Number Patterns

Chapter 7 Study Notes · O-Level 4052

Section 3 of 6
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=4n1\displaystyle T_n = 4n - 1, find (i) T4\displaystyle T_4 and (ii) T6T2\displaystyle T_6 - T_2.

Show solution
(i) T4=4(4)1T_4 = 4(4) - 1
=161\displaystyle = 16 - 1
=15\displaystyle = 15
(ii) T6=4(6)1T_6 = 4(6) - 1
=241\displaystyle = 24 - 1
=23\displaystyle = 23
T2\displaystyle T_2
=4(2)1\displaystyle = 4(2) - 1
=81\displaystyle = 8 - 1
=7\displaystyle = 7
T6T2\displaystyle T_6 - T_2
=237\displaystyle = 23 - 7
=16\displaystyle = 16

Now you try

Given Tn=6n+4T_n = 6n + 4, find T1T_1.

Show solution
Tn\displaystyle T_n
=6n+4\displaystyle = 6n + 4
T1\displaystyle T_1
=6(1)+4substitute n=1\displaystyle = 6(1) + 4 \quad \scriptsize\textit{substitute } n = 1
=6+4\displaystyle = 6 + 4
=10\displaystyle = 10

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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)+8=35T_9 = 3(9) + 8 = 35

Reverse problem: find nn given TnT_n

The n\displaystyle nth term of a sequence is Tn=5n3\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:
5n3\displaystyle 5n - 3
=72\displaystyle = 72
5n\displaystyle 5n
=75\displaystyle = 75
n\displaystyle n
=15\displaystyle = 15
Check: T15=5(15)3=753=72T_{15} = 5(15) - 3 = 75 - 3 = 72

Now you try

The nth term of a sequence is Tn=n25n2T_n = n^2 - 5n - 2. Find the smallest value of n for which Tn>12T_n > 12.

Show solution
Tn=n25n2T_n = n^2 - 5n - 2
n25n2>12set up inequalityn^2 - 5n - 2 > 12 \quad \scriptsize\textit{set up inequality}
n25n14>0rearrangen^2 - 5n - 14 > 0 \quad \scriptsize\textit{rearrange}
T7\displaystyle T_{7}
=12check boundary\displaystyle = 12 \quad \scriptsize\textit{check boundary}
T8\displaystyle T_{8}
=22>12first term exceeding threshold\displaystyle = 22 > 12 \quad \scriptsize\textit{first term exceeding threshold}
n\displaystyle n
=8\displaystyle = 8

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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
Substituting n=3n = 3: 3a+b=143a + b = 14 \quad (1)
Substituting n=7n = 7: 7a+b=307a + b = 30 \quad (2)
Subtract (1) from (2): 4a=164a = 16
a=4a = 4
Substitute into (1): 3(4)+b=143(4) + 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

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