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The sum of the last 3 terms of a geometric progression having n terms is 1024 times the sum of the first 3 terms of the progression. If the third term is 5 find the last term. 

OldTimer  Nov 10, 2017
 #1
avatar+26814 
+3

"The sum of the last 3 terms of a geometric progression having n terms is 1024 times the sum of the first 3 terms of the progression. If the third term is 5 find the last term. "

 

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Alan  Nov 10, 2017
 #2
avatar+87604 
+2

Sum of the last three terms is       ar^(n - 1) + ar^(n - 2) + ar^(n - 3)  ⇒

ar^(n - 3)  [ r^2 + r  + 1 ]  = ar^(n - 3) [ 1 + r + r^2]

 

Sum of the first three terms  is     a + ar + ar^2   =  a [ 1 + r + r^2 ]

 

And the 3rd term is  5  ⇒  ar^2  = 5   ⇒  a = 5 / r^2 

 

And we're told that  the sum of the lst 3 terms  =  1024 times the sum of the 1st three

 

So we have that

 

1024a  [ 1 + r + r^2 ]  = ar^( n-3)  [ 1 + r + r^2]

 

1024  =  r^(n - 3)

 

Note that  1024  = 2 ^10....which implies that

 

2^10  = r^(n - 3)

 

So    r  = 2      and  n = 13

 

And the first term is 5 / 2^2  =  5/4

 

And the third term  is    ar^2  =   (5/4)*2^2  =  (5 /4) * 4  =   5

 

So.... the last term is

 

ar ^ (n -1)  = (5/4) (2)^(13 - 1)  =  (5/4) *2^12  =  5120

 

 

cool cool cool

CPhill  Nov 10, 2017

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