<p>Hi,</p>
<p>I don't understand how to solve this question (from Barron's) test 1, would be great is someone could help me with this.</p>
<ol>
<li>Each term of a sequence, after the first, is inversely proportional to the term preceding it. If the first two terms are 2 and 6, what is the 12th term?</li>
</ol>
<p>Thanks!</p>
<p>According to Barron’s…</p>
<p>“t(n)<em>t(n+1)=k. 2</em>6=k. Therefore, 6*t(3)=12, and so t(3)=2. Continuing the process gives all odd terms to be 2 and all even terms to be 6.”</p>
<p>t(n) is “t subscript n”, not a function…</p>
<p>I read the explanation but I don’t understand that t(subscript 3) thing…What does inversely proportional mean exactly anyway? Can you list out all the terms for me so I can realize what I’m doing wrong?</p>
<p>Thanks!</p>
<p>I dont think a question like this would come up on the Math 2, Barrons is notorious for giving harder and excess material.</p>
<p>here t1= 2 and t2 = 6
t1= preceding term=tn
t2= term = t(n+1)
condition is that they are inversely proportional to each other
t1 ∝ 1/t2 or tn ∝ 1/t(n+1)
putting constant ‘k’ to equate
tn = k/t(n+1) or tn * t(n+1) = k
or t1<em>t2=k further k=2</em>6=12
t2*t3=k put value of t2 and k to get t3=6
go till t12 or u can just guess its 6 even terms will be 6 and odd terms will be 2</p>