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.

Question

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.

Uniteasy Admin · Accepted Answer

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

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