Answer:
As
[tex]a_3=-8[/tex] ; [tex]b_3=1[/tex]
Therefore,
[tex]a_3+b_3=-8+1=-7[/tex]
Step-by-step explanation:
Considering the first two terms of a sequence
[tex]a_1=4\:;\:a_2=-2[/tex]
Finding [tex]a_3[/tex] when the sequence is arithmetic
Let [tex]a_3[/tex] be the third term when the sequence is arithmetic.
The common difference = d = -2 - 4 = -6
The n-th term of Arithmetic sequence is:
[tex]a_n=a_1+\left(n-1\right)d[/tex]
Putting n = 3 in the n-th term to find [tex]a_3[/tex] .
[tex]a_n=a_1+\left(n-1\right)d[/tex]
[tex]a_3=4+\left(3-1\right)\left(-6\right)[/tex]
[tex]a_3=4-12[/tex]
[tex]a_3=-8[/tex]
Finding [tex]b_3[/tex] when the sequence is geometric
Let [tex]b_3[/tex] be the third term when the sequence is geometric.
The common ratio r is:
[tex]r=-\frac{2}{4}=-\frac{1}{2}[/tex]
The n-th term of a geometric sequence with initial value b and common ratio r is given by
[tex]b_{n}=b\,r^{n-1}[/tex]
Putting n = 3 in the n-th term to find [tex]b_3[/tex] .
[tex]b_3=4\cdot \left(-\frac{1}{2}\right)^{3-1}[/tex]
[tex]b_3=4\cdot \frac{1}{2^2}[/tex] ∵ [tex]\left(-\frac{1}{2}\right)^{3-1}=\frac{1}{2^2}[/tex]
[tex]b_3=\frac{1\cdot \:4}{2^2}[/tex]
[tex]b_3=\frac{2^2}{2^2}[/tex]
[tex]b_3=1[/tex]
So,
[tex]a_3=-8[/tex] ; [tex]b_3=1[/tex]
Therefore,
[tex]a_3+b_3=-8+1=-7[/tex]
