A series whose terms are in geometric sequence is called geometric series.
Let a, ar, ar2 , ...arn-1 , ... be the geometric sequence.
Sum of the first n terms of a geometric sequence :
When r ≠ 1 Sn = a(rn - 1)/(r - 1) |
When r = 1 Sn = na |
S∞ = a/(1 - r)
Question 1 :
Find the sum of first n terms of the G.P
(i) 5, -3, 9/5, -27/25, ...............
Solution :
a = 5, r = t2/t1 = -3/5 ≠ 1
Sn = a(rn - 1)/(r - 1)
= 5((-3/5)n - 1)/((-3/5) - 1)
= 5((-3/5)n - 1)/((-3-5)/5)
= 5((-3/5)n - 1)/(-8/5)
= (-25/8)((-3/5)n - 1)
Sn = (25/8)(1 - (-3/5)n)
(ii) 256, 64, 16,…
Solution :
a = 256, r = t2/t1 = 64/256 = 1/4 ≠ 1
Sn = a(rn - 1)/(r - 1)
= 256((1/4)n - 1)/((1/4) - 1)
= 256((1/4)n - 1)/(-3/4)
= 256(-4/3)((1/4)n - 1)
Sn = (-1024/3)((1/4)n - 1)
Sn = (1024/3)(1 - (1/4)n)
Question 2 :
Find the sum of first six terms of the G.P. 5, 15, 45, …
Solution :
n = 6, a = 5, r = 15/5 = 3 ≠ 1
Sn = a(rn - 1)/(r - 1)
= 5(36 - 1)/(3 - 1)
= 5(729 - 1)/2
= 5(728/2)
= 1820
Question 3 :
Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Solution :
S6 = 46872
Sn = a(rn - 1)/(r - 1)
r = 5 and n = 6
S6 = a(56 - 1)/(5 - 1)
46872(4) = a(56 - 1)
46872(4) / 15624 = a
a = 12
Hence the first term is 12.
Question 4 :
Find the sum to infinity of
(i) 9 + 3 + 1 + .....
Solution :
Sum of infinite series :
S∞ = a/(1-r)
a = 9, r = 3/9 = 1/3
S∞ = 9/(1-(1/3))
= 9/(2/3)
= 27/2
Hence the required sum is 27/2.
(ii) 21 + 14 + 28/3 + .............
Solution :
a = 21, r = 14/21 = 2/3
Sum of infinite series :
S∞ = a/(1-r)
S∞ = 21/(1-(2/3))
= 21/(1/3)
= 63
Hence the required sum is 63.
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