Answer:
F(n)=3(n-1)+25; F(12)=58
Step-by-step explanation:
A sequence is arithemetic if the difference of any two consecutive terms in the sequence is always the same constant term.A recursive way to construct an arimethic sequence is to take an initial number [tex]a_1[/tex] an then recursively sum a positive number (distance) [tex]d[/tex] to get each term, for example take [tex]a_1=1[/tex] and [tex]d=2[/tex], we get the terms of the arithmetic sequence:
[tex]a_1=1\\a_2=1+2\\a_3=1+2+2=1+2\cdot2\\a_4=1+2+2+2=1+2\cdot 3\\a_5=1+2+2+2+2=1+2\cdot 4[/tex]
and in general the n-th term of the sequence is given by the equation
[tex]a_n=1+2\cdot(n-1)=a_1+d\cdot(n-1)[/tex].
Note that g(n)=3(n-1) is already an arithmetic sequence with [tex]a_1=0[/tex] and [tex]d=3[/tex] and f(n) is the constant function with constant value equals to 25.
Hence, if we take
[tex]F(n)=f(n)+g(n)=25+3\cdot(n-1)[/tex]
we get an arithmetic sequence with intial value [tex]a_1[/tex] equals to 25 and distance [tex]d[/tex] equals to 3.
