(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 ...
Question 1 Report
(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.
Answer Details
(a) The recurrence is \(T_n = 4T_{n-1} - 3\). Express the terms through \(T_1\):
\[T_2 = 4T_1 - 3,\qquad T_3 = 4T_2 - 3 = 4(4T_1 - 3) - 3 = 16T_1 - 15.\]
Given \(2T_3 = 5T_2\):
\[2(16T_1 - 15) = 5(4T_1 - 3) \Rightarrow 32T_1 - 30 = 20T_1 - 15 \Rightarrow 12T_1 = 15 \Rightarrow T_1 = \tfrac{5}{4}.\]
Then \(T_2 = 4(\tfrac54) - 3 = 2\) and \(T_3 = 4(2) - 3 = 5\).
The first three terms are \(\tfrac{5}{4},\ 2,\ 5\).
(b) Multiplying the matrices row by row:
Row 1: \(2(1) + 0\cdot m + 1\cdot r = 2 + r = k\).
Row 2: \(5(1) - 3m + r = 2 \Rightarrow -3m + r = -3\).
Row 3: \(0 + 4m + 6r = 26\).
From Row 2, \(r = 3m - 3\). Substitute into Row 3:
\[4m + 6(3m - 3) = 26 \Rightarrow 22m - 18 = 26 \Rightarrow m = 2,\]
so \(r = 3(2) - 3 = 3\) and \(k = 2 + r = 5\).
Answer: \(k = 5,\; m = 2,\; r = 3\).
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