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SAT II Math I : Sequences

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Example Question #11 : Sequences

The first and third terms of a geometric sequence comprising only positive elements are $\sqrt{20}$ $\displaystyle \sqrt{20}$ and $\sqrt{180}$ $\displaystyle \sqrt{180}$, respectively. In simplest form, which of the following is its second term?

Possible Answers:

$\displaystyle 10$

$4 \sqrt{10}$ $\displaystyle 4 \sqrt{10}$

$2 \sqrt{15}$ $\displaystyle 2 \sqrt{15}$

None of these

$6 \sqrt{5}$ $\displaystyle 6 \sqrt{5}$

Correct answer:

$2 \sqrt{15}$ $\displaystyle 2 \sqrt{15}$

Explanation:

Let $\displaystyle r$ be the common ratio of the geometric sequence. Then

$a_{2} = r a_{1}$ $\displaystyle a_{2} = r a_{1}$

and

$a_{3} = r a_{2}= r \cdot r a_{1} = r ^{2}a_{1}$ $\displaystyle a_{3} = r a_{2}= r \cdot r a_{1} = r ^{2}a_{1}$

Therefore,

$r ^{2} = \frac{a_{3}}{a_{1}}$ $\displaystyle r ^{2} = \frac{a_{3}}{a_{1}}$,

Setting $a_{1}= \sqrt{20}, a_{3} = \sqrt{180}$ $\displaystyle a_{1}= \sqrt{20}, a_{3} = \sqrt{180}$, and applying the Quotient of Radicals Rule:

$r ^{2} = \frac{a_{3}}{a_{1}} = \frac{\sqrt{180}}{\sqrt{20}} = \sqrt{9} = 3$ $\displaystyle r ^{2} = \frac{a_{3}}{a_{1}} = \frac{\sqrt{180}}{\sqrt{20}} = \sqrt{9} = 3$

Taking the square root of both sides:

$r = \pm \sqrt{3}$ $\displaystyle r = \pm \sqrt{3}$

Substituting, and applying the Product of Radicals Rule:

$a_{2} = r a_{1}$ $\displaystyle a_{2} = r a_{1}$

$a_{2} = \pm \sqrt{3} \cdot \sqrt{20 } =\pm \sqrt{60 } = \pm \sqrt{4} \cdot \sqrt{15}= \pm 2 \sqrt{15}$ $\displaystyle a_{2} = \pm \sqrt{3} \cdot \sqrt{20 } =\pm \sqrt{60 } = \pm \sqrt{4} \cdot \sqrt{15}= \pm 2 \sqrt{15}$

Since all elements of the sequence are positive, $a_{2} = 2 \sqrt{15}$ $\displaystyle a_{2} = 2 \sqrt{15}$.

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SAT II Math I : Sequences

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Example Question #11 : Sequences

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