Answer:
Step-by-step explanation:
Geometric Sequence:
A geometric sequence is a series of number that follows an order.
It follows a rule such as the below:
Xₙ = ar⁽ⁿ⁻¹⁾
Where:
n = is the nth term
a = is the first terms
r = is the common ratio
Now that we have the formula for the geometric sequence, we can now calculate the following geometric sequences:
1) 3,12,48,__ , __
First we need to determine what is the common ratio "r"
looking at the sequence, the common ration is 4 because
3 = first digit
3 x 4 = 12 (the second digit)
12 x 4 = 48 (the third digit)
Hence,
r = 4
The missing digit in this sequence is the 4th and 5th digit, thus using the formula given above:
For the 4th digit:
Xₙ = ar⁽ⁿ⁻¹⁾
X₄ = 3(4)⁽⁴⁻¹⁾
X₄ = 3(4)⁽³⁾
X₄ = 3(64)
X₄ = 192
For the 5th digit:
Xₙ = ar⁽ⁿ⁻¹⁾
X₅ = 3(5)⁽⁵⁻¹⁾
X₅ = 3(4)⁽⁴⁾
X₄ = 3(256)
X₄ = 768
Therefore the geometric sequence is:
3, 12, 48, 192 , 768
2) __,__,32,64,128
First find out "r"
From the 3rd to 5th digit:
the common ratio is 2
(because 128 ÷ 64 = 2, similarly 64 ÷ 32 = 2) thus, r = 2
For the 1st digit "a":
We first need to determine a from the already given sequence, let's say the 3rd sequence:
32 = a(2)⁽³⁻¹⁾
32 = a(2)⁽²⁾
32 = a(4)
a = 32 ÷ 4
Thus the 1st digit is:
a = 8
For the 2nd digit
Xₙ = ar⁽ⁿ⁻¹⁾
X₂ = 8(2)⁽²⁻¹⁾
X₂ = 3(4)⁽¹⁾
X₂ = 3(4)
X₂ = 12
Therefore the geometric sequence is:
8,12,32,64,128
3) 120,60,30,__,__,__
Find r:
From the standard formula for geometric sequence:
Xₙ = ar⁽ⁿ⁻¹⁾
Lets use the 2nd digit from the sequence, 60
60 = 120r⁽²⁻¹⁾
60 = 120r⁽¹⁾
[tex]\frac{60}{120}[/tex] = r
r = [tex]\frac{1}{2}[/tex] or 0.5
Thus finding the 4th, 5th and 6th digits would be:
X₄ = 120(0.5)⁽⁴⁻¹⁾
X₄ = 120(0.5)⁽³⁾
X₄ = 120(0.125)
X₄ = 15 → 4th digit
--
X₅ = 120(0.5)⁽⁵⁻¹⁾
X₅ = 120(0.5)⁽⁴⁾
X₅ = 120(0.0625)
X₅ = 7.5 → 5th digit
--
X₆ = 120(0.5)⁽⁶⁻¹⁾
X₆ = 120(0.5)⁽⁵⁾
X₅ = 120(0.03125)
X₆ = 3.75 → 5th digit
Thus the geometric sequence would be:
120 ,60 ,30 ,15, 7.5 , 3.75
For items 4 to 9 geometric sequence, I will not be showing the full solution as the above solutions would be enough for you to be able to solve the remaining problems, I will only give the common ratio "r" and the complete geometric sequences:
4) 5,__,20,40,__,__
Ans:
r = 2
5, 10, 20, 40, 80, 120
5)__,4,12,40,__,__
Lets find r:
Xₙ = ar⁽ⁿ⁻¹⁾
4 = ar⁽²⁻¹⁾
4 = ar⁽¹⁾
4=ar
thus;
a = [tex]\frac{4}{r}[/tex]
For the third term
12 = [tex]\frac{4}{r}[/tex](r)⁽³⁻¹⁾
12 = [tex]\frac{4}{r}[/tex] r²
in this case the donominator "r" will be cancelled.
12= 4r
thus r = 3
To find the first term lets use the 3rd digit from the sequence
12 = a(3)⁽³⁻¹⁾
12 = a(3)²
12 = a(9)
thus
a = [tex]\frac{12}{9}[/tex]
or
a= 1 [tex]\frac{1}{3}[/tex] → 1.3333
For the 5th term:
X₅ = 1.333(3)⁽⁵⁻¹⁾
X₅ = 1.333 (3)⁽⁴⁾
X₅ = 1.333(81)
X₅ = 108
For the 6th term:
X₆ = 1.333(3)⁽⁶⁻¹⁾
X₆ = 1.333 (3)⁽⁵⁾
X₆ = 1.333(243)
X₆ = 324
Thus the geometric sequence would be:
1 1/3 ,4 ,12 ,40,108,324
6)-2,__,__,-16,-32,-64
Ans:
r = 2
-2,-4, -8 ,-16,-32,-64
7) 256,__,__,-32,16
Ans:
r = 0.5
256,-128,64,-32,16
8) 27,9,__,__,1/3
Ans:
r = 9/27
27 , 9 , 3 , 1 , 1/3
9)1/4,__,__,__,64,256
Ans:
r = 4
1/4 , 1 , 4 , 16 ,64,256
B. Insert 3 terms between 2 and 32 of geometric sequence.
Thus,
2, __, __, __, 32
To find r,
Xₙ = ar⁽ⁿ⁻¹⁾
32 = 2r⁽⁵⁻¹⁾
32 = 2r⁴
r⁴ = 32/2
r = [tex]\sqrt[4]{16}[/tex]
r = 2
Xₙ = ar⁽ⁿ⁻¹⁾
X₂ = 2(2)⁽²⁻¹⁾
X₂ = 2(2)⁽¹⁾
X₂ = 2(2)
X₂ = 4
__
Xₙ = ar⁽ⁿ⁻¹⁾
X₃ = 2(2)⁽³⁻¹⁾
X₃ = 2(2)⁽²⁾
X₃ = 2(4)
X₃ = 8
__
Xₙ = ar⁽ⁿ⁻¹⁾
X₄ = 2(2)⁽⁴⁻¹⁾
X₃ = 2(2)⁽³⁾
X₃ = 2(8)
X₃ = 16
Thus the geometric sequence would be:
2, 4 , 8 , 16 , 32
For more information regarding geometric sequence visit the link below:
https://brainly.ph/question/1583842
_
#BetterWithBrainly
