Algebra
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Q.

The sequences $x_1,x_2.....$ and $y_1,y_2.....$are in arithmetic progressions such that $x_1+y_1=100$ and $x_{22}-x_{21}=y_{99}-y_{100}$ . Find the sum of the first 100 terms of the progression, $(x_1+y_1),(x_2+y_2).....$

 A.

0

 B.

9900

 C.

10,000

 D.

11,000

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Solution:
Option(C) is correct

The condition $x_{22}−x_{21}=y_{99}−y_{100}$ implies that the common difference of the two arithmetic progressions are the negatives of each other. 

Therefore, the series $(x_1+y_1),(x_2+y_2)$, is a constant series , in which each term is equal to $(x_1+y_1)=100$

Thus

Required Sum

$=100\times 100=10,000$ 

Thank you Pratap for pointing out in the discussion, made correction in the question.


(2) Comment(s)


Pratap
 ()

In question, I believe it should be $x_1 + y_1 = 100$ instead of $x_1 + x_2 = 100$.


Deepak
 ()

Correction made, thank you for bringing the anomaly to attention.