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# Number Theory

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This number 15^5 has a long sequence of positive consecutive numbers. The arithmetic mean of 162nd, 163rd and 164th terms equals 10x, or 10 times, the first term. What are the first term, last term and the number of terms of this sequence? Thank you for help.

Aug 10, 2019

#1
+23788
+2

This number $$15^5$$ has a long sequence of positive consecutive numbers.
The arithmetic mean of 162nd, 163rd and 164th terms equals 10x, or 10 times, the first term.
What are the first term, last term and
the number of terms of this sequence?

$$\text{Let the first term =a_1 } \\ \text{Let the number of terms =n } \\ \text{Let the last term =a_n } \\ \text{Let a_{163} = a_{162} + 1  } \\ \text{Let a_{164} = a_{162} + 2  } \\ \text{Let the sum of the sequence  15^5=a_1+a_2+\ldots +a_{n-1}+a_n  }$$

$$\mathbf{a_1=\ ?}$$

$$\begin{array}{|rcll|} \hline \dfrac{a_{162}+a_{163}+a_{164}} {3} &=& 10a_1 \\ \dfrac{a_{162}+(a_{162} + 1)+(a_{162} + 2)} {3} &=& 10a_1 \\ \dfrac{3a_{162}+3} {3} &=& 10a_1 \\ a_{162}+1 &=& 10a_1 \\ && \mathbf{a_n = a_1+(n-1)d} \quad | \quad d = 1 \\ && \boxed{a_n = a_1+ n-1} \\ && a_{162} = a_1+162-1 \\ && \mathbf{a_{162} = a_1 +161} \\ a_{162}+1 &=& 10a_1 \\ (a_1 +161)+1 &=& 10a_1 \\ a_1 +162 &=& 10a_1 \\ 9a_1 &=& 162 \\ \mathbf{a_1} &=& \mathbf{18} \\ \hline \end{array}$$

$$\mathbf{n=\ ?}$$

$$\begin{array}{|rcll|} \hline 15^5 &=& a_1+a_2+\ldots a_{n-1}+a_n \\ 15^5 &=& a_1+(a_1+1)+(a_1+2)+\ldots +(~a_1+(n-2)~)+(~a_1+(n-1)~) \\ 15^5 &=& na_1+\dfrac{\Big(1+(n-1)\Big)}{2}(n-1) \\ 15^5 &=& na_1+\dfrac{ n(n-1) }{2} \quad | \quad \cdot 2 \\ 2\cdot15^5 &=& 2na_1+ n(n-1) \\ \mathbf{2na_1+ n(n-1)} &=& \mathbf{2\cdot15^5} \quad | \quad a_1 = 18 \\ 2\cdot 18n+ n^2-n &=& 2\cdot15^5 \\ n^2 + 35 n -2\cdot 15^5 &=& 0 \\\\ n &=& \dfrac{-35\pm \sqrt{35^2-4\cdot(-2\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{35^2+8\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{1225+6075000)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{6076225 }} {2} \\ n &=& \dfrac{-35\pm 2465} {2} \quad | \quad n>0!\\\\ n &=& \dfrac{-35+ 2465} {2} \\ \mathbf{ n} &=& \mathbf{1215} \\ \hline \end{array}$$

$$\mathbf{a_n=\ ?}$$

$$\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{a_1+n-1} \\ a_n &=& 18+1215-1 \\ \mathbf{ a_n} &=& \mathbf{1232} \\ \hline \end{array}$$

$$\mathbf{check:}$$

$$\begin{array}{|rcll|} \hline 15^5 &=& 18 + 19 +20 + \ldots +1231+1232 \\ 15^5 &=& \dfrac{(18+1232)}{2}\cdot 1215 \\ 15^5 &=& \dfrac{1250}{2}\cdot 1215 \\ 15^5 &=& 625\cdot 1215 \\ 759375&=& 759375\ \checkmark \\ \hline a_{162} &=& 18 + 162-1 \\ a_{162} &=& 179 \\\\ \dfrac{179+180+181}{3} &=& 10 \cdot 18 \\ \dfrac{540}{3} &=& 180 \\ 180 &=& 180\ \checkmark \\ \hline \end{array}$$

