How do I generalize the nth term in an Arithmetic Progression?

Question

Sequence is $18, 14, 10,...$ So $n^{th}$ term is $18-4(n-1)$. Now put $n=16$, it gives $-42$. First term of the series is $18$ but $-42$ is the first term in the series that is a multiple of $21$. If first term multiple of $21$ in this series is $-42$ then what is the $N^{th}$ term that is a multiple of $21$? I need a general way to find $N^{th}$ term that is a multiple of $21$.

Answer 1

HINT: Notice that we are always subtracting $4$ so if $-42$ is the first term divisible by $21$, the other terms should be $-42-4k$ where $k$ is divisible by $21$ (If we know the first term divisible by $21$).

In order to find the first term (because after finding it, we can proceed as in the hint), define $a_n = 18-4n$, $n \in \mathbb{N}$. Then notice that all of the terms are even for this sequence so we can seek the first term divisible by $42$ instead of $21$. Say $18-4n = 42m$, where $n \in \mathbb{N}$ and $m \in \mathbb{Z}$. Here notice that since $n$ cannot be negative, $m$ must be negative or zero. It is easy to see that for $m=0$, $n$ is not a natural number so we can try it for $m = -1$ to see $n = 15$ and therefore $a_{15}$ is the first term divisible by $21$.

Answer 2

First of all, the expression $18-4(k-1)$ will become cumbersome. So we simplify it to $22-4k$. Your sequence, then, is $t_k = 22-4k$. If you are looking for the terms that are multiples of $21$, then you have to solve

\begin{align} 22-4k &\equiv 0 \pmod{21} \\ 4k &\equiv 22 \pmod{21} \\ 16 \cdot 4k &\equiv 16 \cdot 22 \pmod{21} \\ k &\equiv 16 \pmod{21} \\ \end{align}

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If you are not comfortable with congruence equations, you can try to solve

\begin{align} 22-4k &= 21\ell \\ 4k+21\ell &= 22 \\ \end{align}

You found that $k=16$ is a solution. For that particular case, then, you can solve for $\ell$. $22-4(16) = 21\ell$ gives us $\ell=-2$. To find the general solution, we compare and subtract.

\begin{array}{c} 4k &+ &21\ell &= 22 \\ 4(16) &+ &21(-2) &= 22 \\ \hline 4(k-16) &+ &21(\ell-2) &= 0 \\ \end{array}

And so $4(k-16) = 21(2 - \ell)$. Since $4$ clearly divides $4(k-16)$, it must also divide $21(2 - \ell)$. Since $4$ and $21$ are relatively prime, then $4$ must divide $2-\ell$. Say $2-\ell = 4m$. And then

\begin{align} 4(k-16) &= 21(2 - \ell) \\ 4(k-16) &= 21(4m) \\ k-16 &= 21m \end{align}

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Either way, we get $k = 16 + 21m$ for all integers $m$. So the n$^\text{th}$ value of $k$ that gives you a multiple of $21$ is $k = 16 + 21(n-1)$ and the n$^\text{th}$ number in your sequence that is a multiple of $21$ is

$22-4((16 + 21(n-1)) - 1) = 42-84n = 21(2-4n)$