1. ## Geometric Progression

A geometric progression has first term and common ratio both equal to a, a>1. Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms, find the exact value of a.

$\displaystyle S_{12}=28S_6$

$\displaystyle \frac{a(1-r^{12})}{1-r}=28\frac{a(1-r^6)}{1-r}$

$\displaystyle 1-r^{12}=28(1-r^6)$

$\displaystyle r^{12}-28r^6+27=0$

I would have to plot a graph to find the answer, but this would not give me an exact answer. How could I solve it in a way which would give me an exact answer?

2. Originally Posted by Punch
A geometric progression has first term and common ratio both equal to a, a>1. Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms, find the exact value of a.

$\displaystyle S_{12}=28S_6$

$\displaystyle \frac{a(1-r^{12})}{1-r}=28\frac{a(1-r^6)}{1-r}$

$\displaystyle 1-r^{12}=28(1-r^6)$

$\displaystyle r^{12}-28r^6+27=0$

I would have to plot a graph to find the answer, but this would not give me an exact answer. How could I solve it in a way which would give me an exact answer?
The equation is quadratic in form and can we factored as follows

$\displaystyle (r^6)^2-28r^6+27=0 \iff (r^6-1)(r^6-27)=0$

Note that

$\displaystyle r^6=(r^2)^3$

Use the difference of cubes...

Can you finish from here?

3. Originally Posted by TheEmptySet
The equation is quadratic in form and can we factored as follows

$\displaystyle (r^6)^2-28r^6+27=0 \iff (r^6-1)(r^6-27)=0$

Note that

$\displaystyle r^6=(r^2)^3$

Use the difference of cubes...

Can you finish from here?
could there be another way to complete this? because difference of cubes is something i have not learnt

4. Originally Posted by Punch
could there be another way to complete this? because difference of cubes is something i have not learnt
Then you can consider it something new you're learning now.

Sum of Two Cubes: $\displaystyle \displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.

Difference of Two Cubes: $\displaystyle \displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

5. okay, i have solved the first part of the question and now comes another part which i can't solve:

Hence evaluate $\displaystyle log_{3}(\frac{3}{2}a^2+a^4+a^6+....+a^{58})$, giving your answer in the form $\displaystyle A-log_3B$, where $\displaystyle A$ and $\displaystyle B$ are positive integers to be determined.