# Several terms of a sequence {a_n}_(n=1)^oo are given. a. Find the next two terms of the sequence. b. Find a recurrence relation that generates the seq

Several terms of a sequence ${{a}_{n}}_{\left(}n=1{\right)}^{ထ}$
are given. a. Find the next two terms of the sequence. b. Find a recurrence relation that generates the sequence (supply the initial value of the index and the first term of the sequence). c. Find an explicit formula for the nth term of the sequence. {1, 3, 9, 27, 81, ...}

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pivonie8
Concept Used- ${n}^{th}$ term of the geometric sequence is given by,
${T}_{n}=a+\left(n-1\right)d$, where a is the first term and d is the common ratio.
a) The new term of sequence can be find out by multiplying previous term by 3.
So, the next two terms are 243, 729.
b) For recurrence relation, we can write as,
${a}_{1}=1$
${a}_{n+1}=3{a}_{n}$
c) As the nth term for the sequence $\left(a,ar,a{r}^{2},a{r}^{3},\dots .\right)$ is ${T}_{n}=a·{r}^{n-1}$, where a is the first term and r is the common ratio that can be find out by $r=\frac{{T}_{2}}{{T}_{1}}=\frac{{T}_{3}}{{T}_{2}}=\frac{{T}_{4}}{{T}_{3}}=\dots$
Now, for the sequence given $\left\{1,3,9,27,81,\dots .\right\}$, we can write as,
${T}_{n}=1·{3}^{n-1}$
${T}_{n}={3}^{n-1}$
(a) Hence, the next two terms are 243,729.
(b) The recurrence relation is ${a}_{n+1}=3{a}_{n}.$
(c) And explicit formula for the nth term of the sequence is ${T}_{n}={3}^{n-1}.$