Answer:
See Below.
Step-by-step explanation:
The sum of an A.P. is given by:
[tex]\displaystyle S_n=\frac{n}{2}(a+x_n)[/tex]
Where n is the number of terms, a is the initial term, and xₙ is the last term.
Therefore:
[tex]\displaystyle S_{30}=\frac{30}{2}\Big(a+x_{30}\Big)=15(a+x_{30})[/tex]
And likewise:
[tex]\displaystyle S_{20}=\frac{20}{2}\Big(a+x_{20}\Big)=10(a+x_{20})\\\\ S_{10}=\frac{10}{2}\Big(a+x_{10}\Big)=5(a+x_{10})[/tex]
Substitute:
[tex]15(a + x_{30}) = 3(10 (a + x_{20}) - 5(a + x_{10} ))[/tex]
Distribute:
[tex]15a+15x_{30}=30a+30x_{20}-15a-15x_{10}[/tex]
Simplify:
[tex]15x_{30}=30x_{20}-15x_{10}[/tex]
Simplify:
[tex]x_{30}=2x_{20}-x_{10}[/tex]
The nth term for an A.P. is:
[tex]x_n=a+d(n-1)[/tex]
Where a is the initial term and d is the common difference.
So, it follows that:
[tex]x_{30}=a+d(30-1)=29d+a\\ x_{20}=a+d(20-1)=19d+a\\ x_{10}=a+d(10-1)=9d+a[/tex]
Therefore:
[tex]29d+a=2(19d+a)-9d-a[/tex]
Which follows that:
[tex]29d+a=38d+2a-9d-a\\ \\ 29d=29d \\\\ d\stackrel{\checkmark}{=}d[/tex]
QED.
