# 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

- PuggyLv 71 decade agoFavorite 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.

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- ironduke8159Lv 71 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.

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- sahsjingLv 71 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.

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- kizzeeLv 43 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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