Simplify the following: (i) (ii) (iii) Note: Assume that the denominators are not equal to 0.

Understand the Question The multiples of between and form an Arithmetic Progression (AP) with a common difference . The first multiple greater than is (first term ), and the last multiple less than is (last term ). To find the total number of multiples , use the term formula: . Step 1 · Identify the Arithmetic Progression To find the first term , find the smallest multiple of greater than : To find the last term , find the largest multiple of less than : The multiples of form an AP: Here Step 2 · Calculate the Number of Terms Using the term formula of an AP Substitute the values Answer > 60 Common Mistakes Boundary Inclusion Error: Using or directly as the first or last terms without verifying whether they are divisible by . Off-by-One Error: Forgetting to add at the final step, mistakenly giving instead of .

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Question 4

Simplify the following:

(i) $\dfrac{4x^2 + 4x + 1}{4x^2 - 1}$

(ii) $\dfrac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$

(iii) $\dfrac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$

Note: Assume that the denominators are not equal to 0.

Solution

Understand the Question

The multiples of $4$ between $10$ and $250$ form an Arithmetic Progression (AP) with a common difference $d = 4$ .
The first multiple greater than $10$ is $12$ (first term $a = 12$ ), and the last multiple less than $250$ is $248$ (last term $t_n = 248$ ).
To find the total number of multiples $n$ , use the $n^{\text{th}}$ term formula: $t_n = a + (n - 1)d$ .

Step 1 · Identify the Arithmetic Progression

To find the first term $a$ , find the smallest multiple of $4$ greater than $10$ : $10 \div 4 = 2 \text{ remainder } 2$ $a = 4 \times 3 = 12$

To find the last term $t_n$ , find the largest multiple of $4$ less than $250$ : $250 \div 4 = 62 \text{ remainder } 2$ $t_n = 4 \times 62 = 248$ Diagram 1

The multiples of $4$ form an AP: $12, 16, 20, \dots, 248$

Here

\begin{aligned} a &= 12 \\ d &= 4 \\ t_n &= 248 \end{aligned}

Step 2 · Calculate the Number of Terms

Using the $n^{\text{th}}$ term formula of an AP $t_n = a + (n - 1)d$

Substitute the values $248 = 12 + (n - 1)4$

\begin{aligned} 248 - 12 &= (n - 1)4 \\[0.6em] 236 &= (n - 1)4 \\[0.6em] \dfrac{236}{4} &= n - 1 \\[0.6em] 59 &= n - 1 \\[0.6em] n &= 59 + 1 \\[0.6em] n &= 60 \end{aligned}

Answer

60

Common Mistakes

Boundary Inclusion Error: Using $10$ or $250$ directly as the first or last terms without verifying whether they are divisible by $4$ .
Off-by-One Error: Forgetting to add $1$ at the final step, mistakenly giving $n = 59$ instead of $n = 59 + 1 = 60$ .

Question 4

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