1. A geometric sequence has first term 3 and common ratio -2. Find the 10^{th} term.
2. The second term of a geometric sequence is 6 and the fourth term is 96. Find the possible values of the first term and the common ratio.
3. Find the number of terms in the geometric sequence :
1, 2, 4, 8, ......., 512
4. Find the sum of 10 terms of the geometric sequence :
1, 2, 4, 8, ..........
5. Find the sum of first 8 terms of a geometric sequence whose n^{th} term 3^{2n-1}.
6. Find the first term of a geometric sequence whose common ratio is 5 and sum to first 6 terms is 46872.
7. Find the sum of the geometric series :
2 + 6 + 18 + ........ + 13122
8. Find the sum of the geometric series :
5 + 5 + 5 + ........ to 27 terms
9. Find the sum to infinity of 9 + 3 + 1 + ........
10. Peterson writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs $2 to mail one letter, find the amount spent on postage when 8^{th} set of letters is mailed.
1. Answer :
Formula for n^{th} term of a geometric sequence :
a_{n} = a_{1}r^{n - 1}
Substitute n = 10, a_{1} = 3 and d = -2.
a_{10} = 3(-2)^{10 - 1}
= 3(-2)^{9}
= 3(-512)
= -1536
2. Answer :
a_{2} = 6 and a_{4} = 96
a_{2} = 6 a_{1}r^{2 - 1} = 6 a_{1}r = 6 ----(1) |
a_{4} = 96 a_{1}r^{4 - 1} = 96 a_{1}r^{3} = 96 ----(2) |
(2) ÷ (1) :
r^{2} = 16
Take square root on both sides.
r = ±4
The possible values of the common ratio are -4 and 4.
Substitute r = -4 in (1). a_{1}(-4) = 6 a_{1} = -3/2 |
Substitute r = 4 in (1). a_{1}(4) = 6 a_{1} = 3/2 |
The possible values of the first term are -3/2 and 3/2.
3. Answer :
1, 2, 4, 8, ......., 512
This is a geometric sequence with the first term 1 and common ratio 2.
Let a_{n} = 512.
a_{1}r^{n - 1 }= 512
Substitute a_{1} = 1 and r = 2.
1(2)^{n - 1 }= 512
Write 512 as a power of 2.
2^{n - 1 }= 2^{9}
n - 1 = 9
n = 10
4. Answer :
1, 2, 4, 8, ..........
This is a geometric sequence with a_{1} = 1 and r = 2.
Formula for the sum of first n terms of a geometric sequence.
S_{n} = a_{1}(1 - r^{n})/(1 - r)
Substitute n = 10, a_{1} = 1 and r = 2.
S_{10} = 1(1 - 2^{10})/(1 - 2)
= 1(1 - 1024)/(-1)
= -1023/(-1)
= 1023
5. Answer :
a_{n} = 3^{2n-1}
a_{1} = 3^{2(1) - 1}
= 3^{2 - 1}
= 3^{1}
= 3
a_{2} = 3^{2(2)-1}
= 3^{4 - 1}
= 3^{3}
= 27
Common ratio :
r = a_{2}/a_{1}
r = 27/3
r = 9
Formula for the sum of first n terms of a geometric sequence.
S_{n} = a_{1}(1 - r^{n})/(1 - r)
Substitute n = 8, a_{1} = 3 and r = 9.
S_{8} = 3(1 - 3^{8})/(1 - 3)
= 3(1 - 6561)/(-2)
= 3(-6560)/(-2)
= 3(3280)
= 9840
6. Answer :
S_{6} = 46872
a_{1}(1 - r^{6})/(1 - r) = 46872
Substitute r = 5.
a_{1}(1 - 5^{6})/(1 - 5) = 46872
a_{1}(1 - 15625)/(-4) = 46872
a_{1}(-15624)/(-4) = 46872
3906a_{1} = 46872
Divide each side by 3906.
a_{1} = 12
7. Answer :
2 + 6 + 18 + ........ + 13122
This is a geometric series with a_{1} = 2 and r = 3.
Let a_{n }= 13122.
a_{n} = 13122
a_{1}r^{n - 1} = 13122
Substitute a_{1 }= 2 and r = 3.
2(3)^{n - 1} = 13122
3^{n - 1} = 6561
Write 6561 as a power of 3.
3^{n - 1} = 3^{8}
n - 1 = 8
n = 9
Formula for the sum of first n terms of a geometric sequence.
S_{n} = a_{1}(1 - r^{n})/(1 - r)
Substitute n = 9, a_{1} = 2 and r = 3.
S_{9} = 2(1 - 3^{9})/(1 - 3)
= 2(1 - 19683)/(-2)
= 2(-19682)/(-2)
= 19682
8. Answer :
5 + 5 + 5 + ........ to 27 terms
This is a geometric series with a_{1} = 5 and r = 1.
S_{n} = na_{1}
Substitute n = 27 and a_{1 }= 5.
S_{n} = 27(5)
= 135
9. Answer :
9 + 3 + 1 + ........
This is a geometric series with a_{1} = 9 and r = 1/3.
S_{∞ }= a_{1}/(1 - r)
Substitute a_{1 }= 9 and r = 1/3.
S_{∞ }= 9/(1 - 1/3)
= 9/(2/3)
= 9 ⋅ 3/2
= 27/2
10. Answer :
Amount spent when the first set of letters is mailed :
= 4 ⋅ 2
= $8
Amount spent when the second set of letters is mailed :
= 4 ⋅ 4 ⋅ 2
= $32
Amount spent when the third set of letters is mailed :
= 4 ⋅ 4 ⋅ 4 ⋅ 2
= $128
If this pattern continues, we will have a geometric sequence with the first term 8 and common ratio 4 as shown below.
8, 32, 128, ............. to 8 terms
Find the sum of the terms in the above geometric sequence.
S_{n} = a_{1}(1 - r^{n})/(1 - r)
Substitute n = 8, a_{1} = 8 and r = 4.
S_{8} = 8(1 - 4^{8})/(1 - 4)
= 8(1 - 65536)/(-3)
= 8(-65535)/(-3)
= 8(21845)
= $174,760
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