# Thread: Sequence and series question

1. ## Sequence and series question

I am given the common difference/ratio and tn = 48 and Sn = 33, I need to find the first term but I am not given any terms how do I do this?

2. ## Re: Sequence and series question

The common difference or the common ratio? I would need to know to determine if this is arithmetic or geometric!

3. ## Re: Sequence and series question

Hello, Cryptnar!

I've made some progress . . .

$\text{I am given the common difference/ratio and }t_n = 48\text{ and }S_n = 33.$

$\text{I need to find the first term, but I am not given any terms. \; How do I do this?}$

Suppose we have a Geometric Progression.

We are given: . $\begin{Bmatrix}t_n &=& ar^{n-1} &=& 48 & [1] \\ \\[-3mm] S_n &=& a\dfrac{1-r^n}{1-r} &=& 33 & [2] \end{Bmatrix}$

From [2]: . $a(1-r^n) \:=\:33(1-r) \quad\Rightarrow\quad a - ar^n \:=\:33 - 33r\;[3]$

Multiply [1] by $r\!:\;ar^n \:=\:48r$

Substitute into [3]: . $a - 48r \:=\:33 - 33r$

. . Therefore: . $a \:=\:33 + 15r\;\hdots\;\text{ and we are given }r.$

4. ## Re: Sequence and series question

Originally Posted by Soroban
Hello, Cryptnar!

I've made some progress . . .

Suppose we have a Geometric Progression.

We are given: . $\begin{Bmatrix}t_n &=& ar^{n-1} &=& 48 & [1] \\ \\[-3mm] S_n &=& a\dfrac{1-r^n}{1-r} &=& 33 & [2] \end{Bmatrix}$

From [2]: . $a(1-r^n) \:=\:33(1-r) \quad\Rightarrow\quad a - ar^n \:=\:33 - 33r\;[3]$

Multiply [1] by $r\!:\;ar^n \:=\:48r$

Substitute into [3]: . $a - 48r \:=\:33 - 33r$

. . Therefore: . $a \:=\:33 + 15r\;\hdots\;\text{ and we are given }r.$
Hey, I don't really understand the a-arn thing what does the "a-" represent?

5. ## Re: Sequence and series question

It's the first term in the sequence