+0  
 
0
46
2
avatar+17 

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
Sort: 

2+0 Answers

 #1
avatar+26329 
+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. "

 

.

Alan  Nov 10, 2017
 #2
avatar+78744 
+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

7 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details