The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

Mathematics

2 answers:

emmainna [20.7K]2 years ago

7 0

The answer will be 53

lidiya [134]2 years ago

6 0

A+4d=9
a+31d=-84
-27d=93
d=(-31/9)

a+4(-31/9)=9
a=205/9

23rd term=a+22d
=(205/9)+22(-31/9)
=-53

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Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$ >0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that $|a_n-A|n_0$ .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we denote the first series as $b_n=1$ , we will have that $\lim_{k \to \infty} b_k=1$ .

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$ .

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$ .

Since those limits of sub-series are different, we can never find such $n_0\\$ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose $\epsilon =1$ , all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$ .