Number Patterns

Chapter 7 Study Notes · O-Level 4052

Section 3 of 7
Contents▾

7.2General Term of a Sequence

The general term TnT_n gives any term straight from its position, so the 500th needs no list.

A sequence is a function of position: each nn gives exactly one TnT_n.

Evaluating the kkth term
To find Tk, substitute n=k into the formula for Tn\displaystyle \text{To find } T_k \text{, substitute } n = k \text{ into the formula for } T_n

Evaluating terms from a formula

Given Tn=4n−1\displaystyle T_n = 4n - 1, find (i) T4\displaystyle T_4 and (ii) T6−T2\displaystyle T_6 - T_2.

Show solution▾
(i) T4=4(4)−1T_4 = 4(4) - 1
=16−1\displaystyle = 16 - 1
=15\displaystyle = 15
(ii) T6=4(6)−1T_6 = 4(6) - 1
=24−1\displaystyle = 24 - 1
=23\displaystyle = 23
T2\displaystyle T_2
=4(2)−1\displaystyle = 4(2) - 1
=8−1\displaystyle = 8 - 1
=7\displaystyle = 7
T6−T2\displaystyle T_6 - T_2
=23−7\displaystyle = 23 - 7
=16\displaystyle = 16

Now you try

Given Tn=5n−4T_n = 5n - 4, find T7T_{7}.

Show solution▾
Tn\displaystyle T_n
=5n−4\displaystyle = 5n - 4
T7\displaystyle T_{7}
=5(7)−4substitute n=7\displaystyle = 5(7) - 4 \quad \scriptsize\textit{substitute } n = 7
=35−4\displaystyle = 35 - 4
=31\displaystyle = 31

Want it marked, with hints, and a new question each time? Practise this →

Finding which term equals a given value

Given Tn=3n+8\displaystyle T_n = 3n + 8, find the value of n\displaystyle n for which Tn=35\displaystyle T_n = 35.

Show solution▾
3n+8\displaystyle 3n + 8
=35\displaystyle = 35
3n\displaystyle 3n
=27\displaystyle = 27
n\displaystyle n
=9\displaystyle = 9
The 9th term equals 35.
Check: T9=3(9)+8T_9 = 3(9) + 8
=35= 35 ✓

Reverse problem: find nn given TnT_n

The n\displaystyle nth term of a sequence is Tn=5n−3\displaystyle T_n = 5n - 3. Find the value of n\displaystyle n for which Tn=72\displaystyle T_n = 72.

Show solution▾
Set Tn=72T_n = 72:
5n−3\displaystyle 5n - 3
=72\displaystyle = 72
5n\displaystyle 5n
=75\displaystyle = 75
n\displaystyle n
=15\displaystyle = 15
Check: T15=5(15)−3T_{15} = 5(15) - 3
=75−3= 75 - 3
=72= 72 ✓

Now you try

Given Tn=2n2−5n+3T_n = 2n^2 - 5n + 3, find the positive integer value of n such that Tn=15T_n = 15.

Show solution▾
2n2−5n+3=152n^2 - 5n + 3 = 15
We want Tn=15T_n = 15, so try values of n in turn.
T2\displaystyle T_{2}
=1\displaystyle = 1
T3\displaystyle T_{3}
=6\displaystyle = 6
T4=15T_{4} = 15, which is the value we want.
∴n=4\therefore n = 4

Want it marked, with hints, and a new question each time? Practise this →

Reverse problem: find aa and bb from two known terms

A sequence has general term Tn=an+b\displaystyle T_n = an + b. Given T3=14\displaystyle T_3 = 14 and T7=30\displaystyle T_7 = 30, find a\displaystyle a and b\displaystyle b, and hence state the general term.

Show solution▾
Going from T3T_3 to T7T_7 is 7−3=47 - 3 = 4 steps along the sequence.
Over those steps the value rises by 30−14=1630 - 14 = 16.
In Tn=an+bT_n = an + b, each step adds aa, so a=164a = \dfrac{16}{4}
a=4a = 4
Substituting n=3n = 3: 4(3)+b=144(3) + b = 14
12+b\displaystyle 12 + b
=14\displaystyle = 14
b\displaystyle b
=2\displaystyle = 2
Tn\displaystyle T_n
=4n+2\displaystyle = 4n + 2
Check: T3=4(3)+2=14T_3 = 4(3) + 2 = 14 ✓, T7=4(7)+2=30T_7 = 4(7) + 2 = 30 ✓

Recognising common general terms

Write the general term for each sequence: (a) 5,10,15,20,25,…\displaystyle 5, 10, 15, 20, 25, \ldots (b) 1,4,9,16,25,…\displaystyle 1, 4, 9, 16, 25, \ldots (c) 2,8,18,32,50,…\displaystyle 2, 8, 18, 32, 50, \ldots

Show solution▾
(a) Positive multiples of 5: Tn=5nT_n = 5n
Check: T1=5(1)=5T_1 = 5(1) = 5 ✓, T4=5(4)=20T_4 = 5(4) = 20 ✓
(b) Perfect squares: Tn=n2T_n = n^2
Check: T3=32=9T_3 = 3^2 = 9 ✓, T5=52=25T_5 = 5^2 = 25 ✓
(c) Twice the perfect squares: Tn=2n2T_n = 2n^2
Check: T2=2(22)=8T_2 = 2(2^2) = 8 ✓, T5=2(52)=50T_5 = 2(5^2) = 50 ✓

Practise this

Questions on Number Patterns, marked as you go, with the working shown.

Practise Number Patterns