Question 142221
find the explicit formula 
a<sub>7</sub>=-10 and a<sub>13</sub>=-48
<pre><font size = 4 color = "indigo"><b>
The problem is to find a formula that we can use to 
fill in the empty blanks of this sequence:

___,___,___,___,___,___,<u>-10</u>,___,___,___,___,___,<u>-48</u>


Let's assume it's an arithmetic sequence.  Then we use

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

Substitute n=7

a<sub>n</sub> = a<sub>1</sub> + (7-1)d
a<sub>7</sub> = a<sub>1</sub> + 6d

when n=7, a<sub>n</sub> = -10, so a<sub>7</sub> = -10

Substituting a<sub>7</sub> = -10

-10 = a<sub>1</sub> + 6d

Also,

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

Substitute n=13

a<sub>13</sub> = a<sub>1</sub> + (13-1)d
a<sub>13</sub> = a<sub>1</sub> + 12d

when n=13, a<sub>n</sub> = -48, so a<sub>13</sub> = -48

Substituting a<sub>13</sub> = -48

-48 = a<sub>1</sub> + 12d

So we have the system of two equations and
two unknowns

-10 = a<sub>1</sub> + 6d
-48 = a<sub>1</sub> + 12d

So we solve that system and get:

a<sub>1</sub> = 28, d = -19/3

So to get the formula, we take

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

And substitute those values of a<sub>1</sub> and d

a<sub>n</sub> = {{{28 + (n-1)(-19/3)}}}

a<sub>n</sub> = {{{28 -(19/3)n + 19/3}}}

a<sub>n</sub> = {{{84/3 - (19n)/3 + 19/3}}}

a<sub>n</sub> = {{{(103-19n)/3}}}

That's the formula.  We can now fill in the blanks:

<u>28</u>,<u>65/3</u>,<u>46/3</u>,<u>9</u>,<u>8/3</u>,<u>-11/3</u>,<u>-10</u>,<u>-49/3</u>,<u>-68/3</u>,<u>-29</u>,<u>-106/3</u>,<u>-48</u>.

-------------------------------------
That's the solution your teacher wants, most likely.

However, it is also possible that it could be a 
geometric sequence as well:

Let's assume it's a geometric sequence.  Then we use

a<sub>n</sub> = a<sub>1</sub>r<sup>n-1</sup>

Substitute n=7

a<sub>7</sub> = a<sub>1</sub>r<sup>7-1</sup>
a<sub>7</sub> = a<sub>1</sub>r<sup>6</sup>

when n=7, a<sub>n</sub> = -10, so a<sub>7</sub> = -10

Substituting a<sub>7</sub> = -10

-10 = a<sub>1</sub>r<sup>6</sup>

Also,

a<sub>n</sub> = a<sub>1</sub>r<sup>n-1</sup>

Substitute n=13

a<sub>13</sub> = a<sub>1</sub>r<sup>13-1</sup>
a<sub>13</sub> = a<sub>1</sub>r<sup>12</sup>

when n=13, a<sub>n</sub> = -48, so a<sub>13</sub> = -48

Substituting a<sub>13</sub> = -48

-48 = a<sub>1</sub>r<sup>12</sup>

So we have the system of two equations and
two unknowns

-10 = a<sub>1</sub>r<sup>6</sup>
-48 = a<sub>1</sub>r<sup>12</sup>

So we solve that system and get:

a<sub>1</sub> = {{{-25/12}}}, d = {{{root(6,(48))}}}

So to get the formula, we take

a<sub>n</sub> = a1r<sup>n-1</sup>

And substitute those values of a<sub>1</sub> and d

a<sub>n</sub> = {{{(-25/12)(root(6,48))^(n-1)}}}

That's the formula.  We can now fill in the blanks,
but I'll just give their decimal values:

<u>-2.083</u>,<u>-2.706</u>,<u>-3.514</u>,<u>-4.564</u>,<u>-5.928</u>,<u>-7.699</u>,<u>
-10</u>,<u>-12.99</u>,<u>-16.87</u>,<u>-21.91</u>,<u>-28.46</u>,<u>-36.96</u>,<u>-48</u>.

Actually, arithmetic and geometric sequences are only 2 kinds of
many sequences.  So technically there are infinitely many answers to
this problem, but only one arithmetic sequence solution, and only one
geometric sequence solution. 

Edwin</pre>