Joan took out a mortgage from hel local bank. Each monthly

Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is $328000, how many months will it take Joan to pay back her mortgage?
A. 6
B. 8
C. 10
D. 11
E. 13

Since her first payment is $100 and all subsequent payments must be three times as many as the payment for the previous month, then her monthly payments will be:
$100;
$300;
$900;
$2,700;
...

As you can see we have a geometric progression with the first term of 100 and the common ratio of 3. The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}-1)}{r-1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)).

So, \(\frac{100*(3^{n}-1)}{3-1}=$328,000\) --> \(3^n-1=6,560\) --> \(3^n=6,561\). Now, the unit's digit of 3 in a positive integer power repeats in pattern of 4: {3, 9, 7, 1}. Since the last digit of \(3^n\) is 1 then \(n\) can be 4 (3^4=81), 8, 12, ... only 8 is present among options.

Answer: B.

P.S. Please always post answer choices for PS problems and do not reword or shorten the questions.
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followed long manual method of calculating the amount
1st month = 100
2: 300
3: 900
4: 2700
5: 8100
6: 24300
7: 72900
8: 217800
9: more than 6L

hence if we add 8 to 1 months amount it should be more than 328000

hence Answer is B: 8

An alternate way of solving this is as follows - Brunei plz confirm it.

Its geometric progression defined as Cn = 100 x (3 ^ (n-1) ) (100 times 3 raised to power of n-1)

100+300+900+....+100 x 3 ^ (n-1) = 3280 x 100

3^0 + 3 + 3^2 +....+3^(n-2) + 3^(n-1) = 3280 ..... Equation A
3 + 3^2 +....+3^(n-2) + 3^(n-1) = 3280 - 1 = 3279
3(1 + 3 +.....+3^(n-2)) = 3279
1 + 3 +.....+3^(n-2) = 1093

Putting above value in equation A

1093+3^(n-1) = 3280
3(n-1) = 2187 = 3 x 729 = 3 x 81 x 9 = 3^7

n-1 = 7 so n =8

I apologize. I wasn't able to copy and paste from the online application so I must've missed something from typing it. Thank you for your and everyone's answers though. I really appreciate it.

I'm not sure how I got 6 now. Was just really bugging me how I didn't see 8 last night. Thanks again, team.

Is that really a sub level 600 question ?!

No, it's not. I'd say it's around 650 level question.
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Hi Bunuel,
I solved this question in 10 minutes by logic,
but how do you get geometric progression formula, could you please give me any explainations?
Thanks in advance

Check here: sequences-progressions-101891.html It might help.
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Joan starts off with 100 $ .. which is to be tripled every month

Her monthly payments look like this :

100 , 300 , 900 , 2700 ......... Upto 328000

This can be re written as :

100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 ...... 100 x 3280

So we have 1 , 3 , 9 , 27 ..... 32800 in GP

We know that a =1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula Tn = a3^n-1 ...)

Therefore to find the Sum of n terms of a GP we use this formula :

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

Using this and plugging in the information we get ...

3280 = 1 - 3^n / 1-3 ; 1-3^n / -2

Cross multiplying we get

3280 x -2 = 1- 3^n

- 6560 = 1 - 3^n

- 6561 = - 3 ^n

6561 = 3 ^n (negatives cancel out)

6561 can also be re written as 3 ^ 8

Therefore ; 3 ^8 = 3 ^n

Thus n = 8 (B)
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bunuel,
do we really need to know ap gp & hp for gmat?
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I came to this 100 * 3^t = 328000 --> 3^t = 32800, is this correct and how do I calculate t from this? Can somebody please help. Thanks in advance

No, there are other ways:

My way:
Mouth - Payment - Sum
1st - 100 - 100 (4^0*100)
2nd - 300 - 400 (4^1*100)
3rd - 1200 - 1600 (4^2*100)
4th - 4800 - 6200 (4^3*100)
...
So we can say:
Sum = 4^(mouth-1)*100
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Check out this post on Geometric progressions: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... gressions/

Let \(x = 100\). As per the question,

\(328,000 = x + 3x + 9x + ....................\).. Its an GP Series.
Sum of GP Series = \([a(r^n - 1)/r-1\). So,
\(328,000 = 100 (3^n - 1)/2\) --> \(3^n = 6561.\)
Now, the logic here is we dont have to factorize 6561... All cyclicity of 3 is 4 for numbers ending with 1.
From, the option only B satisfies.

We are given that her first mortgage payment is $100. Since her successive monthly payments have to be triple the previous month’s payment, her second would be $300, and the third would be $900 and so on. Therefore, we have 100 + 300 + 900 + … + P = 328,000 where P is the final monthly payment and with the total payments being $328,000.

Notice that the left hand side of the equation is a geometric series, with a_1 = 100, a_2 = 3(100) , a_3 = 9(100), and the last payment P = a_n = 3^(n - 1)*(100).

In this geometric series, we see that the first term a_1 = 100, and the common ratio r = 3. Recall that the sum of the first n terms of a geometric series with first term a_1 and common ratio r is:

S_n = a_1(1 - r^n)/(1 - r)

So we can say:

328,000 = 100(1 - 3^n)/(1 - 3)

3280 = (1 - 3^n)/(-2)

3280 = (3^n - 1)/2

6560 = 3^n - 1

3^n = 6561

Since 3^8 = 6561, n = 8.

Answer: B
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