Question 77523
<pre><font size = 5><b>
Can someone please help me with this problem? 
Find a<sub>1</sub> in a geometric series for which 
S<sub>n</sub>=189, r=1/2, and a<sub>n</sub>=3.

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

a<sub>n</sub> = a<sub>1</sub>(1/2)<sup>n-1</sup>

3 = a<sub>1</sub>(1/2<sup>n-1</sup>)

3(2<sup>n-1</sup>) = a<sub>1</sub>


      a<sub>1</sub>(1 - r<sup>n</sup>)
S<sub>n</sub> = ------------
        1 - r  

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

       a<sub>1</sub>[1 - (1/2)<sup>n</sup>]
189 = ----------------
         1 - 1/2

Simplifying the second

       a<sub>1</sub>[1 - 1/2<sup>n</sup>]
189 = --------------
           1/2

       a<sub>1</sub>(1 - 2<sup>-n</sup>)
189 = -------------
           1/2

Multiply the numerator and 
denominator on the right by 2


189 = 2a<sub>1</sub>(1 - 2<sup>-n</sup>) 

Since 3(2<sup>n-1</sup>) = a<sub>1</sub>, substitute 3(2<sup>n-1</sup>) for a<sub>1</sub> 

189 = 2[3(2<sup>n-1</sup>)](1 - 2<sup>-n</sup>)

189 = 3(2<sup>1</sup>)(2<sup>n-1</sup>)(1 - 2<sup>-n</sup>)

189 = 3(2<sup>n</sup>)(1 - 2<sup>-n</sup>)

Divide both sides by 3

63 = 2<sup>n</sup>(1 - 2<sup>-n</sup>)

Distribute

63 = 2<sup>n</sup> - 2<sup>n-n</sup>

63 = 2<sup>n</sup> - 2<sup>0</sup>

63 = 2<sup>n</sup> - 1

64 = 2<sup>n</sup>

2<sup>6</sup> = 2<sup>n</sup>

6 = n

3(2<sup>n-1</sup>) = a<sub>1</sub>
3(2<sup>6-1</sup>) = a<sub>1</sub>
3(2<sup>5</sup>) = a<sub>1</sub>
  3(32) = a<sub>1</sub>
     96 = a<sub>1</sub>

To check, start with 96 and multiply by 1/2 until 
we have 6 terms:

96, 48, 24, 12, 6, 3

So the nth or 6th term, a<sub>n</sub> = a<sub>6</sub> = 3.  That checks.

S<sub>n</sub> = S<sub>6</sub> = 96+48+24+12+6+3 = 189.  That checks.

Edwin</pre>