I'm having a problem with the question stated below. I stumbled upon it during my revision And I was hoping one of you guys could help me solve it and better yet Understand how to go about it.
**A geometric Progression has the first term a, common ratio r and sum to infinity 6. A second geometric Progression has the first term 2a, common ratio r^2 and sum to infinity 7. What are the values of a and r
Using the formula for infinite sums of geometric progressions you obtain the following system of equations:
$$\begin{cases} \frac{a}{1-r}=6\\ \frac{2a}{1-r^2}=7 \end{cases}$$
According to my calculations, the result is $r=\frac{5}{7}$ and $a=\frac{12}{7}$.
A geometric progression is a sequence of numbers related in some manner. The hints are "first term" and "common ratio".
This sequence has a series, the infinite form of which has a well known closed form. This is known as the geometric series, and if you are revising studies covering this, you should have encountered it in your notes.
$$\sum_{k=0}^\infty \alpha\cdotp \rho^k= \dfrac{\alpha}{1-\rho}\qquad\text{if }\lvert \rho\rvert<1~\vee~\alpha=0$$
so using the formula of sum of infinity... we will get 2 equation. i) 6 = a/1-r ii) 7 = 2a/1-r^2
then can do simultaneous equation but i did a much easier way by dividing equation ii) with equation i).....then 'a' will cancel each other and we will get an algebraic expression of 'r'---> r will have 2 values r=5/7 , r=1(ignored coz doesnt make sense XD) then substitute r to either of the i) or ii) to find a, there you goo. r = 5/7 , a = 12/7 ;D
