Answer:
[tex]\textsf{(a)}\quad a_n=2n-1[/tex]
[tex]\textsf{b)}\quad \sf Day\;51[/tex]
Step-by-step explanation:
Part (a)
An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is constant.
The general formula for the nth term of an arithmetic sequence is:
[tex]\large\boxed{a_n=a+(n-1)d}[/tex]
where:
- [tex]a_n[/tex] is the nth term of the sequence.
- [tex]a[/tex] is the first term of the sequence.
- [tex]n[/tex] is the position of the term in the sequence.
- [tex]d[/tex] is the common difference between consecutive terms.
Given that on the 6th day of the program, Susan performed 11 sit-ups, and on the 15th day she did 29 sit-ups, then:
[tex]a_6=11[/tex]
[tex]a_{15}=29[/tex]
Substitute these values into the formula to create two equations, and rearrange to isolate a:
[tex]\begin{aligned}a_6=a+(6-1)d&=11\\a+5d&=11\\a&=11-5d\end{aligned}[/tex]
[tex]\begin{aligned}a_{15}=a+(15-1)d&=29\\a+14d&=29\\a&=29-14d\end{aligned}[/tex]
Substitute the first equation into the second equation and solve for d:
[tex]\begin{aligned}11-5d&=29-14d\\11-5d+14d&=29-14d+14d\\11+9d&=29\\11+9d-11&=29-11\\9d&=18\\d&=2\end{aligned}[/tex]
Substitute the found value of d back into one of the equations for a, and solve for a:
[tex]\begin{aligned}a&=11-5(2)\\a&=11-10\\a&=1\end{aligned}[/tex]
Substitute the values of a and d into the general formula:
[tex]a_n=1+(n-1)2[/tex]
[tex]a_n=1+2n-2[/tex]
[tex]a_n=2n-1[/tex]
Therefore, the equation that relates the number of sit-ups (aₙ) to the number of days (n) is:
[tex]\Large\boxed{\boxed{a_n=2n-1}}[/tex]
[tex]\hrulefill[/tex]
Part (b)
To determine which day Susan will accomplish 100 sit-ups, we can substitute aₙ = 100 into the equation from part (a) and solve for n:
[tex]\begin{aligned}2n-1&=100\\2n-1+1&=100+1\\2n&=101\\n&=50.5\end{aligned}[/tex]
As we cannot have a part day, we need to round up to 51. Therefore, Susan will be able to accomplish 100 sit-ups on day 51 of her program (when she will actually perform 101 sit-ups).
