Answer:
a) [tex]S_n=(0.4+0.1n)n[/tex]
b) 15
Step-by-step explanation:
The first time Miguel trains, he runs 0.5 mile, so [tex]a_1=0.5[/tex]
Each subsequent time he trains, he runs 0.2 mile farther than he did the previous time, so [tex]d=0.2[/tex]
Use the formula for nth term of arithmetic sequence
[tex]a_n=a_1+(n-1)d\\ \\a_n=0.5+0.2(n-1)=0.5+0.2n-0.2=0.3+0.2n[/tex]
a) The sum of n terms of the arithmetic sequence is
[tex]S_n=\dfrac{a_1+a_n}{2}\cdot n\\ \\S_n=\dfrac{0.5+0.3+0.2n}{2}\cdot n=(0.4+0.1n)n[/tex]
b) A marathon is 26.2 miles, then
[tex](0.4+0.1n)n\ge 26.2\\ \\0.1n^2+0.4n-26.2\ge 0\\ \\n^2+4n-262\ge 0\\ \\D=4^2-4\cdot (-262)=16+1048=1064\\ \\n_{1,2}=\dfrac{-4\pm\sqrt{1064}}{2}=-2\pm \sqrt{266} \\ \\n\in (-\infty,-2-\sqrt{266}]\cup[-2+\sqrt{266},\infty)[/tex]
Since n is positive and [tex]\sqrt{266}\approx 16.31[/tex], the least number of times Miguel must run for his total distance run during training to exceed the distance of a marathon is -2+17=15
Answer:
A) the bottom left is the correct answer (0.3+0.2k)
B) 15 times
Step-by-step explanation
