find the value of $d$.

determine the value of $\sum\limits_{r = 1}^N {{u_r}}$.

use of ${u_n} = {u_1} + (n - 1)d$     M1

${(1 + 2d)^2} = (1 + d)(1 + 5d)$ (or equivalent)     M1A1

$d = - 2$     A1

[4 marks]

$1 + (N - 1) \times - 2 = - 15$

$N = 9$     (A1)

$\sum\limits_{r = 1}^9 {{u_r}} = \frac{9}{2}(2 + 8 \times - 2)$     (M1)

$= - 63$     A1

[3 marks]