# Finding the first term with the second and fourth terms

I don't really know how to explain it as I am not too good at maths vocabulary and mildly good at maths but there is always a question which stumps us and this is mine.

I have two terms, the second(7) and the fourth(43) and the rule which says "the next term in another sequence is to multiply the previous term by 3 and then subtract x, where x is an integer". It wants me to find the first term of this sequence.

I've looked it up on the internet and perhaps I didn't search it right, but all I got was answers to simple sequences which had basic addition to each term with no rules.

Since the answer is needed tomorrow(next-day homework question) and I've tried all resources except this one available to me, I have resorted to this. I ask you, what is the first term and how did you get it? Many thanks in advance.

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It looks like a case of confusing word choice. "the next term in another sequence is to multiply the previous term by 3 and then subtract x, where x is an integer" is not very clear, which is not your fault.

Suppose you have a sequence $a_1, a_2, a_3, a_4, ...$. You're given that the rule for each $a_n$ is $a_n = 3a_{n-1} - x$, where $x$ is some integer that's the same for all of them.

Then, given that $a_2 = 7$ and $a_4 = 43$, find $a_1$.

So what you want to do here is find $a_4$ in terms of $a_2$ and solve for $x$.

We have that:

$a_2 = 7 \\ a_3 = 3(7) - x = 21 - x \\ a_4 = 3(21 - x) - x = 63 - 4x$

Since $a_4 = 43$, this means $4x = 20$ and $x = 5$.

Now, work backwards and solve for $a_1$ given $x$:

$a_2 = 3(a_1) - 5\\ 7 = 3(a_1) - 5\\ 3(a_1) = 12\\ a_1 = 4$

which is the answer you want.

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By the way, if you want to check that you've done it right, here's a different problem for you to solve: Same as above, except the second term is 8 and the fourth term is 56. What is the first term? (It should also be 4.) – Joe Z. Jan 3 '13 at 17:21
Very clear answer which describes both the question and the solution, much appreciated! – PwnageAtPwn Jan 3 '13 at 17:26

Hint: Solve the equation $$3\cdot(7)-x=\frac{43+x}{3}.$$ This gives you the value of $x$. After you find $x$, you can set up the equation $a_2=3\cdot a_1-x$.

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Hint: Let $y$ be the first element in the series. You are told that $7=3y-x$ from the rule. Then we apply the rule twice to find the fourth term. The third term is $3\cdot 7-x=21-x$. The fourth term is then $43=3(21-x)-x=63-4x$ From this you can find $x$, then substitute into the first to find $y$.

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Convert what you're told into mathematical language. Say the $n$th term in your sequence is $a_n$. You're given that $a_2=7$ and $a_4=43$. You're also told that $a_n = 3a_{n-1} - x$ for each $n$. Your task is to find $a_1$.

Using this formula you have $a_2 = 3a_1 - x$. You know what $a_2$ is, so as soon as you find $x$, all you have to do is rearrange this and you're done.

How do you find $x$? Well you know

$$a_3 = 3a_2 - x$$

and

$$a_4 = 3a_3 - x$$

You know $a_4$ and $a_2$, so substitute the first of these equations into the second and solve for $x$.

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It's 43, not 47. – Joe Z. Jan 3 '13 at 17:16
@JoeZeng: Thanks - fix'd. Dunno how that happened. – Clive Newstead Jan 3 '13 at 17:17