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A sequence of positive integers with \(a_1 = 1\)  and a9 + a10 =646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n >= 1, the terms  a_{2-n}, a_{2n}, a_{2n+1} are in geometric progression, and the terms a_{2n}, a_{2n+1}, a_{2n+2} and  are in arithmetic progression. Let a_n be the greatest term in this sequence that is less than 1000. Find a_n.

waffles  Feb 9, 2018

Best Answer 

 #1
avatar+20025 
+1

I assume the question is:

\(\mathbf{\text{A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed $\\$ so that the first three terms are in geometric progression, $\\$ the second, third, and fourth terms are in arithmetic progression, $\\$ and, in general, for all $n\ge1,$ $\\$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, $\\$ and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. $\\$ Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $.}}\)

 

Let AP = arithmetic progression,

Let GP = geometric progression

 

\(\text{AP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i(j-k) + t_j(k-i) + t_k(i-j) = 0 \\\\ & i = 2n & t_i = a_{2n} \\ & j = 2n+1 & t_j = a_{2n+1} \\ & k = 2n+2 & t_k = a_{2n+2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t_i(j-k) + t_j(k-i) + t_k(i-j) &=& 0 \\ a_{2n}(2n+1-(2n+2)) + a_{2n+1}(2n+2-2n) + a_{2n+2}(2n-(2n+1)) &=& 0 \\ \mathbf{-a_{2n} + 2a_{2n+1} - a_{2n+2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array} \)

 

\(\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ & i = 2n-1 & t_i = a_{2n-1} \\ & j = 2n & t_j = a_{2n} \\ & k = 2n+1 & t_k = a_{2n+1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ a_{2n-1}^{2n-(2n+1)} \times a_{2n}^{2n+1-(2n-1)} \times a_{2n+1}^{2n-1-(2n)} = 1 \\\\ \mathbf{a_{2n-1}^{-1} \times 2a_{2n}^{2} \times a_{2n+1}^{-1} } &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

 

So we have:

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{a_{2n+2}} &\mathbf{=}& \mathbf{2a_{2n+1} -a_{2n} } \\ (2) & \mathbf{a_{2n+1} } &\mathbf{=}& \mathbf{\dfrac{ a_{2n}^{2} }{a_{2n-1} } } \\ \hline \end{array}\)

 

\(\mathbf{a_1=1 }\\ \begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_2^2}{a_1} \quad a_1 = 1 & | \quad \text{ Formula $(2)$} \\ \mathbf{a_3} & \mathbf{=} & \mathbf{ a_2^2 } \\ \hline a_4 &=& 2a_3-a_2 \quad a_3 = a_2^2 & | \quad \text{ Formula $(1)$} \\ &=& 2a_2^2-a_2 \\ \mathbf{a_4} & \mathbf{=} & \mathbf{ a_2(2a_2-1) } \\ \hline a_5 &=& \dfrac{a_4^2}{a_3} & | \quad \text{ Formula $(2)$} \\ a_5 &=& \dfrac{a_2^2(2a_2-1)^2}{a_2^2} \\ \mathbf{a_5} & \mathbf{=} & \mathbf{ (2a_2-1)^2 } \\ \hline a_6 &=& 2a_5-a_4 & | \quad \text{ Formula $(1)$} \\ &=& 2(2a_2-1)^2-a_2(2a_2-1) \\ \mathbf{a_6} & \mathbf{=} & \mathbf{ (2a_2-1)(3a_2-2) } \\ \hline a_7 &=& \dfrac{a_6^2}{a_5} & | \quad \text{ Formula $(2)$} \\ a_7 &=& \dfrac{(2a_2-1)^2(3a_2-2)^2}{(2a_2-1)^2} \\ \mathbf{a_7} & \mathbf{=} & \mathbf{ (3a_2-2)^2 } \\ \hline a_8 &=& 2a_7-a_6 & | \quad \text{ Formula $(1)$} \\ &=& 2(3a_2-2)^2-(2a_2-1)(3a_2-2) \\ \mathbf{a_8} & \mathbf{=} & \mathbf{ (3a_2-2)(4a_2-3) } \\ \hline a_9 &=& \dfrac{a_8^2}{a_7} & | \quad \text{ Formula $(2)$} \\ a_9 &=& \dfrac{(3a_2-2)^2(4a_2-3)^2}{(3a_2-2)^2} \\ \mathbf{a_9} & \mathbf{=} & \mathbf{ (4a_2-3)^2 } \\ \hline a_{10} &=& 2a_9-a_8 & | \quad \text{ Formula $(1)$} \\ &=& 2(4a_2-3)^2-(3a_2-2)(4a_2-3) \\ \mathbf{a_{10}} & \mathbf{=} & \mathbf{ (4a_2-3)(5a_2-4) } \\ \hline \end{array}\)

