}}=={{3}^{n-2}}
Hence substituting n = 8 we get t8=382=36=729{{t}_{8}}={{3}^{8-2}}={{3}^{6}}=729 .
Hence the 8th{{8}^{th}} term of the sequence is 729.

Note: Now note that in the formula to find nth{{n}^{th}} term we raise r by n – 1 and not n. The formula can be easily derived by observation. Also note that if r is not given we can easily find r as r is just the ratio of two consecutive terms. Hence r=tn+1tnr=\dfrac{{{t}_{n+1}}}{{{t}_{n}}}.