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Question

Question: How do you find the explicit formula and calculate term 20 for \(3,9,27,81,243\)?...

How do you find the explicit formula and calculate term 20 for 3,9,27,81,2433,9,27,81,243?

Explanation

Solution

From the given series of geometric sequences, we find the general term of the series. We find the formula for tn{{t}_{n}}, the nth{{n}^{th}} term of the series. From the given sequence we find the common ratio which is the ratio between two consecutive terms. We put the values to get the formula for the general term tn{{t}_{n}}. Then we put the value of consecutive natural numbers for nn to find the solution.

Complete step-by-step solution:
We try to express the given sequence of 3,9,27,81,2433,9,27,81,243 in its general form.
We express the terms as tn{{t}_{n}}, the nth{{n}^{th}} term of the series.
The first term be t1{{t}_{1}} and the common ratio be rr where r=t2t1=t3t2=t4t3r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}.
We can express the general term tn{{t}_{n}} based on the first term and the common ratio.
The formula being tn=t1rn1{{t}_{n}}={{t}_{1}}{{r}^{n-1}}.
The first term is 3. So, t1=3{{t}_{1}}=3. The common difference is r=93=279=8127=3r=\dfrac{9}{3}=\dfrac{27}{9}=\dfrac{81}{27}=3.
We put the values of t1{{t}_{1}} and rr to find the general form.
We express general term tn{{t}_{n}} as tn=t1rn1=3×3n1=3n{{t}_{n}}={{t}_{1}}{{r}^{n-1}}=3\times {{3}^{n-1}}={{3}^{n}}.
Now we place consecutive natural numbers for nn as 20 to get the 20th{{20}^{th}} term as 320{{3}^{20}}.

Note: The sequence is an increasing sequence where the common ratio is a positive number. In case of the ratio being in the interval of r<1\left| r \right|<1, the geometric sequence becomes decreasing. They are also termed as infinite G.P. The formula for sum in that case becomes S=a1rS=\dfrac{a}{1-r}.