Practice Problems (Junior Cert Mathematics): Revision Notes

Practice Problems

Problems

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Problem 1:

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Problem 2:

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Solutions

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Problem 1:

• (a) Find the common difference: The "common difference" is the difference between one term and the next in the sequence. For a linear sequence, this difference is the same each time.

Let's calculate the difference between the terms:

• $8 - 3 = 5$

• $13 - 8 = 5$

• $18 - 13 = 5$ Since the difference is always $5$, this is our common difference.

• (b) Write down the general term $T_n$: To find any term in the sequence, we can use a formula called the "general term." For a linear sequence, the formula is: $T_n = a + (n-1) \cdot d$ where:

  • $a$ is the first term (which is $3$ in this case),
  • $d$ is the common difference (which we found to be $5$), and
  • $n$ is the term number we want to find. Plugging in our values: $T_n = 3 + (n-1) \cdot 5$

Let's simplify this: $T_n = 3 + 5n - 5$ $T_n = 5n - 2$

So, the formula to find any term in the sequence is: $T_n = 5n - 2$

This means that if you want to find the $10th$ term, the $50th$ term, or any term at all, you just plug the number of the term into this formula.

• (c) What is the $20th$ term in the sequence? Now, let's use our formula to find the $20th$ term. We simply substitute $n = 20$ into the general term formula: $T_{20} = 5(20) - 2$ $T_{20} = 100 - 2 = 98$

The $20th$ term in the sequence is $98$.

• (d) Determine which term in the sequence is equal to $73$: Sometimes, you might know the value of a term but not its position in the sequence. To find out which term equals $73$, we set our general term equal to $73$ and solve for $n$: $5n - 2 = 73$ First, add $2$ to both sides to isolate the term with $n$: $5n = 75$ Then, divide both sides by $5$: $n = 15$

So, $73$ is the $15th$ term in the sequence.

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Problem 2:

• (a) Show the sequence is quadratic: To determine if a sequence is quadratic, we need to check the differences between the terms.

First differences (subtract each term from the next): $7 - 4 = 3$ $12 - 7 = 5$ $19 - 12 = 7$ $28 - 19 = 9$

The first differences are not the same, so the sequence is not linear. Let's check the second differences (subtract each first difference from the next): $5 - 3 = 2$ $7 - 5 = 2$ $9 - 7 = 2$

Since the second differences are constant (they are all $2$), this sequence is quadratic.

• (b) Derive the general term $T_n$: A quadratic sequence has a general term in the form: $T_n = an^2 + bn + c$

Our job is to find the values of $a$, $b$, and $c$.

Step 1: Find $a$:

The second difference is always $2a$. We know the second difference is $2$, so: $2a = 2 \quad \Rightarrow \quad a = 1$

Step 2: Find $b$ and $c$:

Use the first and second terms of the sequence to create equations. For $T_1$ (the first term): $T_1 = 1(1)^2 + b(1) + c = 4$ Simplifying: $1 + b + c = 4 \quad \Rightarrow \quad b + c = 3 \quad \text{(Equation 1)}$

Now, for $T_2$ (the second term): $T_2 = 1(2)^2 + b(2) + c = 7$ Simplifying: $4 + 2b + c = 7 \quad \Rightarrow \quad 2b + c = 3 \quad \text{(Equation 2)}$

Step 3: Solve the equations:

Subtract Equation $1$ from Equation $2$ to eliminate $c$: $(2b + c) - (b + c) = 3 - 3$ $b = 0$

Now, substitute $b = 0$ back into Equation $1$: $c = 3$

So the general term is: $T_n = n^2 + 3$

• (c) What is the 10th term in this sequence? Now, let's find the 10th term by substituting $n = 10$ into the general term: $T_{10} = 10^2 + 3 = 100 + 3 = 103$

The $10th$ term in the sequence is $103$.

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