Answer :
Answer:
[tex]\begin{aligned}\textsf{8.1)}\quad T_1&=2b\\T_2&=a+3b\\T_3&=2a+4b\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{8.2)}\quad &a=\dfrac{7}{2}\;\;\textsf{and}\;\;b=-\dfrac{1}{2}\\&a=2\;\;\textsf{and}\;\;b=1\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{8.3)}\quad &-1, 2 \;\textsf{and}\; 5\\&2, 5\;\textsf{and}\;8\end{aligned}[/tex]
Step-by-step explanation:
Part 8.1
The general term of an arithmetic sequence is:
[tex]T_k=(k-1)a + (k+1)b[/tex]
where k is the position of the term.
To find the first 3 terms, we substitute k = 1, k = 2 and k = 3 into the general formula:
First term:
[tex]T_1=(1-1)a + (1+1)b\\\\T_1=(0)a + (2)b\\\\T_1=2b[/tex]
Second term:
[tex]T_2=(2-1)a + (2+1)b\\\\T_2=(1)a + (3)b\\\\T_2=a+3b[/tex]
Third term:
[tex]T_3=(3-1)a + (3+1)b\\\\T_3=(2)a + (4)b\\\\T_3=2a+4b[/tex]
Therefore, the first 3 terms of the given arithmetic sequence in terms of a and b are:
[tex]T_1=2b\\\\T_2=a+3b\\\\T_3=2a+4b[/tex]
[tex]\hrulefill[/tex]
Part 8.2
In an arithmetic sequence, the difference between any two consecutive terms is constant, and is known as the common difference.
If the common difference of the given arithmetic sequence is 3, then the difference between consecutive terms is equal to 3:
[tex]T_3-T_2=3\\\\T_2-T_1=3[/tex]
Substitute the expressions for each term that we found in part 8.1 into the two equations:
[tex]\begin{aligned}T_3-T_2&=3\\(2a+4b)-(a+3b)&=3\\2a+4b-a-3b&=3\\2a-a+4b-3b&=3\\a+b&=3\end{aligned}[/tex]
[tex]\begin{aligned}T_2-T_1&=3\\(a+3b)-(2b)&=3\\a+3b-2b&=3\\a+b&=3\end{aligned}[/tex]
The two equations are the same, since the common difference is constant.
Rearrange the equation to isolate a:
[tex]a=3-b[/tex]
Given that (T₁)² = T₃ - 4, then:
[tex]\begin{aligned}(T_1)^2&=T_3-4\\(2b)^2&=(2a+4b)-4\\4b^2&=2a+4b-4\\4b^2-4b+4&=2a\\2(2b^2-2b+2)&=2a\\2b^2-2b+2&=a\\\end{aligned}[/tex]
Substitute [tex]a = 3 - b[/tex] into this equation, so that we have an equation in terms of b only:
[tex]2b^2-2b+2=3 - b[/tex]
Now, we can rearrange the equation to equal zero:
[tex]\begin{aligned}2b^2-2b+2&=3-b\\2b^2-b+2&=3\\2b^2-b-1&=0\end{aligned}[/tex]
Factor the quadratic:
[tex]\begin{aligned}2b^2-b-1&=0\\2b^2-2b+b-1&=0\\2b(b-1)+1(b-1)&=0\\(2b+1)(b-1)&=0\end{aligned}[/tex]
Apply the zero product property to find the two possible values of b:
[tex]2b+1=0 \implies b=-\dfrac{1}{2}[/tex]
[tex]b-1=0 \implies b=1[/tex]
Now, substitute the two values of b into the [tex]a = 3-b[/tex] equation, and solve for a:
[tex]b=-\dfrac{1}{2} \implies a=3-\dfrac{1}{2}=\dfrac{7}{2}[/tex]
[tex]b=1 \implies a=3-1=2[/tex]
So, the possible values of a and b are:
[tex]a=\dfrac{7}{2}\;\;\textsf{and}\;\;b=-\dfrac{1}{2}[/tex]
[tex]a=2\;\;\textsf{and}\;\;b=1[/tex]
[tex]\hrulefill[/tex]
Part 8.3
To find the sequences, simply substitute the possible values of a and b into the expressions for the first 3 terms found in part 8.1:
[tex]\begin{aligned}a=\dfrac{7}{2}\;\;\textsf{and}\;\;b=-\dfrac{1}{2}:\\\\T_1&=2\left(-\dfrac{1}{2}\right)=-1\\\\T_2&=\dfrac{7}{2}+3\left(-\dfrac{1}{2}\right)=2\\\\T_3&=2\left(\dfrac{7}{2}\right)+4\left(-\dfrac{1}{2}\right)=5\end{aligned}[/tex]
[tex]\begin{aligned}a=2\;\;\textsf{and}\;\;b=1:\\\\T_1&=2\left(1}\right)=2\\\\T_2&=2+3\left(1\right)=5\\\\T_3&=2\left(2\right)+4\left(1\right)=8\end{aligned}[/tex]
Therefore, the sequences are:
- -1, 2 and 5
- 2, 5, and 8