 

\(\mathbf{a_9+a_{10}=646}\\ \begin{array}{|rcll|} \hline \mathbf{a_9+a_{10}} & \mathbf{=}& \mathbf{646} \\ (4a_2-3)^2+(4a_2-3)(5a_2-4) &=& 646 \\ (4a_2-3) (4a_2-3+5a_2-4) &=& 646 \\ (4a_2-3) (9a_2-7) &=& 646 \\ \ldots \\ 36a_2^2-55a_2-625 &=& 0 \\\\ a_2 &=& \dfrac{55\pm\sqrt{55^2-4\cdot36\cdot(-625)} }{2\cdot 36} \\\\ &=& \dfrac{55\pm\sqrt{93025} }{72} \\\\ &=& \dfrac{55\pm305 }{72} \\\\ a_2 &=& \dfrac{55+305 }{72} \\\\ \mathbf{a_2 } & \mathbf{=} & \mathbf{5} \\\\ && \text{or} \\\\ a_2 &=& \dfrac{55-305 }{72} \\\\ a_2 & = & -3.47222222222 \\ && \text{no solution, because $a_n$ is a sequence}\\ && \text{of positive integers!}\\ \hline \end{array}\)

 

\(\text{Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $} \)

\(\begin{array}{|lclcl|} \hline a_1 && &=& 1 \\ a_2 && &=& 5 \\ a_3 &=& 5^2 &=& 25 \\ a_4 &=& 5\cdot 9 &=& 45 \\ a_5 &=& 9^2 &=& 81 \\ a_6 &=& 9\cdot 13 &=& 117 \\ a_7 &=& 13^2 &=& 169 \\ a_8 &=& 13\cdot 17 &=& 221 \\ a_9 &=& 17^2 &=& 289 \\ a_{10} &=& 17\cdot 21 &=& 357 \\ a_{11} &=& 21^2 &=& 441 \\ a_{12} &=& 21\cdot 25 &=& 525 \\ a_{13} &=& 25^2 &=& 625 \\ a_{14} &=& 25\cdot 29 &=& 725 \\ a_{15} &=& 29^2 &=& 841 \\ \color{red}a_{16} &=& 29\cdot 33 &\color{red}=& \color{red}957 \qquad a_n < 1000\\ a_{17} &=& 33^2 &=& 1089 \qquad a_n > 1000 \\ \hline \end{array} \)

 

 

\(\mathbf{\text{$a_n$ is $ 957$}}\)

 

 

\(\text{GP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{ratio} & a_{2n-1} & a_{2n} & a_{2n+1} \\ \hline 1 &5 &1 & 5 & 25 \\ 2&1.8 &25 & 45 & 81 \\ 3&1.4\ldots &81 & 117 & 169 \\ 4&1.30769230769 &169 & 221 & 289 \\ 5&1.23529411765 &289 & 357 & 441 \\ 6&1.19047619048 &441 & 525 & 625 \\ 7&1.16 &625 & 725 & 841 \\ 8&1.13793103448 &841 & 957 & 1089 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}\)

 

\(\text{AP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{common difference} & a_{2n} & a_{2n+1} & a_{2n+2} \\ \hline 1 &20 & 5 & 25 & 45 \\ 2 &36 &45 & 81 & 117 \\ 3 &52 &117 & 169 & 221 \\ 4 &68 & 221 & 289 & 357 \\ 5 &84 &357 & 441 & 525 \\ 6 &100& 525 & 625 & 725 \\ 7 &116& 725 & 841 & 957 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}\)

