# Find, in terms of k... [Arithmetic series]

• May 5th 2009, 03:55 AM
Lonehwolf
Find, in terms of k... [Arithmetic series]
The first term of an arithmetic series is (5+3k) where k is a positive integer. The last term is 17(1+k) and the common difference is 2. Find, in terms of k,

(i) the number of terms,
(ii) the sum of the series.

I'm stuck as early as the starting point, I'm trying to go over all of my notes but I can't seem to find anything that discusses first and last terms. I do recall having done these in the first year in college, but the notes are gone (Worried)

Can someone help me through this one please?
• May 5th 2009, 06:34 AM
mr fantastic
Quote:

Originally Posted by Lonehwolf
The first term of an arithmetic series is (5+3k) where k is a positive integer. The last term is 17(1+k) and the common difference is 2. Find, in terms of k,

(i) the number of terms,
(ii) the sum of the series.

I'm stuck as early as the starting point, I'm trying to go over all of my notes but I can't seem to find anything that discusses first and last terms. I do recall having done these in the first year in college, but the notes are gone (Worried)

Can someone help me through this one please?

$t_n = t_1 + (n-1)d \Rightarrow 17(1 + k) = (5 + 3k) + 2(n-1)$.

Solve for n in terms of k.

Now use the usual formula for the sum of an arithmetic series.
• May 5th 2009, 06:38 AM
red_dog
$a_1=5+3k, \ a_n=17(1+k), \ r=2$

$a_n=a_1+(n-1)r\Rightarrow 17+17k=5+3k+2n-2\Rightarrow n=7k+7$

$S=\frac{(a_1+a_n)n}{2}$

Replace $a_1, \ a_n, \ n$ and find S.
• May 6th 2009, 03:45 AM
Lonehwolf