Sequences - Challenge Problems

Below are 3 challenge problems.

Please choose ONE problem and attempt to solve it.

Then do your best to write a CLEAR EXPLANATION below of how to solve the problem. A classmate who reads your explanation should be able to follow and understand it. You can either type it, or write in your book and upload a photo below.

Even if you are not able to solve the problem, please explain YOUR THINKING PROCESS…

• What did you try?
• Why didn’t it work?
• What questions would you ask to help understand the problem?

Question 1 §

Question 1

Each row and column in the grid below is an arithmetic sequence with 5 terms. Find the value of $X$ in the centre cell.


So, with the current square, we have 5 arithmetic sequences to solve.

• 1…25:

This is the arithmetic sequence at the top, so $t_1=1$ and $t_5=25$

The first term is 1 so $a=1$ 	

$\therefore t_5=1+4d$ 

$24=4d$

$d=6$

$t_n=1+6\left(n-1\right)$

This means the sequence goes $1, 7, 13, 19, 25$

• 1…17:

This is the arithmetic sequence at the left, so $t_1=1$ and $t_5=17$

The first term is 1 so $a=1$ 	

$\therefore t_5=1+4d$ 

$16=4d$

$d=4$

$t_n=1+4\left(n-1\right)$

This means the sequence goes $1, 5, 9, 13, 17$

• 17…81:

This is the arithmetic sequence at the bottom, so $t_1=17$ and $t_5=81$

The first term is 17 so $a=17$ 	

$\therefore t_5=17+4d$ 

$64=4d$

$d=16$

$t_n=17+16\left(n-1\right)$

This means the sequence goes $17, 33, 49, 65, 81$

• 25…81:

This is the arithmetic sequence at the right, so $t_1=25$ and $t_5=81$

The first term is 25 so $a=25$ 

$\therefore t_5=25+4d$ 

$56=4d$

$d=14$

$t_n=25+14\left(n-1\right)$

This means the sequence goes $25, 39, 53, 67, 81$

Now we have the following table:

1	7	13	19	25
5				39
9		X		53
13				67
17	33	49	65	81

We can either solve the vertical or the horizontal arithmetic sequence now, so arbitrarily we can go with the horizontal.

$t_1=13$ and $t_5=49$

The first term is 13 so $a=13$

$\therefore t_5=13+4d$

$36=4d$

$d=9$

$t_n=13+9\left(n-1\right)$

This means the sequence goes $13,22,31,40,49$

∴ $X$ is 31

Question 2 §

Question 2

For each positive integer $k$; let $a_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$: For example, $a_3$ is the sequence $1,4,7,...$ For how many values of $k$ does $a_k$ contain the term $2023$?

Let the nth term of the sequence $a_k$ be $2023$:

$a_n=1+k(n-1)$

$\therefore 2023=1+k(n-1)$

$2022=k(n-1)$

Now we have two terms, $k$ and $(n-1)$ whose product produces the number $2022$. By this logic, we should be able to list the factors of $2022$ to produce all the possible values of k.

$1\times 2022=2022$

$2\times 1011=2022$

$3\times 674=2022$

$6\times 337=2022$

$337\times 6=2022$

$674\times 3=2022$

$1011\times 2=2022$

$2022\times 1=2022$

Therefore there are $8$ possible values of $k$ that contain the term $2023$.

Question 3 §

Question 3

The terms of an arithmetic sequence add to $2023$. The first term of the sequence is increased by $1$, the second term is increased by $2$, the third term is increased by $3$, and so on. The terms of the new sequence add to $2176$. Find the middle term in the original sequence.

The new sequence will have $n$ added onto the end each time

For the sequence $1,2,3,4,5,\dots$

$S_n=\frac{n}{2}(2+(n-1))$

$S_n=\frac{n}{2}(n+1)$

$S_n=\frac{(n^2+n)}{2}$

The difference between $2023$ and $2176$ must equal $\frac{(n^2+n)}{2}$ so we get the equation:

$153=\frac{(n^2+n)}{2}$

$306=n^2+n$

$n^2+n-306=0$

$(n-17)(n+18)=0$

$n>0\therefore n=17$

This means there are $17$ terms in the sequence, and if the sum of all these terms is $2023$, the mean must be $\frac{2023}{17}=119$

Since the sequence has a “middle” term, there must be an odd number of terms in the sequence and this middle term must be equivalent to the mean.

∴ The middle term is $119$