Sequences (Nth Term): Practise Online

Either paper 4 marks

The \(n\)th term of a sequence is \(3n + 8\).

(a)
Write down the first three terms of the sequence.

(b)
One term of this sequence is equal to 35.

Work out the value of \(n\) for that term.
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(a) \(11,\ 14,\ 17\)
(b) \(9\)
Put \(n = 1\), then \(n = 2\), then \(n = 3\) into the rule:

\(3 \times 1 + 8 = 11\)

\(3 \times 2 + 8 = 14\)

\(3 \times 3 + 8 = 17\)
Set the rule equal to 35:

\(3n + 8 = 35\)

\(3n = 27\)

\(n = \frac{ 27 }{ 3 } = 9\)

So 35 is the 9th term.
- (a) M1substituting \(n = 1\), \(n = 2\) or \(n = 3\) into \(3n + 8\) to find a term, e.g. \(11\)
- (a) A1all three terms \(11\), \(14\), \(17\)
- (b) M1setting \(3n + 8 = 35\) and rearranging to \(3n = 27\)
- (b) A1\(n = 9\)
Either paper 1 mark

Here are the first four terms of a sequence.

\(10\), \(14\), \(20\), \(28\)

Write down the next term of this sequence.
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\(38\)
The gaps between the terms are \(4\), \(6\), \(8\) — each gap is two more than the one before it.

The next gap carries on growing the same way, so it is \(10\):

\(28 + 10 = 38\)
Either paper 4 marks
Here is a table of values.

\(x\) 4 8 10 12

\(y\) 11 7 3

The \(x\)-values in the table make a linear sequence. The \(y\)-values in the table make a different linear sequence.
(a)
Complete the table.

(b)
The points \((8, 11)\), \((10, 7)\) and \((12, 3)\) are plotted and joined by a straight line.

Use the graph to write down the value of \(y\) when \(x = 9\).
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(a) \(x = 6\), \(y = 19\) and \(y = 15\)
(b) \(9\)
The \(x\)-values go up by 2 each time, so the missing \(x\)-value is \(6\).

The \(y\)-values go down by 4 each time: \(7 - 11 = -4\).

Continuing the \(y\)-values with that difference, the missing values are \(19\) and \(15\).
Go up from \(x = 9\) on the \(x\)-axis to the line, then across to the \(y\)-axis.

The line gives \(y = 9\).
- (a) B1missing \(x\)-value \(6\)
- (a) M1common difference of the \(y\)-values \(-4\)
- (a) A1missing \(y\)-values \(19\) and \(15\)
Either paper 3 marks

Here are the first three patterns in a sequence. Each pattern is drawn on squared paper and some of its squares are marked with a dot.

Pattern 1

Pattern 2

Pattern 3

(a)
Draw pattern 4 on the grid. Mark its squares with dots.

(b)
How many squares in pattern 7 would be marked with a dot?
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You cannot draw on this page. Draw (a) on paper, then show the answer and check your drawing against the model.
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(a) Pattern 4 with 4 dotted squares on each of the 3 arms and the square where the bar meets the stem, 13 dotted squares in all
(b) \(22\)
(a)

Count the dotted squares along one arm of the T-shape: 1 in pattern 1, 2 in pattern 2 and 3 in pattern 3.

Each pattern, every arm gets one more dotted square, so in pattern 4 each of the 3 arms (left, right and down) has 4.

The square where the bar meets the stem is marked.

The arms hold \(3 \times 4 = 12\) dotted squares, and the square where the bar meets the stem adds one more.

So pattern 4 has 13 dotted squares.
The numbers of dotted squares in patterns 1, 2 and 3 are 4, 7 and 10.

They go up by 3 each time.

Pattern 7 is 4 patterns after pattern 3, so it has 4 more steps of 3.

\(10 + 4 \times 3 = 22\)
- (a) B1pattern 4 drawn with 4 dotted squares on each arm, the square where the bar meets the stem marked: 13 dotted squares in all
- (b) M1dotted squares go up by 3 each pattern (4, 7, 10), continued past pattern 4
- (b) A122 dotted squares in pattern 7
Either paper 2 marks

A pattern of square tiles grows in the same way each time.

Pattern 1 uses \(9\) tiles, pattern 2 uses \(16\) tiles, pattern 3 uses \(23\) tiles and pattern 4 uses \(30\) tiles.

Find an expression, in terms of \(n\), for the number of tiles in pattern \(n\).
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\(7n + 2\)
The terms go up by 7 each time, so the rule starts with \(7n\).

The multiples of 7 are 7, 14, 21, … and the sequence is 9, 16, 23, 30, …

Every term is 2 more than the matching multiple of 7, so

\(n\)th term \(= 7n + 2\)
- M1common difference 7 giving \(7n\)
- A1\(n\)th term \(7n + 2\)
Either paper 2 marks

The \(n\)th term of a sequence is \(8n - 1\)

One term of the sequence is equal to \(295\).

Work out the position of that term in the sequence.
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\(37\)
The term in position \(n\) is \(8n - 1\), so set that equal to 295.

\(8n - 1 = 295\)

\(8n = 296\)

\(n = 296 \div 8 = 37\)

So \(295\) is the 37th term.
- M1sets \(8n - 1 = 295\) and gets \(8n = 296\)
- A1position \(37\)
Either paper 4 marks

A term-to-term rule is shown in a flow diagram. Each term goes in and the next term of the sequence comes out.

