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

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

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

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

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

Can you complete it now?

so im gonna use simultaneous to get the answer ryt?

4. Hello erisse_satg17
Originally Posted by erisse_satg17

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

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

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

Can you finish it now?