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.
Hello erisse_satg17Welcome 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
Hello erisse_satg17Yes - 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