3

In a geometric series, the sum of the first three terms is $304$ and the sum of the first six terms is $1330$. Find the sum of the first seven terms.

$S_3=304$ $S_6=1330$

So how do I find $a$ and $r$? Is there any possible simple way to solve this? I'm so frustrated!

pancini
  • 19,216
user358060
  • 31

3 Answers3

2

HINT:

$S_3=\large \frac {a(1-r^3)}{(1-r)}=304$

and

$S_6=\large \frac {a(1-r^6)}{(1-r)}=1330$

2 equation, 2 unknown - Can you solve ahead?

MonK
  • 1,794
2

You know that a geometric series is of the form $a,ar,ar^2,\ldots$, so you have

$$a+ar+ar^2=m$$ and $$a+ar+ar^2+ar^3+ar^4+ar^5=n.$$ Then we have $$n=a+ar+ar^2+ar^3+ar^4+ar^5=m+r^3(a+ar+ar^2)=m+mr^3$$ or $$\frac{n-m}{m}=r^3$$ so now you know the value of $r$ (in the case, the numbers actually work out very nicely). Then there is only one unknown in the equation $$a+ar+ar^2=m$$ so you can solve for $a$. Once you know both, you can simply add up as many terms as you wish.

pancini
  • 19,216
1

Hint:

The sum $S_n$ of the first $n$ terms of the geometric series $a, ar, ar^2\dots$ is given by

$$S_n=a\frac{1-r^n}{1-r}$$

Now, write down the two equations given by your assignment. Which equations do you get?

5xum
  • 123,496
  • 6
  • 128
  • 204