The common ratio of the terms in a geometric series is \({2^x}\) .

(a)     State the set of values of x for which the sum to infinity of the series exists.

(b)     If the first term of the series is 35, find the value of x for which the sum to infinity is 40.

(a)     \(0 < {2^x} < 1\)     (M1)

\(x < 0\)     A1     N2

(b)     \(\frac{{35}}{{1 - r}} = 40\)     M1

\( \Rightarrow 40 - 40 \times r = 35\)

\( \Rightarrow - 40 \times r = - 5\)     (A1)

\( \Rightarrow r = {2^x} = \frac{1}{8}\)     A1

\( \Rightarrow x = {\log _2}\frac{1}{8}{\text{ }}( = - 3)\)     A1

Note: The substitution \(r = {2^x}\) may be seen at any stage in the solution.

[6 marks]

Part (a) was the first question that a significant majority of candidates struggled with. Only the best candidates were able to find the required set of values. However, it was pleasing to see that the majority of candidates made a meaningful start to part (b). Many candidates gained wholly correct answers to part (b).