# Sequences – CAT Scan 4 ## 12 thoughts on “Sequences – CAT Scan 4”

1. Harsh Vardhan |

In your second question (n-1)d=999 why it has to be that d has to be factors of 999? Is it because any other no. will leave a fraction and we wont get a natural number?

2. catcracker |

In an AP both n and d have to be integers. So if (n-1) * d = 999, and n-1 is an integer, d has to be 999/(n-1) which will be a factor of 999.

regards
J

3. pragdeesh |

in the second one after finding the factors what is the reason to take n =999 ?.And why 1 and 1000 are eliminated ?

• catcracker |

We are not taking n = 999. We are taking (n-1)d = 999. Read it carefully (if need be, revisit the original post about APs)
And we are not eliminating the numbers 1 and 1000. We are eliminating the AP (1, 1000). That is because (1, 1000) is an AP with 2 terms and the question specifically asks “at least 3 terms”.

regards
J

4. pragdeesh |

okay thank you

5. pragdeesh |

got it

6. pragdeesh |

in the last question i can’t understand why 11 to 20 ,10 to 21 or 1 to 30 are zero?

• catcracker |

Think how an AP behaves. The average of the set of terms is the middle term. If the middle is zero, the average is zero. And so the total is also zero.

regards
J

7. pragdeesh |

sir can you explain the last question

8. Mayur |

Regarding Question can we also able to find no. of GPs for the same set?

• Mayur |

I mean to say Question 2.

• catcracker |

Well we could do 1r^n = 1000. This would give us only r as 10 or 1000 as a feasible alternative. But r = 1000 would give a 2 term GP. So there would be only that 1 GP with more than 2 terms, viz 1, 10, 100, 1000.

regards
J