All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Seven Entry Series (Posted on 2002-12-11) Difficulty: 5 of 5
What is the next number in this series:

341, 39, 602, 50, 1003, 64, _______?

See The Solution Submitted by cges    
Rating: 2.2727 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution Comment 9 of 9 |
(In reply to answer by K Sengupta)

Let the nth term of the sequence be S(n), and P(n) be the nth prime number.

Then, we observe that:

S(2n-1) = (3n+8)*P(3n+8),

S(2n) = Average (P(3n+9), P(3n+10))
= (1/2)*(P(3n+9) + P(3n+10))

For example, when n=1 and 2 we have:

S(1) = 11*P(11) = 11*31 = 341, since 31 is the 11th prime number.
S(2) = Average(P(12), P(13)) = Average (37, 41) = 39
S(3) = 14*P(14) = 14*43 = 602.
S(4) = Average(P(15), P(16)) = Average (47,53) = 50

We are required to find S(7), so that substituting n=4, we have:

S(7) = 20*S(20) = 20*71 = 1420.

Consequently, the required missing term is 1420.

Edited on May 10, 2008, 4:12 pm
  Posted by K Sengupta on 2008-04-25 06:59:04

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information