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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Given that the nth term of an arithmetic progression is 10000/n+5(n-1)...?

and that the nth term is less than 500, how can i show that

n^2-101n+2000<0 and then find the largest possible value of n?

4 Answers

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  • Puggy
    Lv 7
    1 decade ago
    Favorite Answer

    So you are given that for an arithmetic progression (I'll use square brackets instead of a subscript)

    a[n] = 10000/n + 5(n - 1)

    And you're given that

    a[n] < 500

    Combine those two facts, and you have

    10000/n + 5(n - 1) < 500

    All we have to do now is solve for this equation. Let's put everything on the left hand side over n.

    10000/n + 5n(n - 1)/n < 500

    Now, let's merge the fractions on the left hand side into one.

    [10000 + 5n(n - 1)]/n < 500

    Simplifying the left hand side, we get

    [10000 + 5n^2 - 5n]/n < 500

    Moving the 500 over to the left hand side, we obtain

    [10000 + 5n^2 - 5n]/n - 500 < 0

    And now, merging the 500 with the rest of the fraction gives

    [10000 + 5n^2 - 5n - 500n]/n < 0

    [10000 + 5n^2 - 505n]/n < 0

    Let's put the numerator in descending power order.

    [5n^2 - 505n + 10000]/n < 0

    And let's factor out a 5.

    [5 (n^2 - 101n + 2000)]/n < 0

    Dividing both sides by 5 will get rid of that lingering 5 on the outside.

    [n^2 - 101n + 2000]/n < 0

    Remember that n is a positive number (since n represents the nth term, and n starts at 1 (i.e a[1], a[2], ..... , a[n]). Therefore, you can multiply both sides by n without risk of changing the sign (since multiplying by a negative number will change the sign), getting:

    n^2 - 101n + 2000 < 0

    To find the largest possible value of n, all you have to do is solve the inequality. It's not factorable, so you have to solve the corresponding equation n^2 - 101n + 2000 = 0 to obtain values for n using the quadratic formula

    n = [101 +/- sqrt (101^2 - 4(2000))]/2

    n = [101 +/- sqrt (10201 - 8000)]/2

    n = [101 +/- sqrt (2201)]/2

    This approximates into two values:

    n ~= { 73.96, 27.042 }

    The inequality holds true when n is in the interval (27.042, 73.96), so the largest possible value of n is the highest possible integer in that interval, 73.

  • 1 decade ago

    Given that the nth term of an arithmetic progression is 10000/n+5(n-1)and that the nth term is less than 500, how can i show that

    n^2-101n+2000<0 and then find the largest possible value of n?

    10000/n+5(n-1)< 500

    10000+5n^2 - 5n < 500n

    5n^2 -505n +10000 < 0

    n^2 - 101n + 2000 < 0

    The above equation is a parabola with axis of symmetry at

    n = 50.5. The function is concave downward and so it has a minimum value at n= 50.5, but no maximum value.

    The roots of n^2- 101n+2000 are found using the quadratic formula to be n= [101 +/- sqrt(101^2-8000)]/2.

    So largest value of n is 101/2 + sqrt(2201)/2 = 73.957....

    So 73 is the largest integer that will make n^2-101n +2000 negative; that is <0.

  • 1 decade ago

    "Given that the nth term of an arithmetic progression is 10000/n+5(n-1) and that the nth term is less than 500," we have

    10000/n+5(n-1) < 500

    10000 + 5n^2 - 5n < 500n (multiply by n)

    n^2 - 101n + 2000 < 0 (divide by 5 and collect all terms in LHS)

    Find boundary n values,

    n = (101±√2201)/2 =27.04, 73.96

    Since the parabola opens up and n is an integer, we have

    27 < n < 74

    The largest possible value of n is 73.

  • kizzee
    Lv 4
    4 years ago

    Given: n(a+T(n))/2=ten thousand=> 2na+10n(n-a million)=20000-----------(a million) T(n)=a+10(n-a million)<500=> 2na+20n(n-a million)<1000n----------(2) putting (2) in (a million) get 1000n-20n(n-a million)+10n(n-a million)>20000=> 1000n-10n(n-a million)>20000=> 100n-n^2+n>2000=> n^2-101n+2000<0 n^2-101n+2000<0=> [n-(a hundred and one+sqrt(2201))/2]* [n-(a hundred and one-sqrt(2201))/2]<0=> -sqrt(2201)/2<n-a hundred and one<sqrt(2201)/2=> 27.0425..<n<seventy 3.9574...=> n=seventy 3 is the main important a threat n

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