# Thread: Arithmetic series

1. ## Arithmetic series

Struggling a little with this one.

The fourth term in an arithmetic series is 3k, where k is a constant and the sum of the first six terms of the series is 7k+9

a) Show that the first term of the series is 9-8k
b) Find an expression for the common difference of the series in terms of k.

Given that the seventh term of the series is 12, calculate:

c) The value of k
d) The sum of the first 20 terms of the series

Any help is appreciated, I think if someone just points me in the right direction I should be able to work it out.

2. Originally Posted by cosine
Struggling a little with this one.

The fourth term in an arithmetic series is 3k, where k is a constant and the sum of the first six terms of the series is 7k+9

a) Show that the first term of the series is 9-8k
b) Find an expression for the common difference of the series in terms of k.

Given that the seventh term of the series is 12, calculate:

c) The value of k
d) The sum of the first 20 terms of the series

Any help is appreciated, I think if someone just points me in the right direction I should be able to work it out.
a)
$\displaystyle t_n = a + (n-1)d$

$\displaystyle t_4 = 3k = a + 3d$

$\displaystyle S_n = \frac{n[2a + (n-1)d]}{2}$

$\displaystyle S_6 = 7k + 9 =3(2a + 5d)$.

So you have

$\displaystyle 3k = a + 3d$ and $\displaystyle 7k + 9 = 6a + 15d$

Solve them simultaneously.

Back substitute to answer b).

3. Lovely stuff, got there and wasn't sure if it would work. thank you very much