Respuesta :
Answer:
[tex]\textsf{a)} \quad T_n=-n^2+n+20[/tex]
[tex]\textsf{b)} \quad T_{12}=-112[/tex]
[tex]\textsf{c)} \quad \sf 8th\;term[/tex]
a) Second difference is 2.
b) First term is 10.
Step-by-step explanation:
The given number pattern is:
- 20, 18, 14, 8, ...
To determine the type of sequence, begin by calculating the first differences between consecutive terms:
[tex]20 \underset{-2}{\longrightarrow} 18 \underset{-4}{\longrightarrow} 14 \underset{-6}{\longrightarrow}8[/tex]
As the first differences are not the same, we need to calculate the second differences (the differences between the first differences):
[tex]-2 \underset{-2}{\longrightarrow} -4 \underset{-2}{\longrightarrow} -6[/tex]
As the second differences are the same, the sequence is quadratic and will contain an n² term.
The coefficient of the n² term is half of the second difference.
As the second difference is -2, the coefficient of the n² term is -1.
Now we need to compare -n² with the given sequence (where n is the position of the term in the sequence).
[tex]\begin{array}{|c|c|c|c|c|}\cline{1-5}n&1&2&3&4\\\cline{1-5}-n^2&-1&-4&-9&-16\\\cline{1-5}\sf operation&+21&+22&+23&+24\\\cline{1-5}\sf sequence&20&18&14&8\\\cline{1-5}\end{array}[/tex]
We can see that the algebraic operation that takes -n² to the terms of the sequence is to add (n + 20).
[tex]\begin{array}{|c|c|c|c|c|}\cline{1-5}n&1&2&3&4\\\cline{1-5}-n^2&-1&-4&-9&-16\\\cline{1-5}+n&0&-2&-6&-12\\\cline{1-5}+20&20&18&14&8\\\cline{1-5}\sf sequence&20&18&14&8\\\cline{1-5}\end{array}[/tex]
Therefore, the expression to find the the nth term of the given quadratic sequence is:
[tex]\boxed{T_n=-n^2+n+20}[/tex]
To find the value of T₁₂, substitute n = 12 into the nth term equation:
[tex]\begin{aligned}T_{12}&=-(12)^2+(12)+20\\&=-144+12+20\\&=-132+20\\&=-112\end{aligned}[/tex]
Therefore, the 12th term of the number pattern is -112.
To find the position of the term that has a value of -36, substitute Tₙ = -36 into the nth term equation and solve for n:
[tex]\begin{aligned}T_n&=-36\\-n^2+n+20&=-36\\-n^2+n+56&=0\\n^2-n-56&=0\\n^2-8n+7n-56&=0\\n(n-8)+7(n-8)&=0\\(n+7)(n-8)&=0\\\\\implies n&=-7\\\implies n&=8\end{aligned}[/tex]
As the position of the term cannot be negative, the term that has a value of -36 is the 8th term.
[tex]\hrulefill[/tex]
Given terms of a quadratic number pattern:
- T₂ = 1
- T₃ = -6
- T₅ = -14
We know the first differences are negative, since the difference between the second and third terms is -7. Label the unknown differences as -a, -b and -c:
[tex]T_1 \underset{-a}{\longrightarrow} 1 \underset{-7}{\longrightarrow} -6 \underset{-b}{\longrightarrow}T_4 \underset{-c}{\longrightarrow} -14[/tex]
From this we can create three equations:
[tex]T_1-a=1[/tex]
[tex]-6-b=T_4[/tex]
[tex]T_4-c=-14[/tex]
The second differences are the same in a quadratic sequence. Let the second difference be x. (As we don't know the sign of the second difference, keep it as positive for now).
[tex]-a \underset{+x}{\longrightarrow} -7\underset{+x}{\longrightarrow} -b \underset{+x}{\longrightarrow}-c[/tex]
From this we can create three equations:
[tex]-a+x=-7[/tex]
[tex]-7+x=-b[/tex]
[tex]-b+x=-c[/tex]
Substitute the equation for -b into the equation for -c to create an equation for -c in terms of x:
[tex]-c=(-7+x)+x[/tex]
[tex]-c=2x-7[/tex]
Substitute the equations for -b and -c (in terms of x) into the second two equations created from the first differences to create two equations for T₄ in terms of x:
[tex]\begin{aligned}-6-b&=T_4\\-6-7+x&=T_4\\T_4&=x-13\end{aligned}[/tex]
[tex]\begin{aligned}T_4-c&=-14\\T_4+2x-7&=-14\\T_4&=-2x-7\\\end{aligned}[/tex]
Solve for x by equating the two equations for T₄:
[tex]\begin{aligned}T_4&=T_4\\x-13&=-2x-7\\3x&=6\\x&=2\end{aligned}[/tex]
Therefore, the second difference is 2.
Substitute the found value of x into the equations for -a, -b and -c to find the first differences:
[tex]-a+2=-7 \implies -a=-9[/tex]
[tex]-7+2=-b \implies -b=-5[/tex]
[tex]-5+2=-c \implies -c=-3[/tex]
Therefore, the first differences are:
[tex]T_1 \underset{-9}{\longrightarrow} 1 \underset{-7}{\longrightarrow} -6 \underset{-5}{\longrightarrow}T_4 \underset{-3}{\longrightarrow} -14[/tex]
Finally, calculate the first term:
[tex]\begin{aligned}T_1-9&=1\\T_1&=1+9\\T_1&=10\end{aligned}[/tex]
Therefore, the first term in the number pattern is 10.
[tex]10 \underset{-9}{\longrightarrow} 1 \underset{-7}{\longrightarrow} -6 \underset{-5}{\longrightarrow}-11 \underset{-3}{\longrightarrow} -14[/tex]
Note: The equation for the nth term is:
[tex]\boxed{T_n=n^2-12n+21}[/tex]