(a)
term → + 1 → × 2 → next term
The 1st term of the sequence is \(8\). Find the 3rd term.

(b) A different sequence has this rule.
term → × 4 → + 12 → next term
The 2nd term of this sequence is \(40\). Find the 1st term.
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(a) \(38\)
(b) \(7\)
Apply the boxes in order, left to right, to get the 2nd term.

\(8\), + 1 gives \(9\), then × 2 gives \(18\).

Repeat with the 2nd term to get the 3rd term.

\(18\), + 1 gives \(19\), then × 2 gives \(38\).

The 3rd term is \(38\).
Work backwards from the 2nd term, undoing the second box first.

Undo + 12 with − 12: \(40\) becomes \(28\).

Undo × 4 with ÷ 4: \(28\) becomes \(7\).

The 1st term is \(7\).
- (a) M1applying both boxes in order to the 1st term to reach the 2nd term \(18\)
- (a) A13rd term \(38\)
- (b) M1undoing the second box first to reach \(28\)
- (b) A11st term \(7\)

Either paper 3 marks

The table shows the first five terms of four sequences. Below it are six descriptions.

Sequence	First five terms	Description letter
1	10, 15, 21, 28, 36
2	25, 36, 49, 64, 81
3	4, 5, 9, 14, 23	F
4	24, 20, 16, 12, 8

Letter	Description
A	Square numbers
B	Triangular numbers
C	Arithmetic, increasing (add the same number each time)
D	Arithmetic, decreasing (subtract the same number each time)
E	Geometric (multiply by the same number each time)
F	Fibonacci-type (each term is the sum of the two before it)

Match each sequence to its description. Sequence 3 has been done for you: it is description F. Your answer (just for you: it is never sent anywhere)

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Sequence 1: B; sequence 2: A; sequence 4: D

Test each sequence by finding the gaps between terms, and the ratios where the gaps do not settle.

Sequence 1: gaps \(5,\ 6,\ 7,\ 8\), so the gaps grow by 1 each time, and each term is a triangular number. This is description B.

Sequence 2: gaps \(11,\ 13,\ 15,\ 17\), so the gaps grow by 2 each time, and each term is a whole number squared. This is description A.

Sequence 3: gaps \(1,\ 4,\ 5,\ 9\), so the gaps are not equal, but each term is the sum of the two before it. This is description F.

Sequence 4: gaps \(-4,\ -4,\ -4,\ -4\), so the gaps are all equal and negative. This is description D.

Descriptions C and E match none of the four sequences.

B1sequence 1 matched to B, because the gaps grow by 1 each time, and each term is a triangular number
B1sequence 2 matched to A, because the gaps grow by 2 each time, and each term is a whole number squared
B1sequence 4 matched to D, because the gaps are all equal and negative

Either paper 3 marks
Here are the first three patterns in a sequence made up of small squares.

Pattern 1 Pattern 2 Pattern 3

■ ■ ■ ■ ■ ■

■ ■ ■ ■ ■

■ ■ ■

and 49 extra squares and 49 extra squares and 49 extra squares

The number of small squares in pattern \(n\) is \(n^2 + 49\)
(a)
Work out the number of small squares in pattern 5.

(b)
Work out the least value of \(n\) for which the number of small squares is greater than 110.
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(a) \(74\)
(b) \(8\)
Put \(n = 5\) into \(n^2 + 49\):

\(5^2 + 49 = 25 + 49 = 74\)
We need \(n^2 + 49 > 110\), so \(n^2\) must be greater than \(61\).

Try \(n = 7\): \(7^2 + 49 = 49 + 49 = 98\), which is less than \(110\), so \(n = 7\) is too small.

Try \(n = 8\): \(8^2 + 49 = 64 + 49 = 113\), which is greater than \(110\).

The least value of \(n\) is \(8\).
- (b) M1trial with a square: \(n = 7\) gives \(98\), which is not greater than \(110\)
- (b) A1\(n = 8\), since \(113\) is greater than \(110\)
Either paper 4 marks

Here are the first four terms of an arithmetic sequence.

\(22,\ 17,\ 12,\ 7\)

(a)
Find an expression, in terms of \(n\), for the \(n\)th term of this sequence.

(b)
Work out the 40th term of this sequence.
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(a) \(-5n + 27\)
(b) \(-173\)
The terms go down by 5 each time, so the rule starts with \(-5n\).

When \(n = 1\) the rule so far gives \(-5\), and the first term is 22, so 27 has to be added:

\(n\)th term \(= -5n + 27\)
Put \(n = 40\) into the rule:

\(-5 \times 40 + 27 = -200 + 27 = -173\)
- (a) M1common difference gives \(-5n\)
- (a) A1\(n\)th term \(-5n + 27\)
- (b) M1substitutes \(n = 40\)
- (b) A1\(-173\)

Pattern 1	Pattern 2	Pattern 3
■	■ ■	■ ■ ■
	■ ■	■ ■ ■
		■ ■ ■
and 49 extra squares	and 49 extra squares	and 49 extra squares

\(x\)	4		8	10	12
\(y\)			11	7	3