Insert m numbers between the two numbers −107 and 21 such that the (m + 2) numbers form an arithmetic sequence. It is given that the common difference of the sequence is 4.

Find the value of m.

Find the first positive term of the sequence

Find the value of m.

Find the first positive term of the sequence

Since the (m+2) numbers form an arithmetic sequence,

a_1=-107

a_{m+2}=a_1+(m+2-1)d

=-107+4(m+1)=21

Solving the function yields

m=-\dfrac{124}{4} =31

The formula for general terms of the sequence is

a_n=a_1+(n-1)d

=-107+4(n-1)

=4n-111

If a_n> 0

4n-111 >0

n>\dfrac{111}{4}

=27\dfrac{3}{4}

So the first positive term happens when n=28

a_{28}=4\cdotp 28-111=1