Question 1118377
What are the first three terms if the 11th term of an 
arithmetic sequence is 60 and the sum of the first 11 
terms is 55?
<pre><b>
<u>a<sub>1</sub></u>,__,__,__,__,__,__,__,__,__,<u>60</u>,...

S<sub>n</sub> = (n/2)(a<sub>1</sub> + a<sub>n</sub>)

S<sub>11</sub> = (11/2)(a<sub>1</sub> + a<sub>11</sub>)

55 = (11/2)(a<sub>1</sub> + 60)

Multiply both sides by 2

110 = 11(a<sub>1</sub> + 60)

110 = 11a<sub>1</sub> + 660

-550 = 11a<sub>1</sub>

-550/11 = a<sub>1</sub>

-50 = a<sub>1</sub> 

So

<u>a<sub>1</sub></u>,__,__,__,__,__,__,__,__,__,<u>60</u>,

becomes

<u>-50</u>,__,__,__,__,__,__,__,__,__,<u>60</u>,

To fill in the blanks, we only need the common difference d:

a<sub>n</sub> = a<sub>1</sub> + (n - 1)d

 60 = -50 + (11 - 1)d
 60 = -50 + 10d
110 = 10d
 11 = d

So the sequence

-50,__,__,__,__,__,__,__,__,__,<u>60</u>,...

becomes

<u>-50</u>,<u>-39</u>,<u>-28</u>,<u>-17</u>,<u>-6</u>,<u>5</u>,<u>16</u>,<u>27</u>,<u>38</u>,<u>49</u>,<u>60</u>,...

So the first three terms are <u>-50</u>,<u>-39</u> and <u>-28</u>

Edwin</pre></b>