For a certain sequence, t_n=-(t_{n-1}+1)^{-1} for all values of n such

Question

For a certain sequence,  t_n=-(t_{n-1}+1)^{-1}  for all values of n such that  n \geq {2} . Given that  t_1=1 , how much greater is the product of the first 200 values in this sequence than the sum of the first 200 values in this sequence?

A. 1
B. 68
C. 98
D. 105
E. 228

I found this question in our archives. The post was from 2004!!! Original author claims that it's from Manhattan GMAT (now Manhattan Prep). Give it a try!

TithiMarodia · Answer

okay i tried to solve this but I am not able to get the answer

so when you plug in values for determining the terms in the sequence 
you get ~ 1, -1/2, -2, 1...... gets repeating.

so the product of first three terms = -1 * -1/2 * -2 = 1 
we need the value of product of first 200 terms so the product of first 198 terms will be 1 itself (1 * 99 = 1)
therefore the product of first 200 terms will be 1 * 1 * -1/2 = -1/2

now for the sum of first three numbers we get 1 - 1/2 - 2 = -3/2
the sum of first 98 numbers will be (-3/2)* 99 = - 297/2
to find the sum of 200 numbers will be -297/2 + 1 - 1/2 = -148

but i'm not able to get the answer with this method. 
Could you help me find where i'm wrong?

UzairSohail · Answer

This pattern keeps repeating:

1, -1/2, -2,...

This shows that the product of consecutive three terms is 1. So, in 200 terms, the product of first 198 terms would be 1. The product of next two terms (following the pattern) will be -1/2. Hence:  P=-1/2=-0.5

We can find the sum using the same pattern as above. The sum of first three terms comes out to be -3/2. This pattern would get repeated till the 198th term, which means it will be repeated 198/3=66 times. Add the leftover 2 terms to find the total sum:

Sum = -3/2(66)+1-1/2=-98.5

P-S=-0.5-(-98.5)-=98

Please feel free to point out any mistakes!

Kinshook · Answer

Given: For a certain sequence,  t_n=-(t_{n-1}+1)^{-1}  for all values of n such that  n \geq {2} .

Asked: Given that  t_1=1 , how much greater is the product of the first 200 values in this sequence than the sum of the first 200 values in this sequence?

t_1=1 
 t_2=-(t_1+1)^{-1} = -(2)^{-1} = -1/2 
 t_3=-(t_2+1)^{-1} = -(1/2)^{-1} = -2 
 t_4=-(t_3+1)^{-1} = -(-1)^{-1} = 1 
 t_5=-(t_4+1)^{-1} = -(2)^{-1} = -1/2 
 t_6=-(t_5+1)^{-1} = -(1/2)^{-1} = -2

t_{198} = -2 
 t_{199} = 1 
 t_{200} = -.5

Sum of first 200 values = 66(1-.5-2) + 1 - .5 = 66(-1.5) + .5 = -99 +.5 = -98.5
Product of first 200 values = {1*(-.5)*(-2)}^66*1*(-.5) = -.5

Product of first 200 values - Sum of first 200 values = -.5 - (-98.5) = 98

IMO C

sujoykrdatta · Answer

t1 = 1
t2 = - (t1 + 1)^(-1) = -1/2
t3 = - (t2 + 1)^(-1) = -2
t4 = - (t3 + 1)^(-1) = 1 - since this value is the same as that pf t1, the pattern will repeat
This is a pattern of 3 places
Product of these 3 terms = 1
Sum of these 3 terms = -3/2

Till 200, there will be 66 cycles and 2 extra terms 
Product = 1^66 x 1 x (-1/2) = -0.5
Sum = (-3/2) x 66 + 1 - 1/2 = -98.5

Difference = -0.5 - (-98.5) = 98
 Ans C

For a certain sequence, t_n=-(t_{n-1}+1)^{-1} for all values of n such

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For a certain sequence, t_n=-(t_{n-1}+1)^{-1} for all values of n such

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards