The nth term of an arithmetic sequence is :
[tex]a_n=a_1+d(n-1)[/tex]From the problem, we have :
[tex]\begin{gathered} a_5=4x-3 \\ a_9=12x+9 \end{gathered}[/tex]Substitute a5 and n = 5 :
[tex]\begin{gathered} a_n=a_1+d(n-1) \\ a_5=a_1+d(5-1) \\ 4x-3=a_1+4d \end{gathered}[/tex]Rewrite the equation as d in terms of x and a1 :
[tex]\begin{gathered} 4x-3=a_1+4d \\ 4x-3-a_1=4d \\ d=\frac{4x-3-a_1}{4} \end{gathered}[/tex]Subsitute a9 and n = 9
[tex]\begin{gathered} a_n=a_1+d(n-1) \\ a_9=a_1+d(9-1) \\ 12x+9=a_1+8d \end{gathered}[/tex]Rewrite the equation as d in terms of x and a1 :
[tex]\begin{gathered} 12x+9=a_1+8d \\ 12x+9-a_1=8d \\ d=\frac{12x+9-a_1}{8} \end{gathered}[/tex]Now, equate two equations of d :
[tex]\begin{gathered} \frac{4x-3-a_1}{4}=\frac{12x+9-a_1}{8} \\ 8(4x-3-a_1)=4(12x+9-a_1) \\ 32x-24-8a_1=48x+36-4a_1 \\ 4a_1-8a_1=48x+36-32x+24 \\ -4a_1=16x+60 \\ a_1=-4x-15 \end{gathered}[/tex]The answer is a1 = -4x-15