 

laugh

heureka  Feb 9, 2018
edited by heureka  Feb 9, 2018
 #1
avatar+20025 
+1
Best Answer

I assume the question is:

\(\mathbf{\text{A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed $\\$ so that the first three terms are in geometric progression, $\\$ the second, third, and fourth terms are in arithmetic progression, $\\$ and, in general, for all $n\ge1,$ $\\$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, $\\$ and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. $\\$ Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $.}}\)

 

Let AP = arithmetic progression,

Let GP = geometric progression

 

\(\text{AP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i(j-k) + t_j(k-i) + t_k(i-j) = 0 \\\\ & i = 2n & t_i = a_{2n} \\ & j = 2n+1 & t_j = a_{2n+1} \\ & k = 2n+2 & t_k = a_{2n+2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t_i(j-k) + t_j(k-i) + t_k(i-j) &=& 0 \\ a_{2n}(2n+1-(2n+2)) + a_{2n+1}(2n+2-2n) + a_{2n+2}(2n-(2n+1)) &=& 0 \\ \mathbf{-a_{2n} + 2a_{2n+1} - a_{2n+2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array} \)

 

\(\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ & i = 2n-1 & t_i = a_{2n-1} \\ & j = 2n & t_j = a_{2n} \\ & k = 2n+1 & t_k = a_{2n+1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ a_{2n-1}^{2n-(2n+1)} \times a_{2n}^{2n+1-(2n-1)} \times a_{2n+1}^{2n-1-(2n)} = 1 \\\\ \mathbf{a_{2n-1}^{-1} \times 2a_{2n}^{2} \times a_{2n+1}^{-1} } &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

 

So we have:

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{a_{2n+2}} &\mathbf{=}& \mathbf{2a_{2n+1} -a_{2n} } \\ (2) & \mathbf{a_{2n+1} } &\mathbf{=}& \mathbf{\dfrac{ a_{2n}^{2} }{a_{2n-1} } } \\ \hline \end{array}\)

 

\(\mathbf{a_1=1 }\\ \begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_2^2}{a_1} \quad a_1 = 1 & | \quad \text{ Formula $(2)$} \\ \mathbf{a_3} & \mathbf{=} & \mathbf{ a_2^2 } \\ \hline a_4 &=& 2a_3-a_2 \quad a_3 = a_2^2 & | \quad \text{ Formula $(1)$} \\ &=& 2a_2^2-a_2 \\ \mathbf{a_4} & \mathbf{=} & \mathbf{ a_2(2a_2-1) } \\ \hline a_5 &=& \dfrac{a_4^2}{a_3} & | \quad \text{ Formula $(2)$} \\ a_5 &=& \dfrac{a_2^2(2a_2-1)^2}{a_2^2} \\ \mathbf{a_5} & \mathbf{=} & \mathbf{ (2a_2-1)^2 } \\ \hline a_6 &=& 2a_5-a_4 & | \quad \text{ Formula $(1)$} \\ &=& 2(2a_2-1)^2-a_2(2a_2-1) \\ \mathbf{a_6} & \mathbf{=} & \mathbf{ (2a_2-1)(3a_2-2) } \\ \hline a_7 &=& \dfrac{a_6^2}{a_5} & | \quad \text{ Formula $(2)$} \\ a_7 &=& \dfrac{(2a_2-1)^2(3a_2-2)^2}{(2a_2-1)^2} \\ \mathbf{a_7} & \mathbf{=} & \mathbf{ (3a_2-2)^2 } \\ \hline a_8 &=& 2a_7-a_6 & | \quad \text{ Formula $(1)$} \\ &=& 2(3a_2-2)^2-(2a_2-1)(3a_2-2) \\ \mathbf{a_8} & \mathbf{=} & \mathbf{ (3a_2-2)(4a_2-3) } \\ \hline a_9 &=& \dfrac{a_8^2}{a_7} & | \quad \text{ Formula $(2)$} \\ a_9 &=& \dfrac{(3a_2-2)^2(4a_2-3)^2}{(3a_2-2)^2} \\ \mathbf{a_9} & \mathbf{=} & \mathbf{ (4a_2-3)^2 } \\ \hline a_{10} &=& 2a_9-a_8 & | \quad \text{ Formula $(1)$} \\ &=& 2(4a_2-3)^2-(3a_2-2)(4a_2-3) \\ \mathbf{a_{10}} & \mathbf{=} & \mathbf{ (4a_2-3)(5a_2-4) } \\ \hline \end{array}\)

