Series, Progressions and the Binomial Expansion: Question 7

Worked solution

Part (a): Distance run in week 15

The weekly distances form an arithmetic progression with first term $a=5$ and common difference $d=1.5$. The $n$th term is

$u_n = a + (n-1)d.$

For week $15$ ($n=15$):

$u_{15} = 5 + (15-1)(1.5) = 5 + 14\times1.5 = 5+21 = 26.$

Check: building the first few terms directly, $5, 6.5, 8, 9.5, 11, \ldots$, each step adding $1.5$; extending this pattern by hand to the $15$th term also lands on $26$, confirming the formula.

Part (b): Total distance over the first 15 weeks

$S_n = \frac{n}{2}\bigl(2a+(n-1)d\bigr)$

$S_{15} = \frac{15}{2}\bigl(2(5)+14(1.5)\bigr) = 7.5\times(10+21) = 7.5\times31 = 232.5.$

Check using the equivalent formula $S_n=\frac{n}{2}(u_1+u_n)$ with $u_1=5$ and $u_{15}=26$ (from part (a)):

$S_{15} = \frac{15}{2}(5+26) = 7.5\times31 = 232.5.$

Both methods agree, so $S_{15}=232.5$ km.

Part (c): Least number of weeks for the total to exceed 300 km

We need the smallest $n$ such that $S_n > 300$:

$\frac{n}{2}\bigl(2(5)+(n-1)(1.5)\bigr) > 300 \implies \frac{n}{2}\bigl(10+1.5n-1.5\bigr) > 300 \implies \frac{n}{2}(8.5+1.5n) > 300.$

Multiplying out and clearing the fraction:

$1.5n^2 + 8.5n > 600 \implies 3n^2 + 17n - 1200 > 0.$

Solving $3n^2+17n-1200=0$ with the quadratic formula:

$n = \frac{-17 \pm \sqrt{17^2 + 4(3)(1200)}}{2(3)} = \frac{-17 \pm \sqrt{289+14400}}{6} = \frac{-17 \pm \sqrt{14689}}{6}.$

Since $\sqrt{14689}\approx121.20$, the positive root is

$n \approx \frac{-17+121.20}{6} \approx \frac{104.20}{6} \approx 17.37.$

Since $n$ must be a whole number of weeks, and the inequality holds for $n$ above this root, we check the two nearest integers directly using the original sum formula rather than relying on the rounded root:

$S_{17} = \frac{17}{2}\bigl(10+16(1.5)\bigr) = \frac{17}{2}(10+24) = \frac{17}{2}(34) = 289,$

$S_{18} = \frac{18}{2}\bigl(10+17(1.5)\bigr) = 9\times(10+25.5) = 9\times35.5 = 319.5.$

Check: $S_{17}=289 < 300$, so $17$ weeks is not enough, while $S_{18}=319.5>300$, so $18$ weeks is enough. This confirms the least number of complete weeks needed is $n=18$.

Final answers

• (a) In week $15$ she runs $\boxed{26}$ km.
• (b) Her total training distance over the first $15$ weeks is $\boxed{232.5}$ km.
• (c) The least number of complete weeks needed for her total distance to exceed $300$ km is $\boxed{18}$.