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Can anyone help me with my maths work on the nth term ?

Just need someone to go over this for me

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7 Answers

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  • 2 months ago

    solving the 1st 4 terms

    an = n(n + 1)

    a₁ = 1(1 + 1) = 2

    a₂ = 2(2 + 1) = 6

    a₃ = 3(3 + 1) = 12

    a₄ = 4(4 + 1) = 20

    The 1st 4 terms are 2,6,12,20

    What is the 100th term

    a₁₀₀ = 100(100 + 1) = 10,100  

    Is  53 a term in the sequence: No

    Which term is 240

    240 = n(n + 1)

    240 = n^2 + n

    n^2+ n - 240 = 0

    (n + 16)(n - 15) = 0

     n = - 16, n = 15

    The 15th term...

    Conclusion

    What are the 1st four terms? 2,6,12,20

    What is the 100th term? 10,000

    Is 53 a term in the sequence? NO.

    which term is 240? The 15th term

  • 2 months ago

    T(n)=n(n+1), n>=1=>

    T(1)=2; T(2)=6; T(3)=12; T(4)=20; T(100)=10100.

    Let n(n+1)=53=>

    n^2+n-53=0=>

    n=[-1+/-sqr(1+4*53)]/2=>

    n=[-1+/-sqr(213)]/2 is not an integer.

    Thus, 53 is not in the sequence.

    n(n+1)=240=>

    n^2+n-240=0=>

    n=[-1+/-sqr(1+4*240)]/2=>

    n=15 or -16(rejected)

    Thus T(15)=240.

  • ?
    Lv 7
    2 months ago

     The nth term of a sequence is n(n + 1)

     The first four terms are 2, 6, 12, and 20.

     The 100th term is 10,100.

     Is 53 a term in the sequence? No

     240 is the 15th term.

  • 2 months ago

    You are told that:

    f(n) = n(n + 1)

    What are the first 4 terms?  Solve for f(1), f(2), f(3), f(4):

    f(1) = 1(1 + 1) and f(2) = 2(2 + 1) and f(3) = 3(3 + 1) and f(4) = 4(4 + 1)

    f(1) = 1(2) and f(2) = 2(3) and f(3) = 3(4) and f(4) = 4(5)

    f(1) = 2 and f(2) = 6 and f(3) = 12 and f(4) = 20

    ----

    What's the 100th term?  Solve for f(100):

    f(n) = n(n + 1)

    f(100) = 100(100 + 1)

    f(100) = 100(101)

    f(100) = 10100

    ----

    Is 53 a term?  Set f(n) = 53 and solve for n.  If n isn't an integer then the answer is no:

    f(n) = n(n + 1)

    53 = n(n + 1)

    53 = n² + n

    0 = n² + n - 53

    n = [ -b ± √(b² - 4ac)] / (2a)

    n = [ -1 ± √(1² - 4(1)(-53))] / (2 * 1)

    n = [ -1 ± √(1 + 212)] / 2

    n = (-1 ± √213) / 2

    213 is not a perfect square so the roots of this quadratic are irrational.  Therefore "53" is not a value in the sequence.

    ----

    Which term is 240?  Same steps as above except set f(n) = 240:

    f(n) = n(n + 1)

    240 = n(n + 1)

    240 = n² + n

    0 = n² + n - 240

    n = [ -b ± √(b² - 4ac)] / (2a)

    n = [ -1 ± √(1² - 4(1)(-240))] / (2 * 1)

    n = [ -1 ± √(1 + 960)] / 2

    n = (-1 ± √961) / 2

    n = (-1 ± 31) / 2

    n = -32/2 and 30/2

    n = -16 and 15

    Throw out the negative value since n must be an integer greater than zero, so:

    n = 15

    240 is the 15th term.

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  • ?
    Lv 7
    2 months ago

    What are the first 4 terms ?

    Take n(n + 1) and substitute n for 1 2 3 and 4

    1(1 + 1) = 1 x 2 = 2

    2(2 + 1) = 2 x 3 = 6

    3(3 + 1) = 3 x 4 = 12

    4(4 + 1) = 4 x 5 = 20

    The first four terms are 2 6 12 and 20

    The 100th term is 100(100 + 1) .. 100 x 101

    Which is 10100

    Is 53 a term in this sequence ?

    NO .. all the terms are even numbers.

    Which term is 240 ?

    n(n + 1) = 240

    15 x 16 = 240

    So it is the 15th term.

  • ?
    Lv 7
    2 months ago

    To get the first four terms we substitute for n = 1, 2, 3 and 4

    i.e. 1(1 + 1) = 2

    2(2 + 1) = 6

    3(3 + 1) = 12

    4(4 + 1) = 20

    The 100th term is when n = 100

    so, 100(100 + 1) = 10100

    All the terms are even, hence 53 being odd is not a term in the sequence.

    If the nth term is 240 then we have n(n + 1) = 240

    i.e. n² + n = 240

    or, n² + n - 240 = 0

    so, (n + 16)(n - 15) = 0

    Hence, n = 15...i.e. 15th term

    :)>

  • 2 months ago

    First four terms are 2, 6, 12, 20.

    100th term is 10100.

    All terms are even numbers, so 53 won't appear.

    240 is the 15th term because 15 x 16 = 240.

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