Aptitude 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.

**Pratap**

*()
*

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

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