1. ## Arithmetic progression help..

An arithmetic progression has first term a and common difference 10. The sum of the first n terms of the progression is 10,000. Express a in terms of n, and show that the nth term of the progression is:

10,000/n + 5(n-1)

Given that the nth term is less than 500, show that n^2 - 101n + 2000 < 0 and hence find the largest possible value of n.

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Let d = 10

I tried it using the formula of sum:
S = n/2 (2a + (n-1)d)
10,000 = n/2 (2a + 10n - 10)
10,000 = na + 5n^2 - 5n
10,000 - 5n^2 + 5n = na

=> a = 10,000/n + 5(1 - n)

The nth term should be: a + (n-1)d => 10,000/n + 5(1-n) + 10n - 10
= 10,000/n - 5 + 5n => 10,000/n + 5(n - 1) (shown)

Now ... I don't know how to solve the rest of it.

2. Originally Posted by struck
An arithmetic progression has first term a and common difference 10. The sum of the first n terms of the progression is 10,000. Express a in terms of n, and show that the nth term of the progression is: 10,000/n + 5(n-1)

Given that the nth term is less than 500, show that n^2 - 101n + 2000 < 0 and hence find the largest possible value of n.

-----
Let d = 10

I tried it using the formula of sum:
S = n/2 (2a + (n-1)d)
10,000 = n/2 (2a + 10n - 10)
10,000 = na + 5n^2 - 5n
10,000 - 5n^2 + 5n = na

=> a = 10,000/n + 5(1 - n)

The nth term should be: a + (n-1)d => 10,000/n + 5(1-n) + 10n - 10
= 10,000/n - 5 + 5n => 10,000/n + 5(n - 1) (shown)

Now ... I don't know how to solve the rest of it.
(Thank you for showing what you've done so far!)

You're told that the n-th term is less than 500, and you've shown that the n-th term is of the form 10,000/n + 5(n - 1). Then:

. . . . . $\frac{10,000}{n}\, +\, 5(n\, -\, 1)\, <\, 500$

Since we're trying to find the largest value of n, we can safely assume that n > 0. This means that we can multiply through by "n" and not have to worry about flipping the inequality sign:

. . . . . $10,000\, +\, 5n(n\, -\, 1)\, <\, 500n$

. . . . . $2,000\, +\, n(n\, -\, 1)\, <\, 100n$

. . . . . $2,000\, +\, n^2\, -\, n\, <\, 100n$

Move everything over to the left-hand side, and simplify. This will be the second "show" that they've asked you to do.

Next, you'll need to solve the quadratic inequality to find the range of n-values such that the inequality is true.

Since the quadratic is not factorable, you'll need to find the zeroes by applying the Quadratic Formula. Then use what you know about the graphs of positive quadratics to find the solution interval.

Then pick the largest whole-number value inside that interval.

3. Hello, struck!

An arithmetic progression has first term $a$ and common difference 10.
The sum of the first $n$ terms of the progression is 10,000.
Express $a$ in terms of $n$.

Let $d = 10$

I tried it using the formula of sum: . $S_n \:=\:\tfrac{n}{2}[2a + (n-1)d]$

$\tfrac{n}{2}(2a + 10n - 10) \:=\:10,\!000 \quad\Rightarrow\quad na + 5n^2 - 5n \:=\:10,\!000$

$na \:=\:10,000 - 5n^2 + 5n \quad\Rightarrow\quad a \:= \:\frac{10,\!000}{n} + 5 -5n$ . . . . Good!

and show that the $n^{th}$ term is: . $\frac{10,\!000}{n} + 5(n-1)$

The $n^{th}$ term should be: . $a + (n-1)10$

. . $a_n \;=\;\frac{10,\!000}{n} + 5-5n + 10n - 10 \:=\: \frac{10,000}{n} + 5n - 5$

Therefore: . $a_n \;=\;\frac{10,\!000}{n} + 5(n-1)$ . . . . Right!

Given that the $n^{th}$ term is less than 500:

(a) Show that: . $n^2 - 101n + 2000 \:<\: 0$

(b) Find the largest possible value of $n.$
stapel gave you the game plan . . .

We have: . $a_n \:<\:500\quad\Rightarrow\quad \frac{10,\!000}{n} + 5n-5 \:<\:500\quad\Rightarrow\quad \frac{10,\!000}{n} + 5n \:<\:505$

Multiply by $\frac{n}{5}\!:\;\;2000 +n^2 \:<\:101 n\quad\Rightarrow\quad n^2 -101n + 2000 \:<\:0$ .(a)

We have a parabola, $y \:=\:x^2-101x + 200$ which opens upward.
When is $y$ negative? .When the graph below the x-axis?
. . Answer: between its x-intercepts.

The intercepts occur when: . $n^2 - 101n + 2000 \:=\:0$

Quadratic Formula: . $n \;=\;\frac{101 \pm\sqrt{2201}}{2} \;\approx\;\begin{Bmatrix}27.042 \\ 73.957\end{Bmatrix}$

Therefore, the largest value is: . $n\,=\,73$ .(b)

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