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anonymous

  • 5 years ago

Below are four consecutive numbers of an arithmetic sequence. The middle two are missing. ..24, _ , _ , 81, .. Find the two missing numbers

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  1. anonymous
    • 5 years ago
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    you made it more complicated aha

  2. anonymous
    • 5 years ago
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    ?

  3. anonymous
    • 5 years ago
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    how did you come up with 24?

  4. anonymous
    • 5 years ago
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    You are given it. That is the question exactly. Help ;P

  5. anonymous
    • 5 years ago
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    If the numbers are 43,62 then you get an arithmetic progression

  6. anonymous
    • 5 years ago
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    24,36,54,81 complete arithmetic progression

  7. anonymous
    • 5 years ago
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    (n x 3)/2 that is the formula for that problem.

  8. watchmath
    • 5 years ago
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    \(24+3d=81\) \(3d=57\) \(d=19\) So the two number in the middle are \(43,62\)

  9. radar
    • 5 years ago
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    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.

  10. watchmath
    • 5 years ago
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    24+19=43 my friend radar :)

  11. radar
    • 5 years ago
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    I like it even better now lol

  12. radar
    • 5 years ago
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    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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