Method 1: Using Arithmetic Sequence Analysis
(I did this version first since the question was asked under "Arithmetic Sequences").
If the initial value, #44# is denoted as #a_0#
then we have:
#color(white)("XXX")a_0=44#
#color(white)("XXX")a_1=?#
#color(white)("XXX")a_2=?#
#color(white)("XXX")a_3=?#
#color(white)("XXX")a_4=92#
but we know that for an arithmetic sequence with initial value #a_0# and arithmetic increment #d#,
the #n^(th)# value in the sequence is:
#color(white)("XXX")a_n=a_0+n * d#
#color(white)("XXXXXX")#Some will denote the initial value #a_1#;
#color(white)("XXXXXX")#in this case the formula becomes:
#color(white)("XXXXXXXXX")a_n=a_1+(n-1) * d#
#color(white)("XXXXXX")#This seems unnecessarily complicated to me.
so
#color(white)("XXX")a_4=92=44+4d#
#color(white)("XXX")rarr d=12#
and therefore
#color(white)("XXX")a_1=44+12=56#
#color(white)("XXX")a_2=56+12=68#
#color(white)("XXX")a_3=68+12=80#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method 2: Just calculate the arithmetic averages
The primary arithmetic mean is the midpoint between #44# and #92#
#color(white)("XXX")(44+92)/2=68#
The arithmetic mean between the initial value and the midpoint is
#color(white)("XXX")(44+68)/2=56#
The arithmetic mean between the midpoint and the final value is
#color(white)("XXX")(68+92)/2=80#
