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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

Note by Emmanuel John Baliwag
11 months, 2 weeks ago

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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 \(ar^2\). Thomas Jones · 11 months, 2 weeks ago

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@Thomas Jones This is a very late response from me but, thank you very much! Emmanuel John Baliwag · 8 months, 2 weeks ago

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