Question #730a0

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Question #730a0

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Answer 1 · 2017-03-30T18:19:59

$27$ $27$

Explanation:

Denote the $n$ $n$ :th element in the sequence $a_{n} = b x^{n - 1}$ $a_n = bx^(n-1)$ .
Then $a_{3} = b x^{2}$ $a_3=bx^2$ and $a_{6} = b x^{5}$ $a_6 = bx^5$ . Knowing that $a_{3} = 3$ $a_3 = 3$ and $a_{6} = \frac{1}{9}$ $a_6=1/9$ , we get that
$3 = b \cdot x^{2}$ $3 = b*x^2$
$\frac{1}{9} = b \cdot x^{5} = (b x^{2}) x^{3}$ $1/9 = b*x^5 = (bx^2)x^3$
Substituting the first line into the second we get that
$\frac{1}{9} = 3 x^{3}$ $1/9 = 3x^3$ , which gives that
$x = \frac{1}{3}$ $x = 1/3$ .
Inserting into the first equation, we get that
$b = \frac{3}{x^{2}} = 27$ $b = 3/(x^2) = 27$ .
Thus the first element $a_{1} = b x^{1 - 1} = b = 27$ $a_1 = bx^(1-1) = b = 27$ .

Answer 2 · 2017-03-30T18:24:00

$27 .$ $27.$

Explanation:

Let the Geom. Seq. be $a,ar, ar^2,...,ar^(n-1),...$

By what is given, $ar^2=3, and, ar^5=1/9.$

$:. (ar^5)/(ar^2)=1/9-:3$ .

$:. r^3=1/27$

$:. r=1/3.$

Now, $ar^2=3, r=1/3 rArr a=3-:r^2=3-:(1/3)^2=3-:1/9=27.$

$:." The First term is "27.$

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