From the exercise, we have been given the first and seventh terms.

t (1) t (7) = 8 = 512 \Rightarrow (1, 8) \Rightarrow (7, 512)

Arithmetic sequence

In an arithmetic sequence we have a common difference between consecutive terms. Therefore, if this is an arithmetic sequence we should be able to add a common difference six times to the first term and get a result of $512,$ our seventh term.

From the diagram, we can create the following equation

8 + 6 d = 512

Let's solve this equation.

8 + 6 d = 512

SubEqn

$LHS - 8 = RHS - 8$

6 d = 504

DivEqn

$LHS / 6 = RHS / 6$

d = 84

As we can see, the sequence can be arithmetic. So far we can write our arithmetic sequence as

t (n) = 84 n + t (0) .

To complete the equation we need the zero term,

t (0) .

We can find this by substituting any of the know terms into our formula and solving for

t (0) .

t (n) = 84 n + t (0)

Substitute

$n = 1$

t (1) = 84 (1) + t (0)

Substitute

$t (1) = 8$

8 = 84 (1) + t (0)

▼

Solve for

t (0)

MultByOne

$a \cdot 1 = a$

8 = 84 + t (0)

SubEqn

$LHS - 84 = RHS - 84$

- 72 = t (0)

RearrangeEqn

Rearrange equation

t (0) = - 72

The arithmetic sequence is

t (n) = 84 n - 72 .

Geometric sequence

In a geometric sequence we have a common ratio $r$ between consecutive terms. Therefore, if this is a geometric sequence we should be able to multiply the first term by a common ratio six times and get a result of $512,$ our seventh term.

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