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An infinite geometric series has common ratio \(-1/5\)  and sum \(16.\) What is the first term of the series?

tertre  Jan 5, 2018
 #1
avatar+88836 
+1

Sum of a geometric series  =

 

a1  /  [ 1  - r ]          where  a1  is the first term   and r is the common ratio....so we have

 

16  =  a1  /  [ 1  -  -1/5]

 

16  =  a1  /  [  6/5]       multiply both sides by 6/5

 

(6/5)16  =  a1   =   

 

96/5

 

 

cool cool cool

CPhill  Jan 5, 2018
edited by CPhill  Jan 5, 2018
 #2
avatar+139 
+1

Solution: Let the first term be \(a\). Because the sum of the series is \(16\) , we have \(16= \frac{a}{1-(-1/5)} = \frac{a}{6/5} = \frac{5a}{6}\) . Therefore, \(a=\boxed{\frac{96}{5}}\) .

azsun  Jan 5, 2018

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