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1) The sequence a_1, a_2, a_3, ... satisfies a_1 = 19, a_9 = 99, and for all n$$\ge$$3, a_n is the arithmetic mean of the first n - 1 terms. Find a_2.

2) Let a, b, and c be the roots of x^3 - 5x + 7 = 0.Find the monic polynomial, in x, whose roots are a - 2, b - 2, and c - 2.

3) Let a, b, c be positive real numbers such that $$\log_a b + \log_b c + \log_c a = 0.$$ Find $$(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3.$$

Jul 18, 2019

#1
+24960
+2

3)

Let a,b,c be positive real numbers such that $$\log_a b + \log_b c + \log_c a = 0$$.
Find $$(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3$$.

$$\text{Let \log_a b = \mathbf{x} } \\ \text{Let \log_b c = \mathbf{y} } \\ \text{Let \log_c a = \mathbf{z} }$$

$$\begin{array}{|rcll|} \hline \log_a b + \log_b c + \log_c a &=& 0 \\ \mathbf{x+y+z} &=& \mathbf{0} \qquad (1) \\ \hline \end{array}$$

$$\begin{array}{|rclrcl|} \hline \log_a b &=&\dfrac{\log_c b}{\log_c a} \quad&| \quad \log_c b &=&\dfrac{\log_b b}{\log_b c} \\\\ \log_a b &=&\dfrac{\log_b b}{\log_b c\log_c a} \quad&| \quad \log_b b = 1 \\\\ \log_a b &=&\dfrac{1}{\log_b c\log_c a} \\\\ \log_a b\log_b c\log_c a &=& 1 \\\\ \mathbf{xyz} &=& \mathbf{1} \qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (x+y+z)^3 &=& (x+y+z)^2(x+y+z) \\ &=& \left(x^2+y^2+z^2+2(xy+yz+xz)\right)(x+y+z) \\ &=& (x^2+y^2+z^2)(x+y+z)+ 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +x^2y+x^2z+y^2x+y^2z+z^2x+z^2y + 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +(x+y+z)(xy+yz+xz) -3xyz + 2(x+y+z)(xy+yz+xz) \\\\ (x+y+z)^3&=& x^3+y^3+z^3 -3xyz + 3(x+y+z)(xy+yz+xz) \\ \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \qquad (3) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \\ && \boxed{x+y+z = 0} \\ x^3+y^3+z^3 &=& 0^3-3\cdot 0\cdot (xy+yz+xz)+3xyz \\ x^3+y^3+z^3 &=& 3xyz \\ && \boxed{xyz = 1} \\ x^3+y^3+z^3 &=& 3\cdot 1 \\ x^3+y^3+z^3 &=& 3 \\ \mathbf{(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3} &=& \mathbf{3} \\ \hline \end{array}$$

Jul 18, 2019
#5
+111328
+2

Thanks, heureka......I really liked this problem and your solution  !!!!!

CPhill  Jul 19, 2019
#6
+24960
+1

Thank you, CPhill !

heureka  Jul 19, 2019
#2
+24960
+3

1)
The sequence $$a_1, a_2, a_3,\ \ldots$$ satisfies $$a_1 = 19,\ a_9 = 99$$, and for all $$n \ge 3$$,

$$a_n$$ is the arithmetic mean of the first $$n - 1$$ terms.
Find a_2.

I assume:

