Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each verbal description of a sequence with its appropriate explicit function. an = 3 · (4) n - 1 an = 4 · (2) n - 1 an = 2 · (3) n - 1 an = 4 + 2 (n - 1) an = 2 + 3 (n - 1) an = 3 + 4 (n - 1) a geometric sequence with first term of 4 and a common ratio of 2 arrowRight an arithmetic sequence with a first term of 2 and a common difference of 3 arrowRight a geometric sequence with first term of 3 and a common ratio of 4 arrowRight an arithmetic sequence with a first term of 3 and a common difference of 4

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.