 

\(\mathbf{a_9+a_{10}=646}\\ \begin{array}{|rcll|} \hline \mathbf{a_9+a_{10}} & \mathbf{=}& \mathbf{646} \\ (4a_2-3)^2+(4a_2-3)(5a_2-4) &=& 646 \\ (4a_2-3) (4a_2-3+5a_2-4) &=& 646 \\ (4a_2-3) (9a_2-7) &=& 646 \\ \ldots \\ 36a_2^2-55a_2-625 &=& 0 \\\\ a_2 &=& \dfrac{55\pm\sqrt{55^2-4\cdot36\cdot(-625)} }{2\cdot 36} \\\\ &=& \dfrac{55\pm\sqrt{93025} }{72} \\\\ &=& \dfrac{55\pm305 }{72} \\\\ a_2 &=& \dfrac{55+305 }{72} \\\\ \mathbf{a_2 } & \mathbf{=} & \mathbf{5} \\\\ && \text{or} \\\\ a_2 &=& \dfrac{55-305 }{72} \\\\ a_2 & = & -3.47222222222 \\ && \text{no solution, because $a_n$ is a sequence}\\ && \text{of positive integers!}\\ \hline \end{array}\)

 

\(\text{Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $} \)

\(\begin{array}{|lclcl|} \hline a_1 && &=& 1 \\ a_2 && &=& 5 \\ a_3 &=& 5^2 &=& 25 \\ a_4 &=& 5\cdot 9 &=& 45 \\ a_5 &=& 9^2 &=& 81 \\ a_6 &=& 9\cdot 13 &=& 117 \\ a_7 &=& 13^2 &=& 169 \\ a_8 &=& 13\cdot 17 &=& 221 \\ a_9 &=& 17^2 &=& 289 \\ a_{10} &=& 17\cdot 21 &=& 357 \\ a_{11} &=& 21^2 &=& 441 \\ a_{12} &=& 21\cdot 25 &=& 525 \\ a_{13} &=& 25^2 &=& 625 \\ a_{14} &=& 25\cdot 29 &=& 725 \\ a_{15} &=& 29^2 &=& 841 \\ \color{red}a_{16} &=& 29\cdot 33 &\color{red}=& \color{red}957 \qquad a_n < 1000\\ a_{17} &=& 33^2 &=& 1089 \qquad a_n > 1000 \\ \hline \end{array} \)

 

 

\(\mathbf{\text{$a_n$ is $ 957$}}\)

 

 

\(\text{GP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{ratio} & a_{2n-1} & a_{2n} & a_{2n+1} \\ \hline 1 &5 &1 & 5 & 25 \\ 2&1.8 &25 & 45 & 81 \\ 3&1.4\ldots &81 & 117 & 169 \\ 4&1.30769230769 &169 & 221 & 289 \\ 5&1.23529411765 &289 & 357 & 441 \\ 6&1.19047619048 &441 & 525 & 625 \\ 7&1.16 &625 & 725 & 841 \\ 8&1.13793103448 &841 & 957 & 1089 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}\)

 

\(\text{AP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{common difference} & a_{2n} & a_{2n+1} & a_{2n+2} \\ \hline 1 &20 & 5 & 25 & 45 \\ 2 &36 &45 & 81 & 117 \\ 3 &52 &117 & 169 & 221 \\ 4 &68 & 221 & 289 & 357 \\ 5 &84 &357 & 441 & 525 \\ 6 &100& 525 & 625 & 725 \\ 7 &116& 725 & 841 & 957 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}\)

 

laugh

heureka  Feb 9, 2018
edited by heureka  Feb 9, 2018

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