Hard questions for 2000s and up

Test Preparation SAT Preparation
1

<p>First Question </p>

<pre><code> 1/(1)(2) , 1/(2)(3) , 1/(3)(4)
</code></pre>

<p>The first three terms of a sequence are given. The nth term of the sequence is 1/(n)(n+1).</p>

<p>Which is equal to 1/(n)-(n-1). What is the sum of the first 50 terms of this sequence?</p>

<pre><code>a) 1

b) 50/51

c) 49/50

d) 24/50

e) 1/(50)(51)
</code></pre>

<p>Second Question </p>

<pre><code> Y=-2(x-2)^2 +3
</code></pre>

<p>In the xy-plane, line L passes through the point (4,-5) and the vertex of the parabola </p>

<p>with the equation above. What is the slope of line L ? </p>

<pre><code> A) -4

 b) -1/4 

 c) 0

  d) 1/4 

  e) 4

</code></pre>

<p>Please I need the answers with persuasive explanation</p>

2

<ol>
<li>The sum of the first 50 terms is</li>
</ol>

<p>(1 - 1/2) + (1/2 - 1/3) + … + (1/50 - 1/51)</p>

<p>Every term except 1 and -1/51 cancels (this is known as “telescoping”). The sum is 1 - 1/51 = 50/51.</p>

<ol>
<li>Check out this thread:
<a href=“http://talk.collegeconfidential.com/sat-preparation/1403799-math-problems.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1403799-math-problems.html&lt;/a&gt;&lt;/li&gt;
</ol>

3

<p>I’m very thankful to you
But I have a question about the firsr solution» how have we come with 1. And -1\51</p>

4

<p>First term is 1 - 1/2 (or 1/1 - 1/2). The 50th term is 1/50 - 1/51. When you telescope, all of the intermediate terms (1/2, 1/3, …, 1/50) cancel out, leaving 1 - 1/51.</p>