<p>I got these questions out of an old SAT exam. They’ll probably be easy for a lot of you out there, which is good for me because I don’t know how to approach it. I know, I’m really, really, really bad with math. Help plz?</p>
<li>S is the sum of the first 100 consecutive positive even integers, and T is the sum of the first 100 consecutive positive integers. S is what percent greater than T?<br></li>
</ol>
<p>a. 100%
b. 50%
c. 10%
d. 2%
e. 1%</p>
<p>So how do you do it avoiding the long way?</p>
<li><p>If kn is not equal to k, and n= 1/k, which of the following expressions is equivalent to (1-k)/(1-n)
a. -n
b. -k
c. l
d. k
e. n</p></li>
<li><p>The first 2 numbers of a sequence are 1 and 3, respectively. The 3rd number is 4, and in general, every number after the 2nd is the sum of the 2 numbers immediately preceding it. How many of the first 1000 numbers in this sequence are odd?</p></li>
</ol>
<p>1) percent is change divided by original.
change = S - T, which is 1 + 2 + 3 + 4... + 100
S: 2 + 4 + 6 + 8 + .. + 200</p>
<h2>T: 1 + 2 + 3 + 4 + .. + 100</h2>
<pre><code>1 + 2 + 3 + 4 ..
</code></pre>
<p>original = T = 1 + 2 + 3 + .. + 100</p>
<p>so, it's 1.00, or 100% (answer choice A)</p>
<p>2) (1-k)/(1-n). since n = 1/k, substitute.
(1-k)/(1-1/k). now multiply top and bottom by k.
(k - k^2)/(k - 1)
-(k^2 - k)/(k - 1)
-k(k - 1)/(k - 1)
-k (answer choice B)</p>
<p>3) let's write out the beginning of the sequence.
1, 3, 4, 7, 11, 18, 29, ...
notice how after every two odd nubmers, there's an even number. (this is because an even number is generated after adding two odds, and odd numbers are generated by adding an odd with an even.)</p>
<p>it's easier to count all the evens, and then subtract it from 1000. since it occurs every third one.. it'll be the 3rd term, 6th term, 9th term... until 999th term. So, take 999/3 = 333 evens.
odds = 1000-333 = 667 (answer choice E)</p>
<p>Thank you so much ahhh! All those answers are correct. I never would have been able to figure out the approach on my own. Thank you, thank you, thank you. I hope you'll continue to help as I post more questions sometime soon.</p>