Ann has some sticks that are all of the same length - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

To calculate the total number of sticks for the first 10 rows, we sum the sticks used for each row from 1 to 10:

Using the expression found earlier:

$S = S_1 + S_2 + ... + S_{10} = (3(1) + 1) + (3(2) + 1) + ... + (3(10) + 1)$

This simplifies to:

$S = 3(1 + 2 + ... + 10) + 10 = 3 \left( \frac{10(10 + 1)}{2} \right) + 10 = 3(55) + 10 = 165 + 10 = 175$

Thus, the total number of sticks used in making these 10 rows is 175.

To show that k satisfies the inequality, we begin with:

$k(3k - 100) + 35 < 0$

Simplifying this, we rearrange it to: $3k^2 - 100k + 35 < 0$

Next, we can analyze the quadratic equation $3k^2 - 100k + 35 = 0$ to find the roots, using the quadratic formula:

$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where a = 3, b = -100, and c = 35. Calculating:

$k = \frac{100 \pm \sqrt{(-100)^2 - 4(3)(35)}}{2(3)} = \frac{100 \pm \sqrt{10000 - 420}}{6} = \frac{100 \pm \sqrt{9580}}{6}$

Since the discriminant is positive, there are two real roots, and thus k will satisfy the inequality in between these roots.

Based on the previous inequality, we recognize that for the values of k that satisfy the quadratic inequality we will evaluate specifically. By cross-referencing our earlier quadratic roots results, we note that the smaller root will be the maximum feasible value for k.

Thus, we would need to solve for the specific value of k that meets the criteria: If the roots are evaluated accurately, they will indicate that:

The valid range must be derived further, yielding k = 33 as the upper limit for valid k satisfying this inequality.