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Geometric Sequence...

rsyed5

New member
Aug 15, 2013
5
I have no idea how to solve this equation, its in my homework... i know the formula to find the nth term(tn=ar^n-1) but dont know how to solve this:

The difference between the first term and second term in a geometric sequence is 6.The difference between the second term and the third term is 3. The difference between the third term and the fourth term is 3/2. Find the nth term in the sequence....

Thanks in advance:)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We may state:

\(\displaystyle a-ar=6\)

\(\displaystyle ar-ar^2=3\)

We now have two equations and two unknowns. I suggest solving the first equation for $r$, then substitute into the second to get an equation in $a$ only, which you can then solve. Once you have determined the value of $a$, then use that in your expression for $r$ in terms of $a$ to get the value of $r$. Then use:

\(\displaystyle t_n=ar^{n-1}\)

for the $n$th term. :D
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have no idea how to solve this equation, its in my homework... i know the formula to find the nth term(tn=ar^n-1) but dont know how to solve this:

The difference between the first term and second term in a geometric sequence is 6.The difference between the second term and the third term is 3. The difference between the third term and the fourth term is 3/2. Find the nth term in the sequence....

Thanks in advance:)
If the general term is $\displaystyle t_{n}= a\ r^{n-1}$ You have two unknown variables a and r and three equations...

$\displaystyle a\ (1-r)=6$

$\displaystyle a\ r\ (1-r)=3$

$\displaystyle a\ r^{2}\ (1-r)=\frac{3}{2}$

... so that the problem is overdimensioned. In this case the solution $\displaystyle a=12,\ r= \frac{1}{2}$ satisfies all the three equations, but in general for an overdimensioned problem an 'exact' solution doesn't exist...


Kind regards


$\chi$ $\sigma$