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Q. A geometric series has first term $4$ and common ratio $r$. Where $0 < r < 1$

The first, second and fourth terms of this geometric series form three successive terms of an arithmetic series.

The sum of infinity of the geometric series is.

A. $\frac12 (\sqrt5 - 1)$ B. $2 (3 - \sqrt5)$ C. $2 (1 + \sqrt5)$ D. $2 (3 + \sqrt5)$

Can anyone help me with the answer and explain the method used?

I am really stuck any help would be really appreciated.

Thanks in advance.

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Kathryn Stiff-Tompkins
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1 Answers1

1

Let the geometric series be: $a, ar, ar^2, ar^3$,...

The terms 1, 2 and 4 form an arithmetic series: $a, ar, ar^3$

since the difference between every 2 consecutive terms in arithmetic series is constant:

$ar-a=ar^3-ar$

$ar^3-2ar+a=0$

$a(r-1)(r^2 + r-1)=0$

this leads to 3 possible values of r (assuming $a$ is not zero)

(a) $r=1$. This value leads to a series with infinite sum. Ignore $r=1$.

(b) $r=(-1-\sqrt5)/2$. This is a negative value violating the condition $r>0$.

(c) $r=(-1+\sqrt5)/2$. This value is about .62 and is positive and less than 1, so it satisfies the constraint on r.

now since you know the value of $r$ and you know the sum of an infinite series such as: $a, ar, ar^2,...$ to be:

$S=a/(1-r)$

you can plug in the value of $r$ from (c) above to get:

$S= \frac{a}{1-(\frac{-1+\sqrt5}{2})}$

$S=\frac{2a}{3-\sqrt 5}$

Since a=4; $S=\frac{8}{3-\sqrt 5}$

$S=S \frac{3+\sqrt 5}{3+\sqrt 5}$

$S=\frac{8(3+\sqrt 5)}{9-5}=2(3+\sqrt 5)$

So the correct answer is (D).

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