# Math Help - Literal Equation help

1. ## Literal Equation help

Okay, so I have this equation:

$a_n=[d(n-b)]+a$

and I need to get to

$a_n=(a/b)+d(n-1)$

I've tried going about this a couple of times, but I always end up looping back to a previous step. If anyone could get me started in the right direction, that'd be great. I'm not even asking for a complete solution, just maybe the first few lines. Is this type of literal equation possible to do? I know how to do these equations, it's just this one that I'm having difficulty with.

Also note that this is part of a larger problem, so if more information is needed, it can be provided.

2. Do you mean that $[d(n-b)]+a = (a/b)+d(n-1)$?

It is not possible.

It would be better if you posted your entire question.

3. ## A further explanation of the origin of this problem

I see your thinking there, but it doesn't work. I'll explain the full problem, then.

I was given a formula that could be used to find the nth term in an arithmetic sequence. This equation was:

$a_n=[d(n-b)]+a$

Where:
$a_n$=answer
n=the term you are looking for (e.g. the 5th term, 12th term, or 78th term)
b=the number of terms shown
d=1st difference
a=the last term of the sequence shown

I was asked to prove this is true in all cases, and to do so, I took the already proven theorem of:

$a_n=a_1 +(n-1)*d$

Where:
$a_n$=answer
$a_1$= first term of the sequence
d=1st difference
n=term you are looking for

$a_n=[d(n-b)]+a$

into

$a_n=a_1 +(n-1)*d$

The first problem I ran into was the fact that there was no $a_1$ in the first equation. I determined (but have yet to prove) that dividing the last number shown (a) by the number of terms shown (b), you always get the first term. Thus, my target equation can be written as

$a_n=(a/b) +(n-1)*d$

I then set about trying to reach that equation from my first, and this is where I got stuck. I may have gone about this wrong, and, if so, please correct me. I hope the extra information helped.

4. Originally Posted by fireballs619
I was given a formula that could be used to find the nth term in an arithmetic sequence. This equation was:

$a_n=[d(n-b)]+a$

Where:
$a_n$=answer
n=the term you are looking for (e.g. the 5th term, 12th term, or 78th term)
b=the number of terms shown
d=1st difference
a=the last term of the sequence shown

I was asked to prove this is true in all cases, and to do so, I took the already proven theorem of:

$a_n=a_1 +(n-1)*d$

Where:
$a_n$=answer
$a_1$= first term of the sequence
d=1st difference
n=term you are looking for
Note: d = COMMON difference: no such thing as a 1st difference (but that's not important)

Let f = first term (less confusing than a1)

You can calculate f from your 1st equation, since we know the bth term = a:
f = a - d(b - 1) [1]

f + d(n - 1) = a + d(n - b)
Simplify:
a - f = d(b - 1) [2]

Substitute [1] in [2]:
a - (a - d(b - 1)) = d(b - 1)
a - a + d(b - 1)) = d(b - 1)
d(b - 1)) = d(b - 1)

Methinks that's the proof you need...(but I'm half-asleep right now!)

By the way:
> b=the number of terms shown
> a=the last term of the sequence shown
that could be stated as:
"you are given term number b and it's value a";

5. Hmm... I'm afraid I don't understand your thinkimg here. You say that b=a, which is not necessarily true. For example, in the sequence 5, 10, 15, 20, 25, 30, b ends up being 6 (there are six terms presented in the sequence), but a=30, since it is the last term shown.

6. Originally Posted by fireballs619
Hmm... I'm afraid I don't understand your thinkimg here. You say that b=a, which is not necessarily true. For example, in the sequence 5, 10, 15, 20, 25, 30, b ends up being 6 (there are six terms presented in the sequence), but a=30, since it is the last term shown.
Read my post again...I'm sure not saying that b=a.
I said: "you are given term number b and its value a";
so in your example: "you are given term number 6 and its value 30" .

PLUS there is no need for it being the last term:
you could be given b=5 and a=25; "last term" is misleading.