Expert: Ahmed Salami - 2/15/2009

Question
Hi. Please help me..
A geometric series has third term 27 and sixth term 8.
-Show that the common ratio is 2/3
-find the first term of the series
-find the sum to infinity of the series
-find, to 3significant figures, the difference between the sum of the first 10terms of the series and the sum to infinity of the series.
thank you

Answer
Hi Hussein,
The nth term of a geometric series is
T(n) = ar^(n-1)
where a is the first term and r is the common ratio.
The sum of the first n terms is
S(n) = a[r^n  -  1]/(r-1)
The sum to infinity is
S(inf) = a/1-r      (only for |r|<1)
Now,
T(3) = ar^2 = 27
T(6) = ar^5 = 8
dividing T(6) by T(3),
ar^5/ar^2 = 8/27
r^3 = 8/27
r = 2/3

From ar^2 = 27
a(2/3)^2 = 27
4a/9 = 27
a = 243/4

S(inf) = a/1-r
      = (243/4)/(1 - 2/3)
      = (243/4)/(1/3)
      = 243/4 x 3
      = 729/4

S(10) = a[r^n  -  1]/(r-1)
     = a[r^10  -  1]/(r-1)
     = (243/4)[(2/3)^10  -  1]/[(2/3) - 1]
     = (243/4)[(1024/59049) - 1]/(-1/3)
     = (243/4)(-58025/59049)x(-3)
     = 42300225/236196

S(inf) - S(10) = 729/4  -  42300225/236196
              = 182.25 - 179.09
              = 3.16  to 3sf

Regards