Sequences and the nth Term: Question 1

Worked solution

Part (a): Find the $n$th term

Step 1: Identify the common difference.

Look at how much the sequence increases from one term to the next:

$11 - 7 = 4, \qquad 15 - 11 = 4, \qquad 19 - 15 = 4$

The terms go up by the same amount each time, so this is a linear (arithmetic) sequence with common difference $d = 4$.

Step 2: Start with the multiplier.

Because the terms increase by $4$ each step, the $n$th term is based on the $4$ times table, so it contains $4n$. Compare $4n$ with the actual sequence:

Step 3: Find the constant to add.

Each actual term is $3$ more than $4n$ (since $7 - 4 = 3$). So we add $3$:

$n\text{th term} = 4n + 3$

Quick check: for $n = 4$, $\;4(4) + 3 = 16 + 3 = 19$ ✓

$\boxed{n\text{th term} = 4n + 3}$

Part (b): Find the 30th term

Substitute $n = 30$ into the formula from part (a):

$4n + 3 = 4 \times 30 + 3 = 120 + 3 = 123$

(Be careful to multiply before adding, following the order of operations.)

$\boxed{\text{30th term} = 123}$