# Moderate Algebra Solved QuestionAptitude Discussion

 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

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.