The sequences x1,x2..... and y1,y2.....are in arithmetic progressions

Question

The sequences  x_1 ,  x_2 , ... and  y_1 ,  y_2 , ... are in arithmetic progressions (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant) such that  x_1+y_1=100  and  x_{22}−x_{21}=y_{99}−y_{100} . What is the sum of the first 100 terms of the progression,  (x_1+y_1) ,  (x_2+y_2) , ....

A. 0

B. 9,900

C. 10,000

D. 11,000

E. 12,000

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sujoykrdatta · Accepted Answer

x1, x2, x3 ... form an AP
y1, y2, y3 ... form an AP

x22 - x21 = common difference of 1st AP = d (say)
y100 - y99 = common difference of 2nd AP = - (y99 - y100) = - (x22 - x21) = - d

Thus, the x-series is an increasing AP and the y-series is a decreasing AP, the magnitude of the common difference being the same

Thus, the APs are: 
x-series: x1, x1 + d, x1 + 2d, ...
y-series: y1, y1 - d, y1 - 2d, ...

We know: x1 + y1 = 100

Thus, the sum of the 100 terms of the AP: (x1 + y1), (x2 + y2) , ... (x100 + y100)
= (x1 + y1) + (x1 - d + y1 + d) + ... + (x1 + 99d + y1 - 99d)
= (x1 + y1) + (x1 + y1) + ... 100 times
= 100 x 100
= 10000

Answer C

CrackverbalGMAT · Answer

SOLUTION:
Since x22-x21=y99-y100
It implies the common difference of the two A.Ps are the negatives of each other.
[As, the common difference of 2nd AP = - (y99 - y100) = - (x22 - x21) = - d]
Thus the series,
(x1+y1),(x2+y2),is a constant series with each term equal to 100 (x1+y1=100)
Thus the required sum is
100*100=10,000

Hope this helps

Devmitra Sen

sekipap · Answer

If it is an arithm seq. , then x22- x21 = x2- x1... the same applies to y... 
x2-x1 =y1- y2 --> x2+y2=x1+y1= 100... the same process can be done for every x and y
so, xi +yi = 100 and the sum of the first 100 terms is 100*100 = 10 000 = C

luisdicampo · Answer

Deconstructing the Question  
Sequence  x  is an Arithmetic Progression (AP) with common difference  d_x .
Sequence  y  is an AP with common difference  d_y .
Given:
1.  x_1 + y_1 = 100 
2.  x_{22} - x_{21} = y_{99} - y_{100}

Target: Sum of the first 100 terms of the sequence  (x_n + y_n) .

Step 1: Analyze the Common Differences  
By definition of AP:
 x_{22} - x_{21} = d_x  (Difference between consecutive terms).
 y_{99} - y_{100} = -(y_{100} - y_{99}) = -d_y .

Substitute these into the given equation:
 d_x = -d_y 
 d_x + d_y = 0 .

Step 2: Analyze the Combined Sequence  
Let the new sequence be  z_n = x_n + y_n .
The common difference of this new sequence ( D ) is the sum of the individual common differences:
 D = d_x + d_y .

From Step 1, we know  D = 0 .
This means the sequence  z_n  is a  constant sequence . Every term is equal to the first term.

Step 3: Calculate the Sum  
First term:  z_1 = x_1 + y_1 = 100 .
Since the sequence is constant, the first 100 terms are all equal to 100.

Sum =  100 	imes (	ext{Value of each term}) 
Sum =  100 	imes 100 = 10,000 .

Answer:  C

The sequences x1,x2..... and y1,y2.....are in arithmetic progressions

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The sequences x1,x2..... and y1,y2.....are in arithmetic progressions

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