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Mathematics · Oct / Nov 2024 · Paper 21 · Question 7

YMA01

As printed in the paper

The question as text

7. Jem pays money into a savings scheme, A, over a period of 300 months. Jem pays £20 into scheme A in month 1, £20.50 in month 2, £21 in month 3 and so on, so that the amounts Jem pays each month form an arithmetic sequence.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme
  • 7(a)[1]

    (

    )

    (

    )

    ( )

    =

    20 99 0.5

    +

    =

    £ 69.50*

    u

    100

  • 7(b)[2]

    1 300

    (

    )

    2 20

    299 0.5

    (

    )

    ...

    S

    =

    ×

    +

    =

    300

    2

    or

    1 300

    (

    )

    S

    =

    20 169.50

    +

    =

    ...

    300

    2

    =

    ( )

    £ 28425

  • 7(c)[3]

    299 250

    (

    )

    20

    ×

    r

    299

    =

    250

    r

    =

    =

    1.008483032...

    20

    (

    )

    20 1

    −

    r

    300

    (

    )

    =

    =

    27362.948...

    S

    300

    1

    −

    r

    28425

    27362.948...

    −

    (£)1060

    (a)

    (

    )

    B1*: Correct method shown e.g.

    20

    +

    99 0.5

    =

    69.50

    is sufficient.

    Note that the £ symbol is not required but

    (

    )

    scores B0

    20

    +

    99 0.5

    =

    69.5

    Condone attempts to work backwards e.g.

    (

    )

    (

    )

    69.5

    =

    20

    +

    n

    −

    1

    ×

    0.5

    49.5

    =

    n

    −

    1

    ×

    0.5

    n

    =

    100

    hence true

    Note that 69.5 can be used in this case but a (minimal) conclusion is required.

    (b)

    1

    (

    )

    (

    )

    M1:

    Attempts

    2 n

    2

    ×

    a

    +

    n

    −

    1

    d

    with n = 300, a = 20 and d = 0.5

    1

    (

    )

    or e.g.

    +

    with n = 300, a = 20 and l = 169.5(0) (or an attempt at 20 299 0.5

    2 n a

    l

    A1:

    Correct value. The £ symbol is not required.

    (c)

    Correct strategy to find r e.g. 299 250.

    ) 299 250

    Award for e.g. (

    M1:

    r =

    or e.g.

    20

    20

    1

    250

    (

    )

    299

    =

    .

    May be implied by awrt 1.01 if no incorrect work is seen e.g. clear

    r

    20

    use of 300 250

    which gives 1.008454636… is M0

    20

    1 log12.5

    May be done via logs e.g.

    r =

    10

    299

    M1:

    Uses their r with a correct GP sum formula with a = 20 and n = 300 and subtracts

    from the answer to part (b) either way round.

    Depends on having attempted the sum of an AP in (b).

    May be implied by their working/value.

    Note this has been seen more than once for the GP sum following 299 250

    20

    20 1 12.5

    (

    )

    −

    S

    = −

    300

    1 1.008483032

    (

    )

    1

    −

    1

    a

    −

    r

    n

    and scores M0 as it suggests

    S

    =

    has been used.

    n

    1

    −

    r

    A1:

    Cao not awrt with or without the “£” symbol.

    Note that the accuracy being used for “r” varies considerably.

    Some rounded values for reference are:

    r

    |Correct (b) −

    |

    S

    S

    300

    300

    1.01

    37576.93252…

    9151.932524…

    1.008

    24796.03972…

    3628.960276…

    1.0085

    27458.42195…

    966.5780531…

    1.00848

    27345.9295…

    1079.070499…

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