Sum of an Infinite Geometric Sequence

The sum of an infinite number of terms in a decreasing geometric sequence is 3, and the sum of their squares is also 3. Find the first three terms of the series.

How do you solve such problem? Can someone please write a solution asap? Just really need this for a friend :3

#Algebra #Asap

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

First, derive the formula for the sum to infinity of a geometric sequence (for $|r| < 1$ ). Let $S_\infty$ be the sum to infinite terms of the sequence, $S^{2}_{\infty}$ be the sum to infinite squared terms of the sequence, $a$ be the first term, and $r$ be the common ratio.

$S_\infty = a + ar + ar^2 + ar^3 + \cdots$

$r S_\infty = ar + ar^2 + ar^3 + ar^4 \cdots$

$S_\infty - r S_\infty = a + ar - ar + ar^2 - ar^2 \cdots$

$S_\infty - r S_\infty = a$

$S_\infty (1 - r) = a$

$S_\infty = \frac{a}{1-r}$

Memorising this result will occasionally be helpful, and you don't need to derive it in your friends homework, but there's the derivation up there in case you want to see where it comes from.

(Or if you forget the formula, like I just did then. This may or may not be the reason I put the derivation in there.)

When the terms are squared, both a and r are squared separately, so,

${S}^{2}_{\infty} = \frac{a^2}{1-r^2}$

And, consequently,

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

From this it shouldn't be hard to find the first three terms of the sequence, which are $a$ , $ar$ , and $ar^2$ .

A Former Brilliant Member - 5 years, 8 months ago

This is a very late response from me but, thank you very much!

Emmanuel John Baliwag - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$