Answer:
I think the question is wrong so, I will try and explain with some right questions
Step-by-step explanation:
We are give 6 sequences to analyse
1. an = 3 · (4)n - 1
2. an = 4 · (2)n - 1
3. an = 2 · (3)n - 1
4. an = 4 + 2(n - 1)
5. an = 2 + 3(n - 1)
6. an = 3 + 4(n - 1)
1. This is the correct sequence
an=3•(4)^(n-1)
If this is an
Let know an+1, the next term
an+1=3•(4)^(n+1-1)
an+1=3•(4)^n
There fore
Common ratio an+1/an
r= 3•(4)^n/3•(4)^n-1
r= (4)^(n-n+1)
r=4^1
r= 4, then the common ratio is 4
Then
First term is when n=1
an=3•(4)^(n-1)
a1=3•(4)^(1-1)
a1=3•(4)^0=3.4^0
a1=3
The first term is 3 and the common ratio is 4, it is a G.P
2. This is the correct sequence
an=4•(2)^(n-1)
Therefore, let find an+1
an+1=4•(2)^(n+1-1)
an+1= 4•2ⁿ
Common ratio=an+1/an
r=4•2ⁿ/4•(2)^(n-1)
r=2^(n-n+1)
r=2¹=2
Then the common ratio is 2,
The first term is when n =1
an=4•(2)^(n-1)
a1=4•(2)^(1-1)
a1=4•(2)^0
a1=4
It is geometric progression with first term 4 and common ratio 2.
3. This is the correct sequence
an=2•(3)^(n-1)
Therefore, let find an+1
an+1=2•(3)^(n+1-1)
an+1= 2•3ⁿ
Common ratio=an+1/an
r=2•3ⁿ/2•(3)^(n-1)
r=3^(n-n+1)
r=3¹=3
Then the common ratio is 3,
The first term is when n =1
an=2•(3)^(n-1)
a1=2•(3)^(1-1)
a1=2•(3)^0
a1=2
It is geometric progression with first term 2 and common ratio 3.
4. I think this correct sequence so we will use it.
an = 4 + 2(n - 1)
Let find an+1
an+1= 4+2(n+1-1)
an+1= 4+2n
This is not GP
Let find common difference(d) which is an+1 - an
d=an+1-an
d=4+2n-(4+2(n-1))
d=4+2n-4-2(n-1)
d=4+2n-4-2n+2
d=2.
The common difference is 2
Now, the first term is when n=1
an=4+2(n-1)
a1=4+2(1-1)
a1=4+2(0)
a1=4
This is an arithmetic progression of common difference 2 and first term 4.
5. I think this correct sequence so we will use it.
an = 2 + 3(n - 1)
Let find an+1
an+1= 2+3(n+1-1)
an+1= 2+3n
This is not GP
Let find common difference(d) which is an+1 - an
d=an+1-an
d=2+3n-(2+3(n-1))
d=2+3n-2-3(n-1)
d=2+3n-2-3n+3
d=3.
The common difference is 3
Now, the first term is when n=1
an=2+3(n-1)
a1=2+3(1-1)
a1=2+3(0)
a1=2
This is an arithmetic progression of common difference 3 and first term 2.
6. I think this correct sequence so we will use it.
an = 3 + 4(n - 1)
Let find an+1
an+1= 3+4(n+1-1)
an+1= 3+4n
This is not GP
Let find common difference(d) which is an+1 - an
d=an+1-an
d=3+4n-(3+4(n-1))
d=3+4n-3-4(n-1)
d=3+4n-3-4n+4
d=4.
The common difference is 4
Now, the first term is when n=1
an=3+4(n-1)
a1=3+4(1-1)
a1=3+4(0)
a1=3
This is an arithmetic progression of common difference 4 and first term 3.
. The problem states that its supplemenbt is nine times this, so it has measure 
.
.
, or 162°.