Arithmetic Progression

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Arithmetic Progression

by vikram.sumer » Wed Apr 04, 2012 6:58 pm
Ques: A person saves each year Rs 100 more than he saved in the preceding year, and he
saves Rs 200 the first year. How many years would it take for his savings, not including
interest, to amount to Rs 23000 ?

OA: 20

I know my strategy is wrong, but could not figure Why? Please advise

My Thoughts: This question is like a series: 200,300,400........23000. We need to find what is term in this series which corresponds to 23000

Sn = 100(n)+x

X= 100

23,000 = 100(n)+ 100

X= 229

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by Anurag@Gurome » Wed Apr 04, 2012 8:19 pm
This is an arithmetic series, so the sum can found using the formula:
Sn = (n/2)[2a + (n - 1)d], where a = first term, d = common difference, n = number of terms

Here, let us assume that n = total number of years
First term, a = 200, common difference, d = 100
Then, 23000 = (n/2)[2 * 200 + (n - 1)100]
23000 = n[200 + (n - 1)50]
23000 = 200n + 50n² - 50n
23000 = 50n² + 150n
n² + 3n - 460 = 0
(n - 20)(n + 23) = 0
n = 20 (n = -23 is not possible, as the no. of years cannot be negative)
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by Pharo » Wed Apr 04, 2012 10:33 pm
vikram.sumer wrote:Ques: A person saves each year Rs 100 more than he saved in the preceding year, and he
saves Rs 200 the first year. How many years would it take for his savings, not including
interest, to amount to Rs 23000 ?

OA: 20

I know my strategy is wrong, but could not figure Why? Please advise

My Thoughts: This question is like a series: 200,300,400........23000. We need to find what is term in this series which corresponds to 23000

Sn = 100(n)+x

X= 100

23,000 = 100(n)+ 100

X= 229
Hey dude!

Your mistake is you are solving for the wrong thing. Yes it is a series but the question is not asking what n (year) he would save 23000. The question is asking how long it will take him to save up to 23000. As you demonstrated above, he will save 23000 on his 229th year (haha good luck living that long :P)

So you have to add 200+300+400+...+((100)nth year + 100) and solve for the n:

Sum ((100)n + 100)) from 1 to x (i.e from year one to x) = 100 * sum(n){1:x} + sum(100){1:x}
= 100*((x *(x+1))/2) + 100x ;factor out the first nasty looking term
= 50x^2 + 50x + 100x
= 50x^2 + 150x ; this is how much he saves in x years. Set this to 23000 to find how many years it will take him to save 23000
; 50x^2 + 150x = 23000; divide everything by 50 to make the numbers smaller (smaller = easier)
x^2 + 3x = 460 ; put everything on one side
x^2 + 3x - 460 = 0; find to factors of 460 such that k*m = -460 AND k-m = 3;
x^2 + 3x + ((-20)*23) = 0; --> and the answer is 20 :)

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by GMATGuruNY » Thu Apr 05, 2012 4:21 am
vikram.sumer wrote:Ques: A person saves each year Rs 100 more than he saved in the preceding year, and he
saves Rs 200 the first year. How many years would it take for his savings, not including
interest, to amount to Rs 23000 ?

OA: 20

I know my strategy is wrong, but could not figure Why? Please advise

My Thoughts: This question is like a series: 200,300,400........23000. We need to find what is term in this series which corresponds to 23000

Sn = 100(n)+x

X= 100

23,000 = 100(n)+ 100

X= 229
The values here are evenly spaced integers.
Given evenly spaced values:
Average = (biggest+smallest)/2.
Sum = number*average.

The GMAT would provide answer choices, which we could plug in for the number of years needed to accrue 23,000 in savings.

Let's say that the answer choices were as follows:
5 10 15 20 30

Answer choice C: 15 years

Biggest value = 200 + 14(100) = 1600.
Average = (1600+200)/2 = 900.
Sum = 15*900 = 13,500.
Too small.
Eliminate A, B and C.

Answer choice D: 20 years
Biggest value = 200 + 19(100) = 2100.
Average = (2100+200)/2 = 1150.
Sum = 20*1150 = 23,000.
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