Question 1039313
ONE WAY TO GET TO THE ANSWER (Probably the expected way):
When you fill in the blanks, the five numbers (or expressions) form an arithmetic sequence.
{{{a[1]=a+2}}} is the firs term of the sequence.
{{{a[5]=5a-10}}} is the fifth element of the sequence.
In an arithmetic sequence each term is the one before plus a common difference, {{{d}}} .
In formulas, that is written as 
{{{a[n+1]=a[n]+d}}} , or {{{a[n]=a[n-1]+d}}} .
As a consequence, any term is related to the first term, and the common difference by the formula
{{{a[n]=a[1]+(n-1)*d}}}
So, for {{{n=5}}} , {{{a[5]=a[1]+(5-1)*d}}} <---> {{{a[5]=a[1]+4d}}}
Since we know {{{a[1]=a+2}}} and {{{a[5]=5a-10}}} , we can substitute those expressions and solve for {{{d}}} :
{{{5a-10=a+2+4d}}} ---> {{{5a-10-a-2=a+2-a-2+4d}}} ---> {{{4a-12=4d}}} ---> {{{(4a-12)/4=d}}} ---> {{{d=a-3}}} .
Now, we can find the missing terms:
The second term is
{{{a[2]=a[1]+d=a+2+a-3=highlight(2a-1)}}} .
The third term is
{{{a[3]=a[2]+d=2a-1+a-3=highlight(3a-4)}}} .
The fourth term is
{{{a[4]=a[3]+d=3a-4+a-3=highlight(4a-7)}}} .


ANOTHER WAY (This works only for sequences with an odd number of terms):
When three numbers are in an arithmetic sequence,
the middle term is the average of the first and last terms.
The same thing happens when in the sequence there are
5 terms, 7 terms, 9 terms, 11 terms, and so on.
So, the middle (third) term is
{{{((a+2)+(5a-10))/2+(6a-8)/2=highlight(3a-4)}}} .
The second number, the one between {{{a+2}}} and {{{3a-4}}} is
{{{((a+2)+(3a-4))/2=(4a-2)/2=highlight(2a-1)}}} .
The fourth number, the one between {{{3a-4}}} and {{{5a-10}}} is
{{{((3a-4)+(5a-10))/2=(8a-14)/2=highlight(4a-7)}}} .