Can someone please help me with this problem? 
Find a1 in a geometric series for which 
Sn=189, r=1/2, and an=3.

an = a1rn-1

an = a1(1/2)n-1

3 = a1(1/2n-1)

3(2n-1) = a1


      a1(1 - rn)
Sn = ------------
        1 - r  

Sn = a1(1/2)n-1

       a1[1 - (1/2)n]
189 = ----------------
         1 - 1/2

Simplifying the second

       a1[1 - 1/2n]
189 = --------------
           1/2

       a1(1 - 2-n)
189 = -------------
           1/2

Multiply the numerator and 
denominator on the right by 2


189 = 2a1(1 - 2-n) 

Since 3(2n-1) = a1, substitute 3(2n-1) for a1 

189 = 2[3(2n-1)](1 - 2-n)

189 = 3(21)(2n-1)(1 - 2-n)

189 = 3(2n)(1 - 2-n)

Divide both sides by 3

63 = 2n(1 - 2-n)

Distribute

63 = 2n - 2n-n

63 = 2n - 20

63 = 2n - 1

64 = 2n

26 = 2n

6 = n

3(2n-1) = a1
3(26-1) = a1
3(25) = a1
  3(32) = a1
     96 = a1

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, an = a6 = 3.  That checks.

Sn = S6 = 96+48+24+12+6+3 = 189.  That checks.

Edwin