Question 1 :
Find the sum of the following geometric series.
1 + 3 + 9 + .......... to 10 terms
Question 2 :
Find the sum of 8 terms of the following geometric series.
1 + (-3) + 9 + (-27) +..........
Question 3 :
Find the sum of 30 terms of the following geometric series.
7 + 7 + 7 + 7 +..........
Question 4 :
Find the sum of terms of the following geometric series.
1 + 2 + 4 + 8 +.......... + 2048
Question 5 :
Find the sum of the following infinite geometric series.
3 + 1 + 1/3 +.......... ∞
Question 6 :
Find the sum of the following infinite geometric series.
-3 + (-12) + (-48) +.......... ∞
Question 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?
1. Answer :
1 + 3 + 9 + .......... to 10 terms
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
2. Answer :
1 + (-3) + 9 + (-27) +.......... to 8 terms
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
3. Answer :
7 + 7 + 7 + 7 +.......... to 30 terms
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
4. Answer :
1 + 2 + 4 + 8 +.......... + 2048
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
5. Answer :
3 + 1 + 1/3 +.......... ∞
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
6. Answer :
-3 + (-12) + (-48) +.......... ∞
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.
7. Answer :
(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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Feb 02, 23 07:46 AM
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