(i) Write an equation for u₆ in terms of d and u₁.

(ii) Write an equation for u₁₀ in terms of d and u₁.

Write down the value of

(i) d ;

(ii) u₁ .

Find the total number of seats in the amphitheatre.

A few years later, a second level was added to increase the amphitheatre’s capacity by another 1600 seats. Each row has four more seats than the previous row. The first row on this level has 70 seats.

Find the number of rows on the second level of the amphitheatre.

Write down the next two terms of this geometric sequence.

Write down the common ratio of this geometric sequence.

Calculate the number of people who will receive the text message at 12:30.

Calculate the total number of people who will have received the text message by 12:30.

Calculate the exact time at which a total of 29 524 people will have received the text message.

(i) u₁ + 5d = 100     (A1)

(ii) u₁ + 9d = 124     (A1)

[2 marks]

(i) 6     (G1)(ft)

(ii) 70     (G1)(ft)

Notes: Follow through from their equations in parts (a) and (b) even if working not seen. Their answers must be integers. Award (M1)(A0) for an attempt to solve two equations analytically.

[2 marks]

\(S_{20} = \frac{20}{2}(2 \times 70 + (20 - 1) \times 6)\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sum of AP formula, (A1)(ft) for their correct substituted values.

= 2540     (A1)(ft)(G2)

Note: Follow through from their part (b).

[3 marks]

\(\frac{n}{2}(2 \times 70 + (n - 1) \times 4) = 1600\)     (M1)(A1)

Note: Award (M1) for substituted sum of AP formula, (A1) for their correct substituted values.

4n² +136n – 3200 = 0     (M1)

Note: Award (M1) for this equation (or other equivalent expanded quadratic) seen, may be implied if correct final answer seen.

n = 16     (A1)(G3)

Note: Do not award the final (A1) for n = 16, – 50 given as final answer, award (G2) if n = 16, – 50 given as final answer without working.

[4 marks]

9, 27     (A1)

[1 mark]

3     (A1)

[1 mark]

1 × 3⁶     (M1)

= 729     (A1)(ft)(G2)

Note: Award (M1) for correctly substituted GP formula. Follow through from their answer to part (b).

[2 marks]

\(\frac{{1(3^7 - 1)}}{(3 - 1)}\)     (M1)

Note: Award (M1) for correctly substituted GP formula. Accept sum 1+ 3 + 9 + 27 + ... + 729. If lists are used, award (M1) for correct list that includes 1093. (1, 4, 13, 40, 121, 364, 1093, 3280…)

= 1093     (A1)(ft)(G2)

Note: Follow through from their answer to part (b). For consistent use of n = 6 from part (c) (243) to part (d) leading
to an answer of 364, treat as double penalty and award (M1)(A1)(ft) if working is shown.

[2 marks]

\(\frac{{1(3^n - 1)}}{(3 - 1)} = 29524\)     (M1)

Note: Award (M1) for correctly substituted GP formula. If lists are used, award (M1) for correct list that includes 29524. (1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573...). Accept alternative methods, for example continuation of sum in part (d).

n = 10     (A1)(ft)

Note: Follow through from their answer to part (b).

Exact time = 12:45     (A1)(ft)(G2)

[3 marks]

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.