$$\begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_1+a_2}{2} &\text{or }\qquad 2a_3 = a_1+a_2 \\ \hline a_4 &=& \dfrac{a_1+a_2}{3} + \dfrac{a_3}{3} &\text{or }\qquad 3a_4 = a_1+a_2+a_3 \\\\ &=& \dfrac{2a_3}{3} + \dfrac{a_3}{3} \\\\ &=& \dfrac{3a_3}{3} \\\\ \mathbf{a_4} &=& \mathbf{a_3} \\ \hline a_5 &=& \dfrac{a_1+a_2+a_3}{4} + \dfrac{a_4}{4} &\text{or }\qquad 4a_5 = a_1+a_2+a_3+a_4 \\\\ &=& \dfrac{3a_4}{4} + \dfrac{a_4}{4} \\\\ &=& \dfrac{4a_4}{4} \\\\ \mathbf{a_5} &=& \mathbf{a_4} \\ \hline a_6 &=& \dfrac{a_1+a_2+a_3+a_4}{5} + \dfrac{a_5}{5} &\text{or }\qquad 5a_6 = a_1+a_2+a_3+a_4+a_5 \\\\ &=& \dfrac{4a_5}{5} + \dfrac{a_5}{5} \\\\ &=& \dfrac{5a_5}{5} \\\\ \mathbf{a_6} &=& \mathbf{a_5} \\ \hline \ldots \\ a_3&=&a_4=a_5=a_6=a_7=a_8=a_9 \\ a_3 &=& a_9 \quad | \quad a_3 = \dfrac{a_1+a_2}{2},\ a_9 = 99 \\\\ \dfrac{a_1+a_2}{2} &=& 99 \quad | \quad a_1 = 19 \\\\ \dfrac{19+a_2}{2} &=& 99 \quad | \quad \cdot 2 \\\\ 19+a_2 &=& 198 \\\\ a_2 &=& 198-19 \\ \mathbf{ a_2 } &=& \mathbf{179} \\ \hline \end{array}$$

The sequence: $$19,\ 179,\ 99,\ 99,\ 99,\ 99,\ 99,\ 99,\ 99,\ \ldots$$

Jul 18, 2019
#3
+24960
+2

2)
Let a, b, and c be the roots of $$x^3 - 5x + 7 = 0$$.
Find the monic polynomial, in $$x$$, whose roots are $$a - 2$$, $$b - 2$$, and $$c - 2$$.

$$\begin{array}{|lrcll|} \hline \text{Let a, b, and c be the roots} \\ & x^3 - 5x + 7 &=& (x-a)(x-b)(x-c) \\\\ \text{vieta:} \\ \text{coefficient}\ x^2: & -(a+b+c) &=& 0 \\ \text{coefficient}\ x^1: & ab+bc+ac &=& -5 \\ \text{coefficient}\ x^0: & -abc &=& 7 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{Let a-2, b-2, and c-2 be the roots} \\ x^3+Ax^2+Bx+C &=& \Big(x-(a-2)\Big)\Big(x-(b-2)\Big)\Big(x-(c-2)\Big) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \Big(x-(a-2)\Big)\Big(x-(b-2)\Big)\Big(x-(c-2)\Big) \\ &=& -a b c + a b x + 2 a b + a c x + 2 a c - a x^2 - 4 a x - 4 a + b c x \\ && + 2 b c - b x^2 - 4 b x - 4 b - c x^2 - 4 c x - 4 c + x^3 + 6 x^2 + 12 x + 8 \quad | \quad \text{Expanded form} \\ &=& x^3+x^2\Big(6-(a+b+c)\Big)+x\Big(ab+bc+ac-4(a+b+c)+12\Big)-abc+2(ab+bc+ac)-4(a+b+c)+8 \\ && \boxed{ a+b+c=0\\ ab+bc+ac=-5 \\ -abc=7 } \\ &=& x^3+x^2(6-0)+x(-5-0+12)+7+2(-5)-0+8 \\ &=& \mathbf{x^3+6x^2+7x+5} \\ \hline \end{array}$$

The monic polynomial, in $$x$$, whose roots are $$a - 2$$, $$b - 2$$, and $$c - 2$$ is $$\mathbf{x^3+6x^2+7x+5}$$

Jul 18, 2019
#4
+24960
+2

2)
Let a, b, and c be the roots of $$x^3 - 5x + 7 = 0$$.
Find the monic polynomial, in $$x$$, whose roots are $$a - 2$$, $$b - 2$$, and $$c - 2$$.

$$\begin{array}{|rcll|} \hline && x'^3 - 5x' + 7 \quad | \quad x'=x+2 \quad \text{moving the polynomial by two units} \\ && (x+2)^3 - 5(x+2) + 7 \\ &=& x^3+3x^2\cdot 2+3x\cdot 2^2+2^3-5x-10+7 \\ &=& x^3+ 6x^2+7x+5 \\ \hline \end{array}$$

Jul 18, 2019