Manhattan GMAT 700+ Challenge Problem - July 24, 2006

[KevinFitzgerald]

Most of our students Manhattan GMAT are trying to break the 700+ barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level Problem Solving problem (I'll post the solution next Monday).


Question
Sequentially Speaking

Sequence S is defined as , for all n > 1.
If S1= 201, then which of the following must be true of Q, the sum of the first 50 terms of S?

(A) 13,000 < Q < 14,000
(B) 12,000 < Q < 13,000
(C) 11,000 < Q < 12,000
(D) 10,000 < Q < 11,000
(E) 9,000 < Q < 10,000

[ahmadazeem777]

Originally Posted by KevinFitzgerald

Sequence S is defined as , for all n > 1.
If S1= 201, then which of the following must be true of Q, the sum of the first 50 terms of S?

(A) 13,000 < Q < 14,000
(B) 12,000 < Q < 13,000
(C) 11,000 < Q < 12,000 Correct Answer
(D) 10,000 < Q < 11,000
(E) 9,000 < Q < 10,000

Thankx
Azeem

[rally]

ANS c)
S1= 201
S2= 202 + 1/202= 202 + 0.0049xyz........ = 202.0049xyz....
S3= (S2+1) + 1/(S2 + 1) = 203.0049......+ 1/203.0049....
1/203.0049... < 1/202
i.e 1/203.0049..< 0.0049...
so the fraction part does significantly add to the total, we can only consider the integer part, and we can also safely drop the decimal part of the no.
S2 + S3 + S4 +..............+ S51 = 202 + 203 + 204 + .........+ 251
u can use AP formula to solve the sum ie 50/2[ 2*202 +49]
or to speed up the calculation
(200 +2) + (200 + 3) + (200 +4) +.......+ (200 +51)= 200 *50 + (2 +3 + 4 +....+ 51)
=> 10,000 + 51*52/2= 10,000 + 51 *26=10,000 + 51(25 +1)= 10,000 + 1275 + 51= 10,000+ 1326 = 11,326

as we require only the range, we don't need to go till the end
therefore it is somewhere between 12,000 and 11,000

[KevinFitzgerald]

Answer
To find each successive term in S, we add 1 to the previous term and add this to the reciprocal of the previous term plus 1.

S1= 201


The question asks to estimate (Q), the sum of the first 50 terms of S. If we look at the endpoints of the intervals in the answer choices, we see have quite a bit of leeway as far as our estimation is concerned. In fact, we can simply ignore the fractional portion of each term. Let’s use S2 ≈ 202, S3 ≈ 203. In this way, the sum of the first 50 terms of S will be approximately equal to the sum of the fifty consecutive integers 201, 202 … 250.

To find the sum of the 50 consecutive integers, we can multiply the mean of the integers by the number of integers since average = sum / (number of terms).

The mean of these 50 integers = (201 + 250) / 2 = 225.5

Therefore, the sum of these 50 integers = 50 x 225.5 = 11,275, which falls between 11,000 and 12,000. The correct answer is C.

[bdsam]

Am I late?

Here is my Approach ... Found it kind of short cut to such Qs ...

S1 = 201

S2 = 202 + 1/202

Now, if we consider the second part of the expression ... it will contribute very little to the Actual sum ..... So lets Ignore it ....
So considering only the fist part is enuff.....

So the total sum can be given as

201 x 50 + [1 + 2 + 3 + ......+ 49]

This Gives a range between 11000 to 12000

[crackmba2006]

11,000 < Q < 12,000

It is like

202+203+.. +252
=11123...

so
11,000 < Q < 12,000