(a) The nth term of a sequence is given by (T_{n} = 4T_{n – 1} – 3). If twice the third term is five times the second term, find the first three terms of the sequence.
(b) Given that (begin{pmatrix} 2 & 0 & 1 \ 5 & -3 & 1 \ 0 & 4 & 6 end{pmatrix} begin{pmatrix} 1 \ m \ r end{pmatrix} = begin{pmatrix} k \ 2 \ 26 end{pmatrix}), find the values of the constants k, m and r.
Explanation
(a) (T_{n} = 4T_{n – 1} – 3)
(T_{3} = 4T_{2} – 3 ; T_{2} = 4T_{1} – 3)
(2T_{3} = 5T_{2} implies T_{3} = frac{5}{2} T_{2})
(frac{5}{2} T_{2} = 4T_{2} – 3 implies 3 = frac{3}{2} T_{2})
(T_{2} = frac{3}{frac{3}{2}} = 2)
(2 = 4T_{1} – 3 implies 5 = 4T_{1})
(T_{1} = frac{5}{4})
(2T_{3} = 5(2) = 10)
(T_{3} = 5)
The first three terms of the sequence are (frac{5}{4}, 2, 5).
(b) (begin{pmatrix} 2 & 0 & 1 \ 5 & -3 & 1 \ 0 & 4 & 6 end{pmatrix} begin{pmatrix} 1 \ m \ r end{pmatrix} = begin{pmatrix} k \ 2 \ 26 end{pmatrix})
(begin{pmatrix} 2(1) + r \ 5(1) – 3m + r \ 4m + 6r end{pmatrix} = begin{pmatrix} k \ 2 \ 26 end{pmatrix})
(2 + r = k … (1))
(5 – 3m + r = 2 implies 3m – r = 3 … (2))
(4m + 6r = 26 … (3))
((2) times 6 : 18m – 6r = 18)
((2) + (3) : 22m = 44 implies m = 2)
(4(2) + 6r = 26 implies 6r = 18)
(r = 3)
(2 + r = k implies 2 + 3 = 5)
((k, m, r) = (5, 2, 3)).