We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
74
3
avatar

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

Best Answer 

 #1
avatar+23071 
+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?

 

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

 

\(\mathbf{a_1=\ ?}\)

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

 

\(\mathbf{n=\ ?}\)

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

 

\(\mathbf{a_n=\ ?}\)

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

 

\(\mathbf{check:}\)

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

 

laugh

 Aug 10, 2019
edited by heureka  Aug 10, 2019
edited by heureka  Aug 10, 2019
 #1
avatar+23071 
+2
Best Answer

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?

 

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

 

\(\mathbf{a_1=\ ?}\)

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

 

\(\mathbf{n=\ ?}\)

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

 

\(\mathbf{a_n=\ ?}\)

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

 

\(\mathbf{check:}\)

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

 

laugh

heureka Aug 10, 2019
edited by heureka  Aug 10, 2019
edited by heureka  Aug 10, 2019
 #2
avatar+103122 
+1

Very nice, hureka!!!!

 

 

cool cool cool

CPhill  Aug 10, 2019
 #3
avatar+23071 
+2

Thank you, CPhill !

 

laugh

heureka  Aug 10, 2019

22 Online Users

avatar
avatar
avatar
avatar