Exploring Algebraic Identities | EOT

Question 6

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) $6a^2 - 24b^2$

(ii) $3ps^2 - 15ps + 12p$

Solution

Understand the Question

A sequence of $n$ consecutive natural numbers starting at $a$ forms an Arithmetic Progression with common difference $d = 1$ .
The sum is given by $S_n = \dfrac{n}{2}(2a + n - 1) = 100 \implies n(2a + n - 1) = 200$ .
This requires $n$ to be a positive factor of $200$ and $a$ to be a natural number ( $a \ge 1$ ).
We determine the maximum possible value of $n$ , list all valid factors, and check which factors yield an integer value for $a$ .

Step 1 · Set Up the Equation and Find the Range for $n$

Let the sum of $n$ consecutive natural numbers starting from $a$ be $100$ .

Using the sum formula for an AP with common difference $d = 1$ $S_n = \dfrac{n}{2}(2a + (n-1)d)$

$100 = \dfrac{n}{2}(2a + n - 1)$

$200 = n(2a + n - 1)$

Expressing $2a$ in terms of $n$ $2a = \dfrac{200}{n} - n + 1$

Since $a$ is a natural number ( $a \ge 1 \implies 2a \ge 2$ ) $\dfrac{200}{n} - n + 1 \ge 2$

$\dfrac{200}{n} - n - 1 \ge 0$

$\dfrac{200 - n^2 - n}{n} \ge 0$

Since $n > 0$ , the numerator must be non-negative $200 - n^2 - n \ge 0$

$n^2 + n - 200 \le 0$

Finding the roots of $n^2 + n - 200 = 0$

\begin{aligned} n &= \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-200)}}{2(1)} \\[0.6em] &= \dfrac{-1 \pm \sqrt{1 + 800}}{2} \\[0.6em] &= \dfrac{-1 \pm \sqrt{801}}{2} \end{aligned}

Using $\sqrt{801} \approx 28.3$ $n \approx \dfrac{-1 + 28.3}{2} \approx \dfrac{27.3}{2} \approx 13.65$

Therefore, $n \le 13$ .

Step 2 · Identify Possible Factor Values for $n$

Since $n$ must be a factor of $200$ and $n \le 13$

Factors of $200$ : $1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200$

Possible values of $n$ $n \in \{1, 2, 4, 5, 8, 10\}$

Step 3 · Test Each Value of $n$

Using $2a = \dfrac{200}{n} - n + 1$ , $a$ is a valid natural number only when $2a$ is a positive even integer.

Case 1: $n = 1$

\begin{aligned} 2a &= \dfrac{200}{1} - 1 + 1 = 200 \\[0.6em] a &= 100 \end{aligned}

$100 = 100$

Case 2: $n = 2$

\begin{aligned} 2a &= \dfrac{200}{2} - 2 + 1 = 99 \end{aligned}

Since $99$ is odd, $a$ is not an integer (invalid).

Case 3: $n = 4$

\begin{aligned} 2a &= \dfrac{200}{4} - 4 + 1 = 47 \end{aligned}

Since $47$ is odd, $a$ is not an integer (invalid).

Case 4: $n = 5$

\begin{aligned} 2a &= \dfrac{200}{5} - 5 + 1 = 36 \\[0.6em] a &= 18 \end{aligned}

$100 = 18 + 19 + 20 + 21 + 22$

Case 5: $n = 8$

\begin{aligned} 2a &= \dfrac{200}{8} - 8 + 1 = 18 \\[0.6em] a &= 9 \end{aligned}

$100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$

Case 6: $n = 10$

\begin{aligned} 2a &= \dfrac{200}{10} - 10 + 1 = 11 \end{aligned}

Since $11$ is odd, $a$ is not an integer (invalid).

Answer

The possible ways of expressing $100$ as the sum of consecutive natural numbers are:

1 term: $100 = 100$
5 terms: $100 = 18 + 19 + 20 + 21 + 22$
8 terms: $100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$

Common Mistakes

Non-Integer Starting Value: Forgetting to check if $2a$ is an even integer. If $2a$ is odd (such as when $n = 2$ gives $2a = 99$ ), then $a = 49.5$ , which is not a natural number.
Ignoring Upper Bound on $n$ : Not establishing $a \ge 1$ , which is needed to constrain $n \le 13$ and avoid checking infinitely many factors.

Question 6

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