Arithmetic Series Word problem

Question

Taking a calculus class and trying to solve a series and sequence word problem but I am struggling on the easiest problem. Please someone talk me thru a baby step on this question please?

1) Xin has been given a 14 day training schedule by her coach. Xin will fun for A minutes on day 1, where A is a constant. She will then increase her running time by (d+1) minutes each day, where d is a constant. a) Show that on day 14, Xin will run for

```(A + 13d +13) minutes```

At this point I think I understand that I can write formula as a+(n-1)d where a = A and d=(d+1) and n=14

2) Yi has also been given a 14 day training schedule by her coach. Yi will run for (A-13) minutes on day 1. She will then increase her running time by (2d-1) minutes each day. Given that Yi and Xin will run for the same length of time on day 14,

b) find the value of d.

My understanding is that the formula be written as

a = (A-13), n = 14, d=(2d-1)

At this point I don't know how to find the answer of question b. Something is telling me its a problem that I can use system of linear equation to find value of d. But I am not certain. Please can someone help me?

Thank you very much.

Answer 1

If you are looking to plug numbers into the formula don't use $d$ for both $d$ and $d+1$.

Your formula for an arithmetic sequence seems to be if $a = $ Value of first day. And $d=$ the amount changed each day then $a_n = $ the value on day $n = a+ (n-1)d$.

This is fine.

But as the variable $d$ is given to mean something arbitrarily different we have to use a different variable in our formula. And variables are just notation we could use anything.

I'll just put the formula with a capital $D$ so that $a_n = $ the value on day $n = a + (n-1)D$ where we have $D= d+1$ and $a=A$.

So the formula is $a_{14} = a + (14-1)D= a+13D = A+13(d+1) = A + 13d + 13.

For 2)

We have $a = (A-13)$ and $D = 2d-1$ and the formula is that on the $14$th day she will run $a + (14-1)D = a+13D = (A-13) +13(2d-1) = A -13 + 26d-13=A+26d - 26$.

So we are told that Xi who ran $A+13d + 13$ on the $14$th day and Yi who ran $A+26d$ on the fourteenth day ran the same ammount on the 14th day.

So $A + 13d + 13 = A + 26d-26$.

Solve for $d$.

Note: We don't need to know what $A$ is. If we subtract $A$ from both sides we have

$A+13d + 13 = A + 26d -26$ so

$A+13d + 13-A =A+26d -26 -A$ so

$(A-A) + 13d + 13 = (A-A) + 26d -26$ so

$13d + 13 = 26d - 26$

continue..