Progression
1.2 Arithmetic progression
(A)
Characteristics of Arithmetic Progression
An arithmetic
progression is a progression in which the difference between any
term and the immediate term before is a constant. The constant is called
the common difference, d.
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d = Tn – Tn-1 or d = Tn+1– Tn
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Smart
TIPS: For an arithmetic progression, you always plus or
minus a fixed number
Solution: :-
(B) The
steps to prove whether a given number sequence is an arithmetic progression
Step 1: List down any three consecutive terms. [Example: T1 , T2 , T3 .]
Step 2:
Calculate the values
of T3 − T2 and T2 − T1 .
Step 3:
If T3 − T2 = T2 − T1 = d, then the
number sequence is an arithmetic progression.
Example
2:
Prove
whether the following number sequence is an arithmetic progression
(a) 7,
10, 13, …
(b) –20,
–15, –9, …
Solution:
(a) 7,
10, 13 ← (Step 1: List down T1 , T2 , T3 )
T3 – T2 = 13 – 10
= 3 ← (Step 2:
Find T3 – T2 and T2 – T1)
T2 – T1 = 10 – 7
= 3 ← (Step 2:
Find T3 – T2 and T2 – T1)
T3 – T2 = T2 – T1
Therefore, this is an
arithmetic progression.
(b) –20, –15, –9
T3 – T2 = –9 – (–15)
= 6
T2 – T1 = –15 – (–20)
= 5
T3 – T2 ≠ T2 – T1
Therefore, this is not an
arithmetic progression.
1.3 The nth term of an Arithmetic Progression
(C) The nth term of an Arithmetic Progression
(C) The nth term of an Arithmetic Progression
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Tn = a + (n −
1) d
where
a = first term
d = common difference
n = the number of term
Tn =
the nth term
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(D) The Number of Terms in an Arithmetic Progression
Smart
TIPS: You can find the number of term in an arithmetic progression if you know the last term
Example
1:
Find the
number of terms for each of the following arithmetic progressions.
(a) 5,
9, 13, 17... , 121
(b) 1,
1.25, 1.5, 1.75,..., 8
Solution:
(a) 5, 9, 13, 17... , 121
AP,
a = 5, d = 9 – 5 = 4
The last term, Tn = 121
a + (n – 1) d = 121
5 + (n – 1) (4) = 121
(n – 1) (4) = 116
(n – 1) = 11641164 =
29
n = 30
(b)
1, 1.25, 1.5, 1.75,..., 8
AP,
a = 1, d = 1.25 – 1 = 0.25
Tn = 8
a + (n – 1) d = 8
1 + (n – 1) (0.25) = 8
(n – 1) (0.25) = 7
(n – 1) = 28
n = 29
(E) The Consecutive Terms of an Arithmetic Progression
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If a, b, c are
three consecutive terms of an arithmetic progression, then
c – b = b – a
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Example
2:
If x +
1, 2x + 3 and 6 are three consecutive terms of an arithmetic progression,
find the value of x and its common difference.
Solution:
x +
1, 2x + 3, 6
c – b = b – a
6 –
(2x + 3) = (2x + 3) – (x + 1)
6 –
2x – 3 = 2x + 3 – x – 1
3 – 2x = x + 2
X = 1/3
1/3 + 1, 2(1/3)+3, 6
4/3, 32/3, 6
d = 32/3 − 4/3 = 21/3
The Number Of Terms In An Arithmetic Progression
Hint
: You can find the number of term in an arithmetic progression if you know the last term
Find
the number of terms for each of the following arithmetic progressions.
(a)
5, 9, 13, 17... , 121
(b)
1, 1.25, 1.5, 1.75,..., 8
Three Consecutive
Terms Of An Arithmetic Progression
If a, b, c are three consecutive
terms of an arithmetic progression, then
c – b = b - a
c – b = b - a
Example:
If x + 1, 2x + 3 and 6 are three consecutive terms of an arithmetic progression, find the value of x and its common difference.
If x + 1, 2x + 3 and 6 are three consecutive terms of an arithmetic progression, find the value of x and its common difference.
Geometric Progression,
Series & Sums
Introduction
A geometric sequence is a sequence such that any element
after the first is obtained by multiplying the preceding element by a constant
called the common ratio which is denoted by r. The common
ratio (r) is obtained by dividing any term by the preceding term, i.e.,
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where
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r
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common
ratio
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a1
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first
term
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a2
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second
term
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a3
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third
term
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an-1
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the
term before the n th term
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an
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the n th term
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The geometric sequence is sometimes called the geometric progression or GP, for short.
For example, the sequence 1, 3, 9, 27, 81 is a geometric
sequence. Note that after the first term, the next term is obtained by
multiplying the preceding element by 3.
The geometric sequence has its sequence formation:
To find the nth term of a geometric sequence we use the formula:
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where
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r
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common
ratio
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a1
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first
term
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an-1
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the
term before the n th term
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n
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number
of terms
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Sum of
Terms in a Geometric Progression
Finding the sum of terms in a geometric progression is easily
obtained by applying the formulas:
nth partial sum of a geometric sequence
sum to infinity
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where
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Sn
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sum
of GP with n terms
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S∞
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sum
of GP with infinitely many terms
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a1
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the
first term
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r
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common
ratio
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n
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number
of terms
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Examples
of Common Problems to Solve
Write down a