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anonymous

  • one year ago

Whats the sum of the first 50 terms? \( \huge { 12 - -\frac{7}{10}n } \) I am getting \ ( \huge -\frac{445}{2} \) is that right because it is stating it is wrong. They stat it is \( \huge -\frac{585}{2} \) I am using \( \huge \frac{n}{2}[2*a_1 + (n-1)d) ] to solve\)

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  1. anonymous
    • one year ago
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    Whats the sum of the first 50 terms? \( \huge { 12 - -\frac{7}{10}n } \) I am getting \ ( \huge -\frac{445}{2} \) is that right because it is stating it is wrong. They stat it is \( \huge -\frac{585}{2} \) I am using \( \huge \frac{n}{2}[2*a_1 + (n-1)d) ] \) to solve

  2. anonymous
    • one year ago
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    \( \huge \frac{50}{2}[2*\frac{127}{10} + (50-1)*-\frac{7}{10} ]\)

  3. anonymous
    • one year ago
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    This \( \huge \huge { 12 - -\frac{7}{10}n }\) SHOULD BE \( \huge\huge { 12 - \frac{7}{10}n } \)

  4. anonymous
    • one year ago
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    Wait, so is the sequence \(a_n=12-\frac{7}{10}n\), \(n>0\)?

  5. anonymous
    • one year ago
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    Yes

  6. anonymous
    • one year ago
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    Ok, if so, I think what you did wrong is that your first term is wrong.\[a_1=12-\frac{7}{10}=\frac{113}{10}\]

  7. anonymous
    • one year ago
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    Ah and I got 127/10. Now how did I get 127...... That was the problem.... I hate mis calculating problems lol Thank you. It came up to be correct.

  8. anonymous
    • one year ago
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    lol np. Everyone makes these kinds of mistakes from time to time. :)

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