(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.
(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.
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Correct Answer: Option n
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) \(\begin{pmatrix} 2 & 0 & 1 \\ 5 & -3 & 1 \\ 0 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 \\ m \\ r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)
\(\begin{pmatrix} 2(1) + r \\ 5(1) - 3m + r \\ 4m + 6r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)
\(2 + r = k ... (1)\)
\(5 - 3m + r = 2 \implies 3m - r = 3 ... (2)\)
\(4m + 6r = 26 ... (3)\)
\((2) \times 6 : 18m - 6r = 18\)
\((2) + (3) : 22m = 44 \implies m = 2\)
\(4(2) + 6r = 26 \implies 6r = 18\)
\(r = 3\)
\(2 + r = k \implies 2 + 3 = 5\)
\((k, m, r) = (5, 2, 3)\).
(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) \(\begin{pmatrix} 2 & 0 & 1 \\ 5 & -3 & 1 \\ 0 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 \\ m \\ r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)
\(\begin{pmatrix} 2(1) + r \\ 5(1) - 3m + r \\ 4m + 6r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)
\(2 + r = k ... (1)\)
\(5 - 3m + r = 2 \implies 3m - r = 3 ... (2)\)
\(4m + 6r = 26 ... (3)\)
\((2) \times 6 : 18m - 6r = 18\)
\((2) + (3) : 22m = 44 \implies m = 2\)
\(4(2) + 6r = 26 \implies 6r = 18\)
\(r = 3\)
\(2 + r = k \implies 2 + 3 = 5\)
\((k, m, r) = (5, 2, 3)\).