Recursive and Explicit Formulas from: category_eng

The second and fourth terms of a geometric sequence are $2$ and $6$ . Which of the following is a possible first term?

$extbf{(A) } -sqrt{3} qquad extbf{(B) } -frac{2sqrt{3}}{3} qquad extbf{(C) } -frac{sqrt{3}}{3} qquad extbf{(D) }...$

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The second and fourth terms of a geometric sequence are $2$ and $6$ . Which of the following is a possible first term?

$extbf{(A) } -sqrt{3} qquad extbf{(B) } -frac{2sqrt{3}}{3} qquad extbf{(C) } -frac{sqrt{3}}{3} qquad extbf{(D) }...$

A sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?

$ext {(A)} 1 qquad ext {(B)} 4 qquad ext {(C)} 36 qquad ext {(D)} 49 qquad ext {(E)} 81$

Positive integers $a$ , $b$ , and $2009$ , with $a<b<2009$ , form a geometric sequence with an integer ratio. What is $a$ ?

$mathrm{(A)} 7qquadmathrm{(B)} 41qquadmathrm{(C)} 49qquadmathrm{(D)} 289qquadmathrm{(E)} 2009$

Consider the set of numbers ${1, 10, 10^2, 10^3, ldots, 10^{10}}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?

$extbf{(A)} 1 qquad extbf{(B)} 9 qquad extbf{(C)} 10 qquad extbf{(D)} 11 qquad extbf{(E)} 101$

1.

Let the first term be $a$ and the common difference be $r$ . Therefore,

$ar=2 (1) qquad ext{and} qquad ar^3=6 (2)$

Dividing $(2)$ by $(1)$ eliminates the $a$ , yielding $r^2=3$ , so $r=pmsqrt{3}$ .

Now, since $ar=2$ , $a=frac{2}{r}$ , so $a=frac{2}{pmsqrt{3}}=pmfrac{2sqrt{3}}{3}$ .

We therefore see that $oxed{ extbf{(B)} -frac{2sqrt{3}}{3}}$ is a possible first term.

3.

Let $d$ be the common difference. Then $9, 9 + d + 2 = 11 + d, 9 + 2d + 20 = 29 + 2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1 mathrm{(A)}$ .

4.

The prime factorization of $2009$ is $2009 = 7cdot 7cdot 41$ . As $a<b<2009$ , the ratio must be positive and larger than $1$ , hence there is only one possibility: the ratio must be $7$ , and then $b=7cdot 41$ , and $a=41Rightarrow ext{(B)}$ .

5.