A gematric series is the one which contains the sum of the terms of a geometric sequence.
Geometric Sequence :
a, ar, ar^{2}, ar^{3}, ..........
Geometric Series :
a + ar + ar^{2 }+ ar^{3 }+ ..........
To find the sum of n terms of a geometric series, we use one of the formulas given below.
S_{n} = a_{1}(r^{n} - 1)/(r - 1), if r ≠ 1
S_{n} = na_{1}, if r = 1
where 'a' is the first term of the series and 'r' is the common ratio.
Formula to find the sum of infinite geometric series :
S_{∞} = a_{1}/(1 - r), if -1 < r < 1
where 'a' is the first term of the series and 'r' is the common ratio.
r = second term/first term or a_{2}/a_{1}
Note :
In an infinite geometric series, if the value of r is not in the interval -1 < r < 1, then the sum does not exist.
Example 1 :
Find the sum of the following geometric series.
1 + 3 + 9 + .......... to 10 terms
Solution :
a_{1} = 1
r = a_{2}/a_{1} = 3/1 = 3 > 1
n = 10
Formula to find sum of n terms of a geometric series where r ≠ 1.
S_{n} = a_{1}(r^{n} - 1)/(r - 1)
Substitute a_{1} = 1 and r = 3.
S_{10} = 1(3^{10} - 1)/(3 - 1)
= (59049 - 1)/2
= 59048/2
= 29524
Example 2 :
Find the sum of 8 terms of the following geometric series.
1 + (-3) + 9 + (-27) +..........
Solution :
a_{1} = 1
r = a_{2}/a_{1} = -3/1 = -3 < 1
n = 10
Formula to find sum of n terms of a geometric series where r > 1.
S_{n} = a_{1}(r^{n} - 1)/(r - 1)
Substitute a_{1} = 1, r = -3 and n = 8.
S_{8} = 1[(-3)^{8} - 1]/(-3 - 1)
= (6561 - 1)/(-4)
= 6560/(-4)
= -1640
Example 3 :
Find the sum of 30 terms of the following geometric series.
7 + 7 + 7 + 7 +..........
Solution :
a_{1} = 7
r = a_{2}/a_{1} = 7/7 = 1
n = 30
Formula to find sum of n terms of a geometric series where r = 1.
S_{n} = na_{1}
Substitute a_{1} = 7 and r = 1.
S_{30} = 30(7)
= 210
Example 4 :
Find the sum of terms of the following geometric series.
1 + 2 + 4 + 8 +.......... + 2048
Solution :
a_{1} = 1
r = a_{2}/a_{1} = 2/1 = 2 > 1
The value of n is not given. So, we have to find the value of n.
Let 2048 be the n^{th} term.
a_{n} = 2048
a_{1}r^{n-1} = 2048
Substitute a_{1} = 1 and r = 2.
1(2)^{n-1} = 2048
2^{n-1} = 2^{11}
n - 1 = 11
n = 12
Formula to find sum of n terms of a geometric series where r ≠ 1.
S_{n} = a_{1}(r^{n} - 1)/(r - 1)
Substitute a_{1} = 1, r = -3 and n = 12.
S_{12} = 1[2^{12} - 1]/(2 - 1)
= (4096 - 1)/1
= 4095
Example 5 :
Find the sum of the following infinite geometric series.
3 + 1 + 1/3 +.......... ∞
Solution :
a_{1} = 3
r = a_{2}/a_{1} = 1/3
The value of r (= 1/3) is in the interval -1 < r < 1.
The sum for the given infinite geometric series exists.
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 3 and r = 1/3.
S_{∞} = 3/(1 - 1/3)
= 3/(2/3)
= 3(3/2)
= 9/2
Example 6 :
Find the sum of the following infinite geometric series.
-3 + (-12) + (-48) +.......... ∞
Solution :
a_{1} = 3
r = a_{2}/a_{1} = -12/(-3) = 4
The value of r (= 4) is not in the interval -1 < r < 1.
So, the sum for the given infinite geometric series does not exist.
Example 7 :
A ball is dropped from a height of 6 m and on each bounce it bounces 2/3 of its previous height.
(i) What is the total length of the downward paths ?
(ii) What is the total length of the upward paths ?
(iii) How far does the ball travel till it stops bouncing?
Solution :
(i) Distance covered in the downward path :
= 6 + 4 + 8/3 + 16/9 + .............
a_{1} = 6
r = a_{2}/a_{1} = 4/6 = 2/3
The value of r (= 2/3) is in the interval -1 < r < 1.
Formula to find sum of infinite geometric series :
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 6 and r = 2/3.
= 6/(1 - 2/3)
= 6/(1/3)
= 6(3/1)
= 18 m
(ii) Distance covered in the upward path :
= 4 + 8/3 + 16/9 + ............
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 4 and r = 2/3.
S_{∞} = 4/(1 - 2/3)
= 4/(1/3)
= 4(3/1)
= 12 m
(iii) Total distance covered :
= 18 + 12
= 30 m
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