# Trying to find $n$ in geometric progression.

I have the following information of a geometric progression ($r$ is positive):

$$S_n = 945$$ $$4(a_1 + a_2) = a_3 + a_4$$ $$S_{n-3} = 105$$

I need to find $n$.

The only thing I succeeded is to find $r$, the ratio. $$4S_2 = S_4 - S_2$$ $$5S_2 = S_4$$ $$5\left(\frac{a_1(r^2 -1)}{r-1}\right) = \frac{a_1(r^4 -1)}{r-1}$$ $$5a_1(r^2 - 1) = a_1(r^4-1)$$ $$5(r^2 - 1) = (r^2-1)(r^2+1)$$ $$5 = r^2+1$$ $$4 = r^2$$ $$r = 2$$ $r \neq -2$ because I wrote above that $r$ is positive.

What can I do more? I tried to solve some equations and to use the other information, but $a_1$ is never simplified, nor didn't find its value.

• what is $S_n$ here? Dec 25 '15 at 20:06
• @Dr.SonnhardGraubner The sum of the progression. $S_{n-3}$ is the sum of progression without counting the last three terms. Dec 25 '15 at 20:08

Using the fact that $S_n = 945$ and $r = 2$, we have:

$$S_n = \dfrac{a_1(r^n-1)}{r-1}$$

$$945 = \dfrac{a_1(2^n-1)}{2-1}$$

$$945 = a_1(2^n-1)$$

Similarly, using the fact that $S_{n-3} = 105$ and $r = 2$, we have:

$$S_{n-3} = \dfrac{a_1(r^{n-3}-1)}{r-1}$$

$$105 = \dfrac{a_1(2^{n-3}-1)}{2-1}$$

$$105 = a_1(2^{n-3}-1)$$

$$105 \cdot 2^3 = a_1(2^{n-3}-1) \cdot 2^3$$

$$840 = a_1(2^n-8)$$

You can use the equations $945 = a_1(2^n-1)$ and $840 = a_1(2^n-8)$ to solve for $a_1$ by subtracting the second equation from the first. Once you have $a_1$, solving for $n$ is easy.

When you subtract the two equations, you get $105 = 7a_1$, so $a_1 = 15$. Plugging that into the first equation, you get $945 = 15(2^n-1)$, i.e. $2^n-1 = 63$. So, the solution is $n = 6$.

• I have in my math book that $S_n = \frac{a_1(r^n - 1)}{r-1}$. Note that you wrote $r^{n+1}$. Dec 25 '15 at 20:19
• ^Thanks for catching that. I've fixed it. Dec 25 '15 at 20:22
• Seeing the fact that you multiplied by $2^3$ in one step was a good idea and from there I succeeded. Thanks. Dec 25 '15 at 20:31

HINT: using the equation $$4(a_1+a_2)=a_3+a_4$$ we get $$4(a_1+a_1q)=a_1q^2+a_1q^3$$ if $a_1\ne 0$ we get for $$q=-2,-1,2$$ can you proceed?

let's call the first term $a_1$.

The sum of the first $n$ terms of a geometric series with ratio $r$ is given by

$$S_n = \frac{a_1 (1-r^n)}{1-r}.$$

In your case $r=2$, and so

$$S_n = \frac{a_1 (1-2^n)}{1-2}= a_1 (2^n-1)=945.$$

And so

$$S_{n-3} = a_1 (2^{n-3}-1)=105.$$

Therefore

$$\frac{a_1 (2^n-1)}{a_1 (2^{n-3}-1)} = \frac {945} {105}$$

Notice that $a_1$ cancels and you get

$$\frac{2^n-1}{2^{n-3}-1} = 9.$$

Solve this equation for $n$. Hint: multiply both sides by $\frac 1 8$.