Answer:
an=9+(50-1)(-3)
Step-by-step explanation:
an= a1 +(n-1)d
a1 is the first term=9
nth term= 50
d is common difference=-3( because it is decreasing it has a negative)
Answer:
[tex]\sf \textsf{nth term} (a_n) = 9 + (n-1)(-3)[/tex]
[tex] a_{50} = -138[/tex]
Step-by-step explanation:
The given sequence is an arithmetic sequence where each term decreases by 3. We can represent this sequence using the formula for the nth term of an arithmetic sequence:
[tex] \boxed{\boxed{\sf a_n = a_1 + (n-1)d}}[/tex]
where:
In this case:
Therefore, the nth term [tex] \sf ( a_n)[/tex] is given by:
[tex]\sf a_n = 9 + (n-1)(-3)[/tex]
Now, to find [tex] \sf a_{50}[/tex] , substitute [tex] \sf n = 50[/tex] into the formula:
[tex]\sf a_{50} = 9 + (50-1)(-3)[/tex]
[tex] \sf a_{50} = 9 + 49(-3)[/tex]
[tex] \sf a_{50} = 9 - 147[/tex]
[tex] \sf a_{50} = -138[/tex]
So, the 50th term [tex]( a_{50})[/tex] in the given arithmetic sequence is [tex] -138[/tex] .
