anonymous
  • anonymous
Below are four consecutive numbers of an arithmetic sequence. The middle two are missing. ..24, _ , _ , 81, .. Find the two missing numbers
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you made it more complicated aha
anonymous
  • anonymous
?
anonymous
  • anonymous
how did you come up with 24?

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anonymous
  • anonymous
You are given it. That is the question exactly. Help ;P
anonymous
  • anonymous
If the numbers are 43,62 then you get an arithmetic progression
anonymous
  • anonymous
24,36,54,81 complete arithmetic progression
anonymous
  • anonymous
(n x 3)/2 that is the formula for that problem.
watchmath
  • watchmath
\(24+3d=81\) \(3d=57\) \(d=19\) So the two number in the middle are \(43,62\)
radar
  • radar
There are several answers posted here, and they are different, obviously some of them must be wrong. So lets check them out. The answer that this is a progression with (n x 3)/2 looks good but if that is the case n for the number 24 would be n=16. Then n for the 81 would be 19 giving (using the formula (3 x 19)/2=57/2=28.5) us not even an integer, and not 81. I don;t see it. Then we have the solution saying it just a difference of 19, then 24, 24+19=53, 53+19=72, 72+19=81. So the sequence would then appear as 24, 53, 72, 91 If this is the case then the post showing the two missing numbers as 43, and 62 would be incorrect, but maybe the d=19 is good and then the missing numbers are 53, and 72. I like that so far.
watchmath
  • watchmath
24+19=43 my friend radar :)
radar
  • radar
I like it even better now lol
radar
  • radar
Looks like the missing numbers are indeed 24, 43, 62, and 81. Good job watchmath, I checked good answer for your post.

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