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22 Oct 2005, 12:47
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If the first number of a series is 420 and every number after the first number is multiplied by a fixed constant and the third and fourth terms of this series are 105 and 52.5, respectively, then which of the following represents the sum of the first eight terms of this sequence?
A) ((420*(1/2^8)+420(1/2^0))/2)*(8)
B) 840*2^7/(1/2^1/3)
C) (8^1/3)*420*((2^7 - 1)/(2^7))/(1/2^8)
D) (7*(8^1/3)*30*(2^8 - 1)/(2^8))/(1/(8^1/3)
E) ((8^1/3)*14*30*(2^8 - 1)/(2^7))/(1/2^1/3)
You got 1.75 min! :snipersmile:
A) ((420*(1/2^8)+420(1/2^0))/2)*(8)
B) 840*2^7/(1/2^1/3)
C) (8^1/3)*420*((2^7 - 1)/(2^7))/(1/2^8)
D) (7*(8^1/3)*30*(2^8 - 1)/(2^8))/(1/(8^1/3)
E) ((8^1/3)*14*30*(2^8 - 1)/(2^7))/(1/2^1/3)
You got 1.75 min! :snipersmile:
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22 Oct 2005, 13:21
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using GM formulae
find r first
an =ar^n-1
a4=a3*r^2-1
52.5 = 105(r^2-1) => r=1/2
So sum = a1*(1-r^n)/1-r = 420*(1-(1/2)^8)/1-(1/2)
= 840(1-(2^-8))
whatever this simplifies to
I think it simplifies to D or E.... I'll pick D
find r first
an =ar^n-1
a4=a3*r^2-1
52.5 = 105(r^2-1) => r=1/2
So sum = a1*(1-r^n)/1-r = 420*(1-(1/2)^8)/1-(1/2)
= 840(1-(2^-8))
whatever this simplifies to
I think it simplifies to D or E.... I'll pick D
22 Oct 2005, 16:35
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I dont know about you but A looks straight fwd...
I got A
I assume number 1 thru 8 is not inclusive?
I got A
I assume number 1 thru 8 is not inclusive?
