The General Term using the Common Difference

Given u_1 and d

In the applet below, you are given the first term, and the common difference. You are then asked to calculate one of the other terms of the sequence.

Notice the calculation that you are repeating each time. Let’s call the first term u_1 and the common difference d. What calculation do you perform each time you wish to calculate

  1. u_3 ?
  2. u_7 ?
  3. u_n ?

The answer to the third question is the a formula that may be used to calculate the general term for an arithmetic sequence.


Example 1

An arithmetic sequence has first term u_1=8 and common difference 3. Find an expression to calculate any term of the sequence, u_n.

Solution

The first five terms of the sequence are 8,\,11,\,14,\,17,\,20,\dots

We know that u_n=u_1+(n-1)d. We can replace u_1 with 8 and we can replace d with 3.

    \[u_n=8+(n-1)\times3\]

It’s helpful to put the 3 before the brackets:

    \begin{align*}u_n&=8+3(n-1)\\&=8+3n-3\\&=8-3+3n\\&=5+3n\end{align*}

We can conclude that u_n=5+3n.

Now we can use this formula to find any term in the sequence: let’s find the fifth term. We replace the value n with 5.

u_5=5+3(5)=5+15 = 20.

We can confirm this calculation with the sequence written out above.


Example 2

An arithmetic sequence has first term u_1=20 and common difference -3. Find an expression to calculate any term of the sequence, u_n.

Solution

The first five terms of the sequence are 20,\,17,\,14,\,11,\,8,\dots

We know that u_n=u_1+(n-1)d. We can replace u_1 with 20 and we can replace d with -3.

    \[u_n=20+(n-1)\times -3\]

It’s helpful to put the (-3) before the brackets, remember to take the negative sign with you:

    \begin{align*}u_n&=20+(-3)(n-1)\\&=20-3(n-1)\\&=20-3n+3\\&=20+3-3n\\&=23-3n\end{align*}

We can conclude that u_n=23-3n.

Now we can use this formula to find any term in the sequence: let’s find the fourth term. We replace the value n with 4.

u_4=23-3(4)=23-12 = 11.

We can verify this calculation with the sequence written out above.


Example 3

A regular non-agon has the first angle positioned at 10 degrees, second angle at 50 degrees, etc.

(a) Write out the sequence of angles for all nine vertices.

(b) Calculate the nth term for 1\leq n \leq 9.

(c) Use the formula to verify that the last angle (vertex number 9) is at 330^{\circ}.

Solution:

(a) The angles are positioned at

    \[10^{\circ}, 50^{\circ}, 90^{\circ}, 130^{\circ}, 170^{\circ}, 210^{\circ}, 250^{\circ}, 290^{\circ}, 330^{\circ}\]

(b) The common difference is d=40.

    \begin{align*}u_n&=10+(40)(n-1)\\&=10+40n-40)\\&=-30+40n\\&=40n-30\end{align*}

(c) The formula says that u_n=40n-30 where in this case, n represents the number of the vertex.

At the ninth vertex, n=9. The formula says u_9=40(9)-30=360-30=330 as required.


Practice

Explore the Arithmetic Sequences problem sets on Transum Math.

Or, find the general term of the problems given here:

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