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Question
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 .
Answer/Explanation
Markscheme
(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]