PROBLEM:
A six-day conference was called by the principal of Tabuco Central School. On the first day there were 250 pupils present. The following day, 50 more pupils attended, and the increase was uniform until the last day. How many pupils were present during the fifth day?
[tex]\red {\overline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}[/tex]
SOLUTION:
- We can apply what we have learned from the lesson about arithmetic sequence since we have the common difference of 50 pupils.
- You're asked to solve for the number of pupils that are present during the fifth day, thus, the 5th term of the sequence.
- Let's use the formula for the arithmetic sequence given the first term, 250 pupils, and the common difference, 50 pupils.
[tex]\large\rm a_n=a_1+(n-1)d[/tex]
[tex]\large\rm a_5=250+(5-1)(50)[/tex]
[tex]\large\rm a_5=250+4(50)[/tex]
[tex]\large\rm a_5=250+200[/tex]
[tex]\large\rm a_5=450[/tex]
[tex]\red {\overline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}[/tex]
ANSWER:
- There were 450 pupils present during the fifth day.
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