Arithmetic Sequences

We take the following arithmetic sequence $2,4,6,8,10...$. Notice that when adding 2 to any term, we will get the next term. So, the common difference between every two successive terms is 2.

Another example is $5,8,11,14,...$ The difference between every two consecutive terms in this sequence is 3.

In the sequence $5,8,11,14,17$, the first term is 5 because it is the first number of the sequence. The first term is usually denoted by $a$.

For the sequence $2,4,6,8,...$ the common difference is 2 throughout because

$8-6=2,6-4=2,4-2=2$.

This is an arithmetic sequence.

For the sequence $,-1,2,4,8,10,...$ the common difference is not constant because

$10-8=2,8-4=4,4-2=2,2-(-1)=2+1=3$

This is not an arithmetic sequence.

Having the first term of an arithmetic sequence as 6 and the common difference as 3, then the sequence will be

$6,9,12,15,...$

Find the next three terms of the arithmetic sequence below.

$4,7,10,13,16,...$

Solution

First, we identify the first term which is 4.

Next, we find the common difference by subtracting two consecutive terms from the sequence.

$7-4=3,10-7=3$

Now that we know the first term and the common difference, we can find the next three terms of the sequence by adding the common difference to the last term in the sequence.

The last known term in the sequence is 16. So, we will add 3 to 16, to get

$16+3=19$

We are to look for the next three terms, so we would do this two more times. The last term is no longer 16 but 19. So, we would add 3 to 19 to get,

$19+3=22$

We will now add 3 to 22 to get,

$22+3=25$

So, the next three terms are 19, 22 and 25.

The arithmetic sequence will be $4,7,10,13,16,19,22,25$,....

Find the next three terms of the arithmetic sequence

$5,3,1,-1,-3,...$

Solution

To solve this, we must know the first term and the common difference. The first term is 5 and the common difference is -2.

We get the common difference by subtracting two consecutive numbers in the sequence to get the difference,

$3-5=-2,1-3=-2$.

Therefore,

$d=-2$.

To get the next three terms, we will add the common difference to the last term.

The last known term is -3. So, the next term will be,

$(-3)+(-2)=-5$.

We are to get three terms, so we will repeat this step twice again.

The last term is now -5. So, the next term will be,

$(-5)+(-2)=-7$.

The next term will be,

The next three terms of the arithmetic sequence are thus $-5,-7,-9$.

Therefore, the arithmetic sequence is: $5,3,1,-1,-3,-5,-7,-9$,....

Find the third term of the arithmetic sequence below,

$7,14,\_\_,28,35$.

Solution

We are asked to find the third term of the sequence. We will use the formula below.

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

Let us define all the parameters of the formula,

$$\begin{gathered} a_n=? \\ a=7(thefirsttermofthesequence) \\ n=3(becausewearelookingforthethirdterm) \\ d=7(because14-7=7,35-28=7) \end{gathered}$$

Now, let's substitute into the formula,

$a_3=a_1+(3-1)d=7+(3-1)7=7+\left(2\right)7=7+14=21$

Therefore, the third term is 21.

Find the missing terms in the arithmetic sequence below.

.$\_\_,\_\_,-20,-12,-4$

Solution

We are asked to find the missing terms in the sequence and usually, we use the first term in the sequence to do this. In this case, the first term is one of the missing terms. So how would we go about this?

We will first find the common difference and think of a way to use it to find the missing terms.

To get the common difference, we subtract two consecutive terms.

$d=(-4)-(-12)=8$.

We can’t use the formula we used in the previous examples because of the absence of the first term. So, let’s look for a pattern we can follow. Let’s label the first term as $x$ and the second as $y$.

So the sequence will be

$x,y,-20,-12,-4$

The relationship between each term is the common difference. If you find the difference between any two consecutive terms, you will get the same number.

Thus, if

$(-20)-y=8;y-x=8$

We take the first equation,

$(-20)-y=8$

We will solve for y by collecting like terms and making it the subject of the formula.

$y=-20-8=-28$

This means that the second term is $-28$.

Recall, $y-x=8$ We now know the value of $y$ as $-28$, thus

$$\begin{gathered} -28-x=8 \\ -x=8+28=36 \\ x=-36 \end{gathered}$$

This means the first term is $-36$.

So, the sequence is $-36,-28,-20,-12,-4$.

Find the 17^th term of the arithmetic sequence below,

$2,8,14,20,26,...$.

Solution

To find the 17^th term of the sequence, we will have to use the formula below.

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

$$\begin{gathered} a_n=? \\ a=2(thefirstterm) \\ n=17(becausewearelookingforthe17thterm) \end{gathered}$$

Let's find the common difference d.

$d=26-20=6$

Now we will find the 17^th term,

$a_{17}=2+(17-1)6=2+\left(16\right)6=2+96=98$

The 17^th term is 98.

Find the sum of the arithmetic sequence $5,9,13,17,...$ up to 10 terms.

Solution

From the question, we are given the first term which is 5 but we do not know the last term. But we know that the number of terms is 10. The formula to use when the last term is not known is.

$S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

$a=5,n=10$

The common difference is,

Now, we will substitute the values in the formula, $S_n=\frac{10}{2}\left[2\left(5\right)+(10-1)4\right]=5\left[10+\left(9\right)4\right]=5\left[10+36\right]=5\left[46\right]=230$

The sum of the terms is 230.

Find the sum of an arithmetic progression of 4 terms whose first term is 4 and its last term is 19.

Solution

Here, we know what the first and the last term are. The formula to use in this case is,

$S_n=\frac{n}{2}\left[a+a_n\right]$

$n=4,a=4,a_4=19$

We will now substitute the values in the formula,

$S_n=\frac{4}{2}\left[4+19\right]=2\left[23\right]=46$

The sum of the 4 terms is 46.