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anonymous

  • 5 years ago

. Find the sum of the given arithmetic series. The sequence whose general term is an=3n represents the positive multiples of 3. Find the sum of the first 102 positive multiples of 3.

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  1. anonymous
    • 5 years ago
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    Sum = (n/2)( 2a +(n-1)d )

  2. anonymous
    • 5 years ago
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    just about remembering formulas

  3. anonymous
    • 5 years ago
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    I dont even need to know what d is ( I could get it by simultaneously eqns ) , because remember that general Tn ( or in your case "an" ) = a +(n-1)d

  4. anonymous
    • 5 years ago
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    so the part inside the bracket is: a+ ( a+(n-1)d )

  5. anonymous
    • 5 years ago
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    ok

  6. anonymous
    • 5 years ago
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    now I will call "an" T(n) , ie T(n) = 3n so our sum becomes: Sum = (n/2) ( a +T(n) )

  7. anonymous
    • 5 years ago
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    Now n= 102 ( because we want first 102 terms ) and T(n) = 3n ( thats given to us )

  8. anonymous
    • 5 years ago
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    so sum = (n/2) ( a+ 3n ) now we just need to know what "a" is , we already know "n" from before ( its 102 ) so if we sub n=1 into our T(n) formula that will give us the first term ( which is "a" ) so T(1) = a = 3(1) = 3

  9. anonymous
    • 5 years ago
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    now sub it in sum = (102/2) ( 3 + 3(102) ) = 51 ( 309) = 15759

  10. anonymous
    • 5 years ago
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    wow i dont know how this comes so easy for u, thank you. i have like 3 more problems ;(

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