Aug 10, 2019
edited by heureka  Aug 10, 2019
edited by heureka  Aug 10, 2019

#1
+23788
+2

This number $$15^5$$ has a long sequence of positive consecutive numbers.
The arithmetic mean of 162nd, 163rd and 164th terms equals 10x, or 10 times, the first term.
What are the first term, last term and
the number of terms of this sequence?

$$\text{Let the first term =a_1 } \\ \text{Let the number of terms =n } \\ \text{Let the last term =a_n } \\ \text{Let a_{163} = a_{162} + 1  } \\ \text{Let a_{164} = a_{162} + 2  } \\ \text{Let the sum of the sequence  15^5=a_1+a_2+\ldots +a_{n-1}+a_n  }$$

$$\mathbf{a_1=\ ?}$$

$$\begin{array}{|rcll|} \hline \dfrac{a_{162}+a_{163}+a_{164}} {3} &=& 10a_1 \\ \dfrac{a_{162}+(a_{162} + 1)+(a_{162} + 2)} {3} &=& 10a_1 \\ \dfrac{3a_{162}+3} {3} &=& 10a_1 \\ a_{162}+1 &=& 10a_1 \\ && \mathbf{a_n = a_1+(n-1)d} \quad | \quad d = 1 \\ && \boxed{a_n = a_1+ n-1} \\ && a_{162} = a_1+162-1 \\ && \mathbf{a_{162} = a_1 +161} \\ a_{162}+1 &=& 10a_1 \\ (a_1 +161)+1 &=& 10a_1 \\ a_1 +162 &=& 10a_1 \\ 9a_1 &=& 162 \\ \mathbf{a_1} &=& \mathbf{18} \\ \hline \end{array}$$

$$\mathbf{n=\ ?}$$

$$\begin{array}{|rcll|} \hline 15^5 &=& a_1+a_2+\ldots a_{n-1}+a_n \\ 15^5 &=& a_1+(a_1+1)+(a_1+2)+\ldots +(~a_1+(n-2)~)+(~a_1+(n-1)~) \\ 15^5 &=& na_1+\dfrac{\Big(1+(n-1)\Big)}{2}(n-1) \\ 15^5 &=& na_1+\dfrac{ n(n-1) }{2} \quad | \quad \cdot 2 \\ 2\cdot15^5 &=& 2na_1+ n(n-1) \\ \mathbf{2na_1+ n(n-1)} &=& \mathbf{2\cdot15^5} \quad | \quad a_1 = 18 \\ 2\cdot 18n+ n^2-n &=& 2\cdot15^5 \\ n^2 + 35 n -2\cdot 15^5 &=& 0 \\\\ n &=& \dfrac{-35\pm \sqrt{35^2-4\cdot(-2\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{35^2+8\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{1225+6075000)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{6076225 }} {2} \\ n &=& \dfrac{-35\pm 2465} {2} \quad | \quad n>0!\\\\ n &=& \dfrac{-35+ 2465} {2} \\ \mathbf{ n} &=& \mathbf{1215} \\ \hline \end{array}$$

$$\mathbf{a_n=\ ?}$$

$$\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{a_1+n-1} \\ a_n &=& 18+1215-1 \\ \mathbf{ a_n} &=& \mathbf{1232} \\ \hline \end{array}$$

$$\mathbf{check:}$$

$$\begin{array}{|rcll|} \hline 15^5 &=& 18 + 19 +20 + \ldots +1231+1232 \\ 15^5 &=& \dfrac{(18+1232)}{2}\cdot 1215 \\ 15^5 &=& \dfrac{1250}{2}\cdot 1215 \\ 15^5 &=& 625\cdot 1215 \\ 759375&=& 759375\ \checkmark \\ \hline a_{162} &=& 18 + 162-1 \\ a_{162} &=& 179 \\\\ \dfrac{179+180+181}{3} &=& 10 \cdot 18 \\ \dfrac{540}{3} &=& 180 \\ 180 &=& 180\ \checkmark \\ \hline \end{array}$$

heureka Aug 10, 2019
edited by heureka  Aug 10, 2019
edited by heureka  Aug 10, 2019
#2
+106515
+1

Very nice, hureka!!!!

CPhill  Aug 10, 2019
#3
+23788
+2

Thank you, CPhill !

heureka  Aug 10, 2019