Write down the common ratio.

Write down the value of a.

Write down the value of b.

The sum of the first n terms is 31.9375. Find the value of n.

0.5     \(\left( {\frac{1}{2}} \right)\)     (A1)    (C1)

[1 mark]

4     (A1)

[1 mark]

1     (A1)     (C2)

\(\frac{{16(1 - {{0.5}^n})}}{{(1 - 0.5)}} = 31.9375\)     (M1)(M1)

Note: Award (M1) for correct substitution in the GP formula, (M1) for equating their sum to 31.9375.

OR

sketch of the function  \(y = \frac{{16(1 - {{0.5}^n})}}{{(1 - 0.5)}} \)     (M1)

indication of point where y = 31.9375     (M1)

OR

16 + 8 + 4 + 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 31.9375     (M1)(M1)

Note: Award (M1) for a list of at least 7 correct terms, (A1) for the sum of the terms equated to 31.9375.

n = 9     (A1)(ft)     (C3)

Note: Follow through from their answer to part (a) but answer mark is lost if n is not a whole number.

[3 marks]

Parts a) and b) of this question were well answered by many of the candidates, although in some cases students wrote ½ instead of 2 for the common ratio in a).

Parts a) and b) of this question were well answered by many of the candidates, although in some cases students wrote ½ instead of 2 for the common ratio in a).

Parts a) and b) of this question were well answered by many of the candidates, although insome cases students wrote ½ instead of 2 for the common ratio in a).

Many candidates were also able to set an equation in c) with a correct expression of the sum of the first n terms equated to 31.9375, for which they gained two more marks. The last mark in many cases was not awarded either because the candidates didn’t know how to solve the equation or/and gave an incorrect answer.