Sequences and nth Term Worked ExamplesIGCSE Maths 0580

Write the terms and find the first differences. A constant first difference means the sequence is linear.

A constant difference of $3$ means the nth term starts $3n$. Compare $3n$ with the sequence to find the constant.

Every term of the sequence is $1$ more than $3n$, so add $1$.

(b) Substitute $n = 25$.

(c) Set the expression equal to $100$ and solve for $n$.

Treat the numerators and the denominators as two separate sequences. Numerators first.

Now the denominators. The first difference is a constant $2$, so the nth term starts $2n$.

Each denominator is $1$ more than $2n$.

Combine the two results into a single fraction.

Substitute $n = 10$ for the tenth term.

Find the first differences, then the second differences.

The second difference is constant, confirming a quadratic. The coefficient of $n^{2}$ is half the second difference.

Substitute $n = 1$ (term $3$) to get one equation in $b$ and $c$.

Substitute $n = 2$ (term $10$) to get a second equation.

Subtract the first equation from the second to eliminate $c$.

Check against a term you did not use — always test $n = 5$.

The first differences are not constant. Test for a common ratio by dividing consecutive terms.

The ratio is constant, so the sequence is geometric with first term $a = 5$ and common ratio $r = 3$.

(b) Substitute $n = 8$. Note the index is $n - 1 = 7$, not $8$.

Evaluate $3^{7}$ first, then multiply.

(a) Set the nth term equal to $0$ — this is just a quadratic equation.

Factorise: two numbers multiplying to $12$ and adding to $-7$ are $-3$ and $-4$.

Both are positive integers, so both are valid positions. The 3rd and 4th terms are each $0$.

(b) Now form an inequality and bring everything to one side.

This does not factorise, so use the quadratic formula to find the critical values.

$\sqrt{201} = 14.177\ldots$, so the positive critical value is $10.588\ldots$ The expression is positive beyond this root.

$n$ must be a positive integer, so take the next one up and verify both sides of the boundary.

(a) We want to show $\frac{3n-1}{n+4} < 3$. Since $n$ is a positive integer, $n + 4 > 0$, so multiplying both sides by $n+4$ does not flip the inequality.

Expand the right-hand side and simplify.

This is true for every $n$, and each step was reversible, so the original inequality holds for every term. Proved.

(b) Set up the inequality for the term exceeding $2.5$.

Multiply through by $n + 4$ (positive, so the sign is unchanged) and expand.

Collect the $n$ terms.

The smallest integer is $n = 23$. Check the boundary: at $n = 22$ the term is exactly $2.5$, which is not greater than $2.5$.

The first difference is constant at $+4$, so the nth term begins $4n$.

Compare $4n$ with the sequence: each term is $3$ more.

Write the nth term.

(b) Substitute $n = 50$.

(a) Each new pattern adds one column of $3$ tiles: $2$ white (top and bottom) and $1$ shaded. So the white tiles go up by $2$ and the total by $3$.

(b) The white tiles go up by $2$ each time, so the nth term starts $2n$. Compare with $2n$: for $n = 1$, $2n = 2$ but there are $8$ white tiles, so add $6$.

(c) Find the total first. The totals go up by $3$ from $9$, so the total is $3n + 6$ (the diagram confirms it: $3$ rows of $n + 2$ tiles). Set this equal to $90$.

Substitute $n = 28$ into the white-tile expression.

(d) $2n + 6 = 2(n + 3)$ is always a multiple of $2$, so the number of white tiles is always even. $75$ is odd. (Equivalently, $2n + 6 = 75$ gives $n = 34.5$, which is not a whole number.)

(a) Count the squares: $1, 3, 6, 10$. The first differences are $2, 3, 4$ and the second difference is a constant $1$, so the sequence is quadratic with $n^{2}$ coefficient $\frac{1}{2} \times 1 = \frac{1}{2}$.

Subtract $\frac{1}{2}n^{2}$ from each term to find the linear part: $1 - 0.5 = 0.5$, $3 - 2 = 1$, $6 - 4.5 = 1.5$, $10 - 8 = 2$. This is $\frac{1}{2}n$.

(b) Set the expression equal to $210$ and multiply by $2$ to clear the fraction.

The pattern number must be positive, so $n = 20$. Check: $\frac{1}{2} \times 20 \times 21 = 210$ ✓.

(c) From the diagram, the perimeters are $4, 8, 12, 16\text{ cm}$. Each staircase has the same perimeter as an $n$ by $n$ square, because the steps just move edges of the square inwards without changing their total length.

(d) Form an equation: squares $= 5 \times$ perimeter. Since $n \neq 0$, divide both sides by $n$.

(a) Each shaded triangle becomes $3$ shaded triangles, so the number is multiplied by $3$ each time: $1, 3, 9, 27, 81$.

(b) This is a geometric sequence with first term $1$ and common ratio $3$, so the nth term is $1 \times 3^{n - 1}$.

(c) Each small triangle made by joining midpoints has half the side length, so a quarter of the area. Each shaded triangle is replaced by $3$ shaded quarters, so the shaded area is multiplied by $\frac{3}{4}$ at every step.

Pattern $n$ has had $n - 1$ steps, so the shaded area is the first term multiplied by $\frac{3}{4}$ a total of $n - 1$ times.

(d) Evaluate the area for successive patterns with a calculator until it falls below $5$. Show the values either side of the boundary.

Pattern $9$ is still above $5\text{ cm}^{2}$ and Pattern $10$ is the first below it.

(a) Find the differences. The first differences are $14, 38, 74, 122$, the second differences are $24, 36, 48$ and the third differences are $12, 12$. A constant third difference means a cubic sequence. For $an^{3}$ the third difference is $6a$, so $6a = 12$ and $a = 2$.

Compare with $2n^{3}$: $2, 16, 54, 128, 250$. Every term of the sequence is $1$ less. So the nth term is $2n^{3} - 1$, and the next (6th) term is $2(6)^{3} - 1$.

(b) Treat the numerators and denominators as two separate linear sequences. Numerators $3, 5, 7, 9$ go up by $2$; denominators $7, 11, 15, 19$ go up by $4$.

Put them back together. The next term has numerator $11$ and denominator $23$.

(c) Each term is $3$ times the one before, so the sequence is geometric with ratio $3$, and the next term is $3 \times 3 = 9$. The third term is $1 = 3^{0}$, so the power is $n - 3$ (check: $n = 1$ gives $3^{-2} = \frac{1}{9}$ ✓).

(a) Keep subtracting $5$ until the terms go below zero.

(b) The common difference is $-5$ (the sequence is going down), so the nth term starts with $-5n$. Work out $-5n$ for $n = 1$: it is $-5$. The first term is $23$, so add $28$.

Check with the second term: $28 - 5(2) = 18$ ✓.

(c) Set the nth term equal to $-52$ and solve. Because $n$ comes out as a positive whole number, $-52$ is in the sequence.

(a) With $n = 1$, the 'sum' is just the first term.

With $n = 2$ the formula gives the first two terms added together, so subtract the first term.

(b) The 6th term is the sum of the first $6$ terms minus the sum of the first $5$ terms.

(Check: the terms run $5, 9, 13, 17, 21, 25$ — a linear sequence going up by $4$, so $25$ fits.)