Exercise 38 Page 205

Since we already know that the sequence is arithmetic, we also know that the difference between consecutive terms is constant. Let's subtract the second term from the first term to get this common difference

d .

d = \frac{4}{7} - \frac{3}{7} \Leftrightarrow d = \frac{1}{7}

Let's now recall the equation for the

n th

term of an arithmetic sequence.

a_{n} = a_{1} + (n - 1) d

Next, we can substitute the first term of the sequence

\frac{3}{7}

and the common difference

\frac{1}{7}

into the above formula. Let's do it!

a_{n} = a_{1} + (n - 1) d

SubstituteII

$a_{1} = \frac{3}{7}$ , $d = \frac{1}{7}$

a_{n} = \frac{3}{7} + (n - 1) \frac{1}{7}

▼

Simplify right-hand side

Distr

Distribute $\frac{1}{7}$

a_{n} = \frac{3}{7} + \frac{1}{7} n - \frac{1}{7}

SubFrac

Subtract fractions

a_{n} = \frac{1}{7} n + \frac{2}{7}

To find the value of

a_{10},

which is the

10 th

term, we can substitute

10

for

n

into the equation we have just obtained.

a_{n} = \frac{2}{7} + \frac{1}{7} n

Substitute

$n = 10$

a_{10} = \frac{2}{7} + \frac{1}{7} (10)

▼

Evaluate right-hand side

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

a_{10} = \frac{2}{7} + \frac{1 0}{7}

AddFrac

Add fractions

a_{10} = \frac{1 2}{7}

We found that the tenth term is

\frac{1 2}{7} .

Extra

To find the common difference, we calculated the difference between the second and the first terms. However, since in an arithmetic sequence the difference between consecutive terms is constant, we can use any two consecutive terms. For example, if we subtract the fourth term from the third term, we get the same common difference.

d = \frac{6}{7} - \frac{5}{7} \Leftrightarrow d = \frac{1}{7}