Find the value of ${u_{20}}$.

Find the sum of the first 10 terms of the sequence.

Now consider the sequence ${v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots$ where

$${v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24$$

This sequence continues in the same manner.

Find the exact value of ${v_{10}}$.

Now consider the sequence ${v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots$ where

$${v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24$$

This sequence continues in the same manner.

Find the sum of the first 8 terms of this sequence.

$k$ is the smallest value of $n$ for which ${v_n}$ is greater than ${u_n}$.

Calculate the value of $k$.

$600 + (20 - 1) \times 17$     (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 20 correct terms seen.

$= 923$     (A1)(G3)

[3 marks]

$\frac{{10}}{2}\left[ {2 \times 600 + (10 - 1) \times 17} \right]$     (M1)(A1)

Note: Award (M1) for substituted arithmetic series formula, (A1) for their correct substitutions. Follow through from part (a). For consistent use of geometric series formula in part (b) with the geometric sequence formula in part (a) award a maximum of (M1)(A1)(A0) since their final answer cannot be an integer.

OR

${u_{10}} = 600 + (10 - 1)17 = 753$     (M1)

${S_{10}} = \frac{{10}}{2}\left( {600 + {\text{their }}{u_{10}}} \right)$     (M1)

Note: Award (M1) for their correctly substituted arithmetic sequence formula, (M1) for their correctly substituted arithmetic series formula. Follow through from part (a) and within part (b).

Note: If a list is used, award (M1) for at least 10 correct terms seen, award (A1) for these terms being added.

$= 6765$   (accept $6770$)     (A1)(ft)(G2)

[3 marks]

$3 \times {2^9}$     (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 8 correct terms seen.

$= 1536$     (A1)(G3)

Note: Exact answer only. If both exact and rounded answer seen, award the final (A1).

[3 marks]

$\frac{{3 \times \left( {{2^8} - 1} \right)}}{{2 - 1}}$     (M1)(A1)(ft)

Note: Award (M1) for substituted geometric series formula, (A1) for their correct substitutions. Follow through from part (c). If a list is used, award (M1) for at least 8 correct terms seen, award (A1) for these 8 correct terms being added. For consistent use of arithmetic series formula in part (d) with the arithmetic sequence formula in part (c) award a maximum of (M1)(A1)(A1).

$= 765$     (A1)(ft)(G2)

[3 marks]

$3 \times {2^{k - 1}} > 600 + (k - 1)(17)$     (M1)

Note: Award (M1) for their correct inequality; allow equation. 

Follow through from parts (a) and (c). Accept sketches of the two functions as a valid method.

$k > 8.93648 \ldots$   (may be implied)     (A1)(ft)

Note: Award (A1) for $8.93648…$ seen. The GDC gives answers of $-34.3$ and $8.936$ to the inequality; award (M1)(A1) if these are seen with working shown.

OR

${v_8} = 384$     ${u_8} = 719$     (M1)

${v_9} = 768$     ${u_9} = 736$     (M1)

Note: Award (M1) for ${v_8}$ and ${u_8}$ both seen, (M1) for ${v_9}$ and ${u_9}$ both seen.

$k = 9$     (A1)(ft)(G2)

Note: Award (G1) for $8.93648…$ and $-34.3$ seen as final answer without working. Accept use of $n$.

[3 marks]