Precalculus, Chapter 9, 9.2, Section 9.2, Problem 36

a_4=16
a_10=46
To determine the first five terms of this arithmetic sequence, consider its nth term formula which is:
a_n=a_1+(n-1)d
To apply this, plug-in the given nth terms.
Plugging in a_4=16, the formula becomes:
16=a_1 + (4-1)d
16=a_1+3d (Let this be EQ1.)
Also, substituting a_10=46, the formula becomes:
46=a_1+(10-1)d
46=a_1+9d (Let this be EQ2.)
Then, use these two equations to solve for the values of a_1 and d. To do so, isolate a_1 in the first equation.
16=a_1+3d
16-3d=a_1
Plug-in this to the second equation.
46=a_1+9d
46=16-3d+9d
46=16+6d
30=6d
5=d
Then, solve for a_1. To do so, plug-in d=5 to the first equation.
16=a_1 + 3d
16=a_1+3(5)
16=a_1+15
1=a_1
Then, plug-in these two values a_1=1 and d=5 to the formula of nth terms of arithmetic sequence.
a_n=a_1+(n-1)d
a_n=1+(n-1)(5)
a_n=1+5n-5
a_n=5n-4
Now that the formula of a_n is known, use this to solve for the values of a_2, a_3 and a_5. (Take note that the values of a_1 and a_4 are already known.)
1st term: a_1=1
2nd term: a_2=5(2)-4=6
3rd term: a_3=5(3)-4=11
4th term: a_4=16
5th term: a_5=5(5)-4=21
Therefore, the first five terms of the arithmetic sequence are {1, 6, 11, 16, 21,...}.