How do you convert between these types of arithmetic explicit equations?

Question

I was just wondering how you convert the formula: "a_n=dn+a₀", to the one seen on the ACT: "a_n=a₁+(n-1)*d". 

It would also be helpful if you could give an example. 

 

Thanks.

Answer 1

The reason people use one or the other formula is because they call the first term of the sequence something different.

 

If you're used to calling the first term of your sequence a_o then you're going to use the first equation. If you call the first term of your equation a₁ instead, you'll use the second equation. Otherwise there's not really a difference. The d will be the same. Whatever you were calling a_o in the first one you will call a₁ in the second. 

 

Here's an example:

 

2, 5, 8, 11, 14, ...

 

If I call the first term in this sequence a_o then I would write the equation

 

a_n = 2 + 3n

 

where 2 is my a_o term and d is 3 because I'm adding three to get the the next term. Also, here a_n is really the n + 1th term of the sequence because I'm starting by calling the first term the 0th.

 

My formula gives me the a_n in a sequence that's going

 

a_o, a₁, a₂, a₃, a₄, ...

 

So a₁ is actually the second term and a₂ is actually the third term, and so on.

 

The reason you might prefer to use this formula though is that it simplifies the expression you're using to find a_n.

 

Let's consider the same sequence again:

 

2, 5, 8, 11, 14, ...

 

This time I'm going to call my first term a₁.

 

Notice then that

 

a₂ = 2 + 3                           and 

 

a₃ = 2 + 3 + 3                     and 

 

a₄ = 2 + 3 + 3 + 3               and so on.

 

The formula I would then get for a_n would have to be

 

a_n = 2 + 3(n - 1)

 

It can't be the same as the first formula because then a₁, which I'm saying is the first term, would be 5 instead of 2.

 

So converting between the two, nothing changes except that you're calling your first term something different and you're multiplying d by n - 1 instead of n.