[tex] \large\underline \bold{{SOLUTION\:1}}[/tex]
Using the Arithmetic Sequence Formula.
[tex]\sf{A_n=A_1+(n-1)d}[/tex]
[tex]\sf{A_n=A_1+(n-1)\frac{3}{4}}[/tex]
[tex]\sf{A_n=A_1+(\frac{3n}{4}-\frac{3}{4})}[/tex]
Now , All we need to do is to find the First term
[tex]\sf{A_1-\frac{3}{4}=10}[/tex]
[tex]\sf{A_1=10+\frac{3}{4}}[/tex]
[tex]\sf{A_=\frac{43}{4}}[/tex]
Thus , the sequence looks like this
[tex]\sf{A_n=A_1+(n-1)d}[/tex]
[tex]\sf{A_1=\frac{43}{4}+(n-1)\frac{3}{4}}[/tex]
Where the common difference is [tex]\boxed{\sf{\frac{3}{4}}}[/tex]
[tex] \large\underline \bold{{SOLUTION\:2}}[/tex]
Using the substitution method:
[tex]\sf{a_n=\frac{3}{4}n-10}[/tex]
[tex]\sf{a_1=\frac{3}{4}(1)-10}[/tex]
[tex]\sf{a_1=\frac{3}{4}-10}[/tex]
[tex]\sf{a_1=\frac{3}{4}n-10}[/tex]
[tex]\sf{a_1=\frac{-37}{4}}[/tex]
[tex]\sf{a_n=\frac{3}{4}n-10}[/tex]
[tex]\sf{a_2=\frac{3}{4}(2)-10}[/tex]
[tex]\sf{a_2=\frac{3}{2}-10}[/tex]
[tex]\sf{a_2=\frac{-17}{2}}[/tex]
[tex]\sf{a_n=\frac{3}{4}n-10}[/tex]
[tex]\sf{a_3=\frac{3}{4}(3)-10}[/tex]
[tex]\sf{a_3=\frac{9}{4}-10}[/tex]
[tex]\sf{a_3=\frac{-31}{4}}[/tex]
To find their common difference , We will subtract each of them.
[tex]\sf{A_2-A_1=\frac{-17}{2}-(\frac{-37}{4})=\boxed{\frac{3}{4}}}[/tex]
[tex]\sf{A_3-A_2=\frac{-31}{4}-(\frac{-17}{2})=\boxed{\frac{3}{4}}}[/tex]
[tex] \large\underline{ \bold{ANSWER}}[/tex]
- Yes , it is an arithmetic sequence with a common difference of 3/4
