The second term of a geometric series is 24 and the fifth term is 3.
a. Show that the common ratio of the series is 1/2.
b. Find the first term of the series.
c. Find the sum to infinity of the series.
That's pretty straightforward- if you know the definitions. The "nth" term of a geometric series with first term a and common ratio r is $\displaystyle ar^{n-1}$. Knowing that the second term is 24 tells you that $\displaystyle ar^{2-1}= ar= 24$ and knowing that the 5th term is 3 tells you that $\displaystyle ar^{5-1}= ar^4= 3$. That gives you two equations to solve for a and r. In particular, if you divide the second equation by the first,
$\displaystyle \frac{ar^4}{ar}= r^3= \frac{3}{24}= \frac{1}{8}$
Once you know r, it is easy to use $\displaystyle ar= 24$ to find a.
And, once you know both a and r, the "sum to infinity" is just
$\displaystyle \frac{a}{1- r}$
Hi ansonbound,
____, 24, ____, ____, 3
Using $\displaystyle a_n=a_1 \cdot r^{n-1}$, we can verify that the common ratio is 1/2
$\displaystyle a_n=3$
$\displaystyle a_1=24$
$\displaystyle n=4$
$\displaystyle 3=24 \cdot r^{4-1}$
$\displaystyle \frac{1}{8}=r^3$
$\displaystyle r=\frac{1}{2}$
Working to the left of 24, the first term if found by multiplying 24 by 2. Thus, the first term is 48.
The sum S of an infinite geometric series with -1 < r < 1 is given by:
$\displaystyle S=\frac{a_1}{1-r}$
Since $\displaystyle r=\frac{1}{2}$, and since $\displaystyle \left|\frac{1}{2}\right| <1$, the sum exists.
$\displaystyle S=\frac{48}{1-\frac{1}{2}}$
$\displaystyle S=\frac{48}{\frac{1}{2}}$
$\displaystyle S=96$
THANKS SO MUCH!!
But im stuck on this question now... only b)
http://www.mathhelpforum.com/math-he...rst-terms.html