How do you write the first five terms of the sequence defined recursively $a}_{1}=14,{a}_{k+1}=(-2){a}_{k$, then how do you write the nth term of the sequence as a function of n?

Dymnembalmese2n
2022-09-22
Answered

How do you write the first five terms of the sequence defined recursively $a}_{1}=14,{a}_{k+1}=(-2){a}_{k$, then how do you write the nth term of the sequence as a function of n?

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Carina Moon

Answered 2022-09-23
Author has **6** answers

We are given ${a}_{1}=14$ and as $a}_{k+1}=(-2){a}_{k$, we have

${a}_{2}=(-2){a}_{1}=(-2)\times 14=-28$

${a}_{3}=(-2){a}_{2}=(-2)\times (-28)=56$

${a}_{4}=(-2)\times {a}_{3}=(-2)\times (56=-112$ and

${a}_{5}=(-2)\times (-112)=224$

It is apparent that it is geometric sequence with first term ${a}_{1}=14$ and common ratio as −2. As such $n}^{th$ term $a}_{n$ is given by

$a}_{n}={a}_{1}\times {(-2)}^{(n-1)}=14{(-2)}^{(n-1)$

${a}_{2}=(-2){a}_{1}=(-2)\times 14=-28$

${a}_{3}=(-2){a}_{2}=(-2)\times (-28)=56$

${a}_{4}=(-2)\times {a}_{3}=(-2)\times (56=-112$ and

${a}_{5}=(-2)\times (-112)=224$

It is apparent that it is geometric sequence with first term ${a}_{1}=14$ and common ratio as −2. As such $n}^{th$ term $a}_{n$ is given by

$a}_{n}={a}_{1}\times {(-2)}^{(n-1)}=14{(-2)}^{(n-1)$

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