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How do I find n (when tn = -20) in the arithmetic series 5 and 1/3, 5, 4 and 2/3... using the formula $${\mathtt{tn}} = {\mathtt{a}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{\mathtt{d}}$$

 May 12, 2015

Best Answer 

 #1
avatar+130516 
+5

I assume you might be wanting to know which term of the series = - 20....if so, we have......

 

-20 = [5 + 1/3] + (-1/3) (n-1)    simplify

-20 = 16/3 + (-1/3)(n-1)   multiply through by 3

-60  = 16 - 1(n - 1)       subtract 16 from both sides

-76 = -n + 1          subtract 1 from both sides

-77  = -n              multiply through by -1

77 = n

 

It will be the 77th term....!!!!

 

  

 May 12, 2015
 #1
avatar+130516 
+5
Best Answer

I assume you might be wanting to know which term of the series = - 20....if so, we have......

 

-20 = [5 + 1/3] + (-1/3) (n-1)    simplify

-20 = 16/3 + (-1/3)(n-1)   multiply through by 3

-60  = 16 - 1(n - 1)       subtract 16 from both sides

-76 = -n + 1          subtract 1 from both sides

-77  = -n              multiply through by -1

77 = n

 

It will be the 77th term....!!!!

 

  

CPhill May 12, 2015

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