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anonymous

  • 5 years ago

How do I find the twelfth term of 5,8,11,14

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  1. anonymous
    • 5 years ago
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    That's an arithmetic progression. The formula for the nth term of an arithmetic progression is:\[u _{n}=a+(n-1)d\]Where a is the first term and d is common difference. So, to find the twelfth term of this sequence:\[u _{12}=a+(12-1)d=5+11\times3=38\]

  2. amistre64
    • 5 years ago
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    its adding 3

  3. amistre64
    • 5 years ago
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    A{n} = A{n-1} +3

  4. anonymous
    • 5 years ago
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    That's the recurrence relation but that's not particularly necessary here.

  5. amistre64
    • 5 years ago
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    i dont operate on necessary lol

  6. anonymous
    • 5 years ago
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    Ok thank you

  7. anonymous
    • 5 years ago
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    its 3n + 2 :D( n = term) for n = 1(first term): 3 + 2 =5 for n= 2(second term): 6 + 2 = 8 for n=20(20th term) 60 + 2 = 62 kk?

  8. anonymous
    • 5 years ago
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    just they way i solve is really easier

  9. amistre64
    • 5 years ago
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    \(A_n = A_{n-1}+3\); and \(A_{n-1} = A_{n-2}+3\) \[A_n = A_{n-2}+3+3 \iff A_n=A_{n-2}+2(3)\] \[A_n = A_{n-r} + r(3)\] \(A_{n-r} = A_1;\) when \(r = n-1\) \[\] \(A_n = A_1 + (n-1)(3);\) and \(A_1 = 5\) \[A_{12} = 5 + (12-1)(3) = 38\] Thats how you learn it in discrete math ...

  10. anonymous
    • 5 years ago
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    97 to go and then i open a bottle of champagne!

  11. amistre64
    • 5 years ago
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    97..years? or seonds lol

  12. anonymous
    • 5 years ago
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    if it happens when i am not here i will be sad

  13. amistre64
    • 5 years ago
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    ill send a carrier pigeon just before it happens to let you know lol

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