Show that $a = - \frac{3}{2}d$ .

Show that the ${{\text{4}}^{{\text{th}}}}$ term of the geometric sequence is the ${\text{1}}{{\text{6}}^{{\text{th}}}}$ term of the arithmetic sequence.

let the first three terms of the geometric sequence be given by ${u_1}$ , ${u_1}r$ , ${u_1}{r^2}$

$\therefore {u_1} = a + 2d$ , ${u_1}r = a + 3d$ and ${u_1}{r^2} = a + 6d$     (M1)

$\frac{{a + 6d}}{{a + 3d}} = \frac{{a + 3d}}{{a + 2d}}$     A1

${a^2} + 8ad + 12{d^2} = {a^2} + 6ad + 9{d^2}$     A1

2a + 3d = 0

$a = - \frac{3}{2}d$     AG

[3 marks]

${u_1} = \frac{d}{2}$ , ${u_1}r = \frac{{3d}}{2}$ , $\left( {{u_1}{r^2} = \frac{{9d}}{2}} \right)$     M1

r = 3     A1

geometric ${{\text{4}}^{{\text{th}}}}$ term ${u_1}{r^3} = \frac{{27d}}{2}$     A1

arithmetic ${\text{1}}{{\text{6}}^{{\text{th}}}}$ term $a + 15d = - \frac{3}{2}d + 15d$     M1

$= \frac{{27d}}{2}$     A1

Note: Accept alternative methods.

[3 marks]

This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid.

This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid.