Asked by Jon
Find a1 in a geometric series for which Sn=189,r=1/2,and an=3.
Sn=a1(1-r^n)/1-r
189=a1(1-1/2^3)/1-1/2
189=a1(-.5)/.5
189/-.25
a1=-756
Sn=a1(1-r^n)/1-r
189=a1(1-1/2^3)/1-1/2
189=a1(-.5)/.5
189/-.25
a1=-756
Answers
Answered by
Reiny
YOur third line is wrong
should be:
189 = a(1 - 1/8)/(1 - 1/2)
189 = a(7/8)/(1/2)
189 = a(7/4)
756 = 7a
108 = a
(check 108+54+27 = 189 !)
should be:
189 = a(1 - 1/8)/(1 - 1/2)
189 = a(7/8)/(1/2)
189 = a(7/4)
756 = 7a
108 = a
(check 108+54+27 = 189 !)