# The first term of an arithmetic series is 4. The nineteenth term is four times larger than the fourth term, how do you find the common difference for this series and its thirtieth term?

Feb 21, 2016

See solution below.

#### Explanation:

Assuming x represents the common ratio, and y ${t}_{4}$.

$4 + x + x + x = y$

$4 + 18 x = 4 y$

Solve by substitution:

$4 + 18 x = 4 \left(3 x + 4\right)$

$4 + 18 x = 12 x + 16$

$18 x - 12 x = 16 - 4$

$6 x = 12$

$x = 2$

$4 + \left(2\right) + \left(2\right) + \left(2\right) = y$

$y = 10$

The common difference is 2. Now, we can use the formula ${t}_{n} = a + \left(n - 1\right) d$ to find ${t}_{30}$.

${t}_{30} = 4 + \left(30 - 1\right) 2$

${t}_{30} = 4 + 58$

${t}_{30} = 62$

The 30th term is 62.

Hopefully this helps!