0
$\begingroup$

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

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Do you remember the definition for "geometric progression" -- or "geometric sequence" as some do call it? $\endgroup$ – Graham Kemp Jun 24 '18 at 6:21
  • $\begingroup$ Yes I do remember $\endgroup$ – RodrigoArts Jun 24 '18 at 6:23
  • $\begingroup$ Demonstrate.... $\endgroup$ – Graham Kemp Jun 24 '18 at 6:23
  • $\begingroup$ 1,2,3..... (To infinity) $\endgroup$ – RodrigoArts Jun 24 '18 at 6:25
  • $\begingroup$ @RodrigoArts A geometric progression is of form $a,ar,ar^2,...$ your example is an arithmetic progression $\endgroup$ – user428700 Jun 24 '18 at 6:26
1
$\begingroup$

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 $a=\frac{12}{7}$.

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

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.

$$\sum_{k=0}^\infty \alpha\cdotp \rho^k= \dfrac{\alpha}{1-\rho}\qquad\text{if }\lvert \rho\rvert<1~\vee~\alpha=0$$

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.