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anonymous

  • one year ago

Help please?? <3 Find S12 for 1 + 6 + 11 + 16 +… 308 322 336 342

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  1. hartnn
    • one year ago
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    so you know the formula for Sum of terms in Arithmetic series?

  2. anonymous
    • one year ago
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    no

  3. hartnn
    • one year ago
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    ok, here you go! \(\Large S_n = n/2 (2a+ (n-1)d) \) where a = First term = 1 d = common difference = .... ? can you find? and n = number of terms = 12, because we need sum of 12 terms

  4. anonymous
    • one year ago
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    how do you find the common difference?

  5. hartnn
    • one year ago
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    d = common difference = difference between next term and current term.. like d = 2nd term - 1st term = 3rd term - 2nd term = ..and so on now can you find the d?

  6. anonymous
    • one year ago
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    so the common difference is 5

  7. hartnn
    • one year ago
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    thats right! d= 5 :) now plug in all the values in the formula :)

  8. anonymous
    • one year ago
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    \[s _{12}=12/2(2\times1+(12-1)5)\] Right?

  9. hartnn
    • one year ago
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    yup, go on!

  10. anonymous
    • one year ago
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    \[s _{12}=12/2(2\times1(11)5)\]

  11. hartnn
    • one year ago
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    so whats your final answer ?

  12. anonymous
    • one year ago
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    i got 12/114 but that's not an option so... I don't know what I did wrong?

  13. hartnn
    • one year ago
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    12/2 is separate which will evaluate to 6 6 (2+ 55) = ... ?

  14. anonymous
    • one year ago
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    oh okay so it 342

  15. hartnn
    • one year ago
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    yes! thats correct :) the whole \(2a + (n-1)d\) is in numerator and not in denominator :)

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