Arithmetic Sequences Examples

Write the first four terms of the arithmetic sequence with a₁ = 4 and d = 5.

To get from one term to the next we add d = 5. So

a₂ = a₁ + 5 = 9

a₃ = a₂ + 5 = 14

a₄ = a₃ + 5 = 19

We were given the first term, a₁ = 4. The first four terms are

4, 9, 14, 19.

If a₁ = 3 and d = 5, find a₅.

To get from a₁ to a₅ takes four steps.

Each step has size d = 5.

If we start at a₁ = 3 and take four steps of size d = 5, we land at

a₅ = a₁ + 4d

= 3 + 4(5)

= 23

If a₁ = 12 and d = -2, find a₁₁.

It takes 1 step to get to a₂, 2 steps to get to a₃, 3 steps to get to a₄, etc. It must take 10 steps to get to a₁₁.

Each step has size d = -2.

If we start at a₁ = 12 and take 10 steps of size d = -2, we land at

a₁₁ = a₁ + 10d

= 12 + 10(-2)

= -8.

An arithmetic sequence has a₅ = 11 and a₆ = 13. Find a₁ and d.

If this is an arithmetic sequence, then d = 2 because that's the size of the step from a₅ to a₆. We must have taken 4 steps to get from a₁ to a₅. Working backwards by stepping down 2 each time, we can fill in a₄, a₃, a₂, and finally a₁. We find that a₁ = 3.

Here's another way to do the example. Whatever a₁ is, we took 4 steps of size 2 to get to a₅.

In symbols,

a₁ + 4(2) = a₅.

Since we know a₅ = 11, we can solve this equation for a₁.

a₁ + 4(2) = 11

a₁ = 3

Thankfully, we got the same answer as before.

An arithmetic sequence has a₁₀ = 52 and a₁₁ = 56. Find a₁ and d.

Since this is an arithmetic sequence, the step size d is

a₁₁ – a₁₀ = 4.

We could work backwards and find a₉, a₈, etc., but that's more steps than we really want to do. Instead, let's just think about working backwards. Whatever a₁ is, we know it takes 9 steps of size 4 to get to a₁₀. In symbols,

a₁ + 9(4) = a₁₀.

Since we know a₁₀ = 52, we can solve this equation for a₁:

a₁ + 9(4) = 52

a₁ = 52 – 36

a₁ = 16

Find a₁ and d for the arithmetic sequence with a₂ = 8 and a₄ = 28

To get from a₂ to a₄ takes two steps.

The difference between a₄ and a₂ is

a₄ – a₂ = 28 – 8 = 20.

In order for this difference of 20 to be split between two equally-sized steps, each step must have size 10.

This means d = 10.

Now we proceed as before. There's one step from a₁ to a₂. If d = 10 and a₂ = 8, then

a₁ + 1d = a₂

a₁ + (10) = 8

a₁ = -2.

Find a₁ and d for the arithmetic sequence with a₅₀ = 14 and a₁₀₀ = -136.

There are

100 – 50 = 50

steps from a₅₀ to a₁₀₀. This means the difference

a₁₀₀ – a₅₀ = -136 – 14 = -150

must be divided into 50 separate steps. We find

Then we proceed as usual. We have 49 steps from a₁ to a₅₀, so

a₁ + 49d = a₅₀

a₁ + 49(-3) = 14

a₁ = 161.

To summarize, if we're given two terms a_m and a_n that aren't consecutive, we

1) find a_n – a_m

2) divide by the number of steps (n – m) to get d

3) proceed as before to find a₁.