You can put this solution on YOUR website! Find a(1) in a geometric series for which S(n) =189, r=1/2, and a(n)=3.
Thank you
AN=A1*R^(N-1)
3=A1*(1/2)^(N-1)=A1*0.5^(N-1)..............................I
SN=A1*{1-R^N}/(1-R)
189=A1*{1-(1/2)^N }/(1-0.5)=2A1*(1-0.5^N).........................II
EQN.II/EQN.I
189/3=2A1(1-0.5^N)/{A1*0.5^(N-1)}=63
2(1-(0.5^(N-1))*(0.5))=63*0.5^(N-1)
LET 0.5^(N-1)=X
2(1-0.5X)=63X
2-X=63X
64X=2
X=2/64=1/32=0.5^(N-1)
0.5^(N-1)=(1/2)^5=0.5^5
N-1=5
N=6
PUTTING IN EQN.I
3=A1*(0.5)^5=A1/32
A1=32*3=96
