TOPIC: Number patterns, sequences and series Sub-topic: Sequences 1. The third term of a geometric sequence is 18 and the fifth term is 162 . No4 Determine the common ratio and the first term. 2. In an arithmetic sequence the second term is 110 and the twentieth term is 56 . Determine the formula \( T_{n} \), the general term in its simplest form. the first five terms of a quadratic number pattern are \( -12 ; y ; 3 ; 15 \). Determine the values of \( x \) and \( y \).

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a+19d=56\\a+d=110\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=56-19d\\a+d=110\end{array}\right.\) - step2: Substitute the value of \(a:\) \(56-19d+d=110\) - step3: Add the terms: \(56-18d=110\) - step4: Move the constant to the right side: \(-18d=110-56\) - step5: Subtract the numbers: \(-18d=54\) - step6: Change the signs: \(18d=-54\) - step7: Divide both sides: \(\frac{18d}{18}=\frac{-54}{18}\) - step8: Divide the numbers: \(d=-3\) - step9: Substitute the value of \(d:\) \(a=56-19\left(-3\right)\) - step10: Simplify: \(a=113\) - step11: Calculate: \(\left\{ \begin{array}{l}a=113\\d=-3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}a=113\\d=-3\end{array}\right.\) - step13: Rewrite: \(\left(a,d\right) = \left(113,-3\right)\) Solve the system of equations \( a r^2 = 18; a r^4 = 162 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}ar^{2}=18\\ar^{4}=162\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=\frac{18}{r^{2}}\\ar^{4}=162\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\frac{18}{r^{2}}\times r^{4}=162\) - step3: Reduce the fraction: \(18r^{2}=162\) - step4: Divide both sides: \(\frac{18r^{2}}{18}=\frac{162}{18}\) - step5: Divide the numbers: \(r^{2}=9\) - step6: Simplify the expression: \(r=\pm \sqrt{9}\) - step7: Simplify: \(r=\pm 3\) - step8: Separate into possible cases: \(r=3\cup r=-3\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}a=\frac{18}{r^{2}}\\r=3\end{array}\right.\cup \left\{ \begin{array}{l}a=\frac{18}{r^{2}}\\r=-3\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}a=2\\r=3\end{array}\right.\cup \left\{ \begin{array}{l}a=2\\r=-3\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}a=2\\r=-3\end{array}\right.\cup \left\{ \begin{array}{l}a=2\\r=3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}a=2\\r=-3\end{array}\right.\cup \left\{ \begin{array}{l}a=2\\r=3\end{array}\right.\) - step13: Rewrite: \(\left(a,r\right) = \left(2,-3\right)\cup \left(a,r\right) = \left(2,3\right)\) Solve the system of equations \( x^2 + bx + c = -12; x^2 + bx + c = y; x^2 + bx + c = 3; x^2 + bx + c = 15 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}+bx+c=-12\\x^{2}+bx+c=y\\x^{2}+bx+c=3\\x^{2}+bx+c=15\end{array}\right.\) - step1: Rewrite the expression: \(\left\{ \begin{array}{l}x^{2}+bx+c=-12\\y=x^{2}+bx+c\\x^{2}+bx+c=3\\x^{2}+bx+c=15\end{array}\right.\) - step2: Substitute the value of \(y:\) \(\left\{ \begin{array}{l}x^{2}+bx+c=-12\\x^{2}+bx+c=3\\x^{2}+bx+c=15\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}c=-12-x^{2}-bx\\x^{2}+bx+c=3\\x^{2}+bx+c=15\end{array}\right.\) - step4: Substitute the value of \(c:\) \(\left\{ \begin{array}{l}x^{2}+bx-12-x^{2}-bx=3\\x^{2}+bx-12-x^{2}-bx=15\end{array}\right.\) - step5: Simplify: \(\left\{ \begin{array}{l}-12=3\\-12=15\end{array}\right.\) - step6: Calculate: \(\left\{ \begin{array}{l}b \in \varnothing \\c \in \varnothing \\x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step7: Rewrite: \((b, c, x, y) \in \varnothing\) Let's solve each part of your problem step by step. ### 1. Geometric Sequence We know: - The third term \( T_3 = a r^2 = 18 \) - The fifth term \( T_5 = a r^4 = 162 \) We can set up the equations: 1. \( a r^2 = 18 \) 2. \( a r^4 = 162 \) To find \( a \) and \( r \), we can divide the second equation by the first: \[ \frac{a r^4}{a r^2} = \frac{162}{18} \] This simplifies to: \[ r^2 = 9 \implies r = 3 \text{ or } r = -3 \] Now, substituting \( r \) back into the first equation to find \( a \): 1. If \( r = 3 \): \[ a (3^2) = 18 \implies a \cdot 9 = 18 \implies a = 2 \] 2. If \( r = -3 \): \[ a (-3^2) = 18 \implies a \cdot 9 = 18 \implies a = 2 \] Thus, the solutions are: - \( (a, r) = (2, 3) \) - \( (a, r) = (2, -3) \) ### 2. Arithmetic Sequence We know: - The second term \( T_2 = a + d = 110 \) - The twentieth term \( T_{20} = a + 19d = 56 \) Setting up the equations: 1. \( a + d = 110 \) 2. \( a + 19d = 56 \) Subtract the first equation from the second: \[ (a + 19d) - (a + d) = 56 - 110 \] \[ 18d = -54 \implies d = -3 \] Now substitute \( d \) back into the first equation: \[ a - 3 = 110 \implies a = 113 \] Thus, the general term \( T_n \) can be expressed as: \[ T_n = a + (n-1)d = 113 + (n-1)(-3) = 113 - 3(n-1) \] ### 3. Quadratic Number Pattern The first five terms are given as: - \( T_1 = -12 \) - \( T_2 = y \) - \( T_3 = 3 \) - \( T_4 = 15 \) Assuming a quadratic pattern, we can express the terms as: \[ T_n = an^2 + bn + c \] Setting up the equations based on the known terms: 1. \( a(1^2) + b(1) + c = -12 \) 2. \( a(2^2) + b(2) + c = y \) 3. \( a(3^2) + b(3) + c = 3 \) 4. \( a(4^2) + b(4) + c = 15 \) However, the results indicate that there is no solution for the quadratic pattern based on the provided terms. This suggests that the terms do not fit a quadratic sequence. ### Summary of Results 1. For the geometric sequence: \( (a, r) = (2, 3) \) or \( (2, -3) \) 2. For the arithmetic sequence: \( T_n = 113 - 3(n-1) \) 3. For the quadratic pattern: No solution exists for the values of \( x \) and \( y \) based on the provided terms.