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by Frankenstein » Thu Jun 09, 2011 10:14 pm
Hi,
First mortgage is 100$
Second will be 100*3$
.
.
.
nth will be 100*3^(n-1)
So, 100+100*3+....100*3^(n-1) = 328,000
Sum of geometric series is a(r^n -)/(r-1). a=100, r=3
So, 100(3^n -1)/(3-1) = 328000 =>3^n = 6561=3^8.
If you find it difficult to find n when u reach the last step, just remember that 3^n ends with 1 if n is a multiple of 4.

Hence, B
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by cans » Fri Jun 10, 2011 6:30 am
p(n) = 3*p(n-1) = 3^(n-1) * P(1)
p(1)=100
P(1)+p(2)+...+ P(n) = 328000
p(1) (1 + 3^1 + 3^2 +..... + 3^(n-1) ) = 328000
1 + 3^1 + 3^2 +..... + 3^(n-1) = 3280
3^n - 1 = 3280*2 = 6560
3^n = 6561
n=8
IMO B
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