Answer:
[tex]\textsf{a)} \quad k=\dfrac{1}{2}, \quad k=-2[/tex]
[tex]\textsf{b)} \quad d = -5,\quad d=15[/tex]
Step-by-step explanation:
The consecutive terms of an arithmetic sequence are given as:
[tex]a_{n}=10k + 1 \\\\ a_{n+1}=2k\\\\a_{n+2}=4k^2 - 5[/tex]
In an arithmetic sequence, the difference between consecutive terms is constant. Hence, we can set up the following equation based on the definition of an arithmetic sequence:
[tex]a_{n+2}-a_{n+1}=a_{n+1}-a_n[/tex]
Substitute the given expressions and rearrange in the form of a quadratic equation:
[tex](4k^2-5)-2k=2k-(10k+1) \\\\4k^2-5-2k=2k-10k-1 \\\\4k^2-2k-5=-8k-1\\\\4k^2-2k-5+8k+1=-8k-1+8k+1\\\\4k^2+6k-4=0[/tex]
Factor the quadratic:
[tex]4k^2 + 8k - 2k -4 = 0 \\\\4k(k+2)-2(k+2)=0 \\\\ (4k-2)(k+2)=0[/tex]
Solve for k:
[tex]4k-2=0 \implies k=\dfrac{2}{4}=\dfrac{1}{2} \\\\\\k+2=0 \implies k=-2[/tex]
Therefore, the possible values of k are:
[tex]\large\boxed{\boxed{k=\dfrac{1}{2}, \quad k=-2}}[/tex]
The common difference (d) is the difference between consecutive terms, so:
[tex]d = a_{n+1}-a_{n} \\\\ d = 2k - (10k + 1) \\\\d=2k-10k-1\\\\d=-8k-1[/tex]
Now, substitute each value of k into the equation for d:
For k = 1/2:
[tex]d=-8\left(\dfrac{1}{2}\right)-1\\\\\\d=-4-1\\\\\\d=-5[/tex]
For k = -2:
[tex]d=-8\left(-2\right)-1\\\\\\d=16-1\\\\\\d=15[/tex]
Therefore, the common differences of the sequence for the two possible values of k are:
[tex]\large\boxed{\boxed{d = -5,\quad d=15}}[/tex]
