# Thread: sum to infinity - GP

1. ## sum to infinity - GP

i cant understand this question... because i was away when they learn it so yea...

Find the 1st 3 terms of a geometric sequence, given that the sum of the 1st three terms is 76/3 and the sum to infinity is 36.

2. ## GP

Hello erisse_satg17
Originally Posted by erisse_satg17
i cant understand this question... because i was away when they learn it so yea...

Find the 1st 3 terms of a geometric sequence, given that the sum of the 1st three terms is 76/3 and the sum to infinity is 36.
Welcome to Math Help Forum.

This is what you need to know. In a GP:

• the first term is denoted by $\displaystyle a$
• the common ratio is denoted by $\displaystyle r$
• the first three terms are $\displaystyle a, ar, ar^2$
• the sum of the first $\displaystyle n$ terms is $\displaystyle \frac{a(1-r^n)}{1-r}$
• the sum to infinity is $\displaystyle \frac{a}{1-r}$, provided $\displaystyle |r| < 1$

So you'll need to plug in the values you've been given:

$\displaystyle \frac{a(1-r^3)}{1-r}=\frac{76}{3}$

and $\displaystyle \frac{a}{1-r}=36$

and solve for $\displaystyle a$ and $\displaystyle r$. (Don't forget to answer the question when you done this.)

Can you complete it now?

Grandad

3. thanks for your help...

so im gonna use simultaneous to get the answer ryt?

4. Hello erisse_satg17
Originally Posted by erisse_satg17
thanks for your help...

so im gonna use simultaneous to get the answer ryt?
Yes - and it's very easy. You can just get rid of the $\displaystyle \frac{a}{1-r}$ in the first equation, by using the second equation to replace it by $\displaystyle 36$. Like this:

$\displaystyle \frac{\color{red}a\color{black}(1-r^3)}{\color{red}1-r}=\frac{76}{3}$

$\displaystyle \Rightarrow \color{red}36\color{black}(1-r^3)=\frac{76}{3}$

Can you finish it now?

Grandad