19N.1.SL.TZ0.T_7

pestleMathematicsAISLPaper 119N· sl-1-3-geometric-sequences-and-seriessource ↗

A geometric sequence has a first term of $\frac{8}{3}$ and a fourth term of $9$ .

Find the common ratio.

[2]

Write down the second term of this sequence.

[1]

The sum of the first $k$ terms is greater than $2500$ .

Find the smallest possible value of $k$ .

[3]

Markscheme / solution

$9 = \left( {\frac{8}{3}} \right){r^3}$ (M1)

Note: Award (M1) for correctly substituted geometric sequence formula equated to $9$ .

$\left( {r = } \right)\,1.5\,\,\,\,\left( {\frac{3}{2}} \right)$ (A1) (C2)

[2 marks]

$4$ (A1)(ft) (C1)

Note: Follow through from part (a).

[1 mark]

$2500 < \frac{{\left( {\frac{8}{3}} \right)\left( {{{\left( {1.5} \right)}^k} - 1} \right)}}{{1.5 - 1}}$ (M1)

Note: Award (M1) for their correctly substituted geometric series formula compared to $2500$ .

$k = 15.2\,\,\left( {15.17319 \ldots } \right)$ (A1)(ft)

$\left( {k = } \right)\,16$ (A1)(ft) (C3)

Note: Answer must be an integer for the final (A1)(ft) to be awarded.
Follow through from part (a).

[3 marks]

Examiners’ report

[N/A]