<p>Hey, so I found this problem on the College Board website, and I'm confused. Help me please?</p>
<p>For some real number t, the first three terms of an arithmetic sequence are 2t,5t−1, and 6t +2. What is the numerical value of the fourth term?
(A) 4 (B) 8 (C) 10 (D) 16 (E) 19</p>
<p>The answer I got was 30. I tried to solve for t by setting (3rd term - 2nd term) equal to (2nd term - 1st term), but then all the numbers in the sequence were too big.</p>
<p>SOS? Thanks. :D</p>
<p>The answer is E....what u did was correct (3rd term - 2nd term) = (2nd term - 1st term). so </p>
<p>6t + 2 - (5t-1) = 5t - 1 - 2t
t + 3 = 3t - 1
t = 2</p>
<p>You plug in 2 for each other terms:
t1= 2(2) = 4
t2= 5(2)-1 = 9
t3= 6(2)+2 = 14</p>
<p>The difference between them is 5 therefore the next term is 14 + 5 = 19</p>
<p>The difference between the 1st and 2nd and 2nd and 3rd terms will be equal since it's arithmetic.</p>
<p>5t - 1 - 2t = 6t + 2 - (5t -1)
3t -1 = t + 2 + 1
2t = 4
t = 2</p>
<p>So it goes
4, 9, 14</p>
<p>Add the common difference, 5 to 14, and you get 19.</p>
<p>Omg, I used 6t+4 instead of 6t+2 for the third term!</p>
<p>Thanks, guys.</p>