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Find a and r such that a+ar+ar^2+ar^3+ar^4+ar^5=378, ar^5+ar^6+ar^7+ar^8+ar^9+ar^10=12096.

 

I'm not sure how to solve this. Could I have some help?

 May 5, 2021
 #1
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What if we divided the 2 equations.

(ar^5+ar^6+ar^7+ar^8+ar^9+ar^10)/(a+ar+ar^2+ar^3+ar^4+ar^5) = r^5 = 12096/378 = 32

r = 2

Now that we have r, we can solve for a using the sum of geometric series. 

a(r^n-1)/(r-1) = sum

a(63)/(1) = 378

a = 6

 

=^._.^=

 May 5, 2021
 #2
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Thanks so much!

 May 5, 2021
 #3
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You're welcome. :))

 

=^._.^=

catmg  May 5, 2021

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