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.  
 A: 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 $a_2=3\cdot a_1-x$.
A: 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 $y$.
A: 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 $x$.
A: 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.
