Consider an infinite geometric sequence with \({u_1} = 40\) and \(r = \frac{1}{2}\) .

(i)     Find \({u_4}\) .

(ii)    Find the sum of the infinite sequence.

Consider an arithmetic sequence with n terms, with first term (\( - 36\)) and eighth term (\( - 8\)) .

(i)     Find the common difference.

(ii)    Show that \({S_n} = 2{n^2} - 38n\) .

The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .

(i) correct approach     (A1)

e.g. \({u_4} = (40){\frac{1}{2}^{(4 - 1)}}\) , listing terms

\({u_4} = 5\)     A1     N2

(ii) correct substitution into formula for infinite sum     (A1)

e.g. \({S_\infty } = \frac{{40}}{{1 - 0.5}}\) , \({S_\infty } = \frac{{40}}{{0.5}}\)

\({S_\infty } = 80\)     A1     N2

[4 marks]

(i) attempt to set up expression for \({u_8}\)     (M1)

e.g. \( - 36 + (8 - 1)d\)

correct working     A1

e.g. \( - 8 = - 36 + (8 - 1)d\) , \(\frac{{ - 8 - ( - 36)}}{7}\)

\(d = 4\)     A1     N2

(ii) correct substitution into formula for sum     (A1)

e.g. \({S_n} = \frac{n}{2}(2( - 36) + (n - 1)4)\)

correct working     A1

e.g. \({S_n} = \frac{n}{2}(4n - 76)\) , \( - 36n + 2{n^2} - 2n\)

\({S_n} = 2{n^2} - 38n\)     AG     N0

[5 marks]

multiplying \({S_n}\) (AP) by 2 or dividing S (infinite GP) by 2     (M1)

e.g. \(2{S_n}\) , \(\frac{{{S_\infty }}}{2}\) , 40

evidence of substituting into \(2{S_n} = {S_\infty }\)     A1

e.g. \(2{n^2} - 38n = 40\) , \(4{n^2} - 76n - 80\) (\( = 0\))

attempt to solve their quadratic (equation)     (M1)

e.g. intersection of graphs, formula

\(n = 20\)     A2     N3

[5 marks]

Most candidates found part (a) straightforward, although a common error in (a)(ii) was to calculate 40 divided by \(\frac{1}{2}\) as 20.

In part (b), some candidates had difficulty with the "show that" and worked backwards from the answer given.

Most candidates obtained the correct equation in part (c), although some did not reject the negative value of n as impossible in this context.