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5 August, 17:50

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

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  1. 5 August, 18:34
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    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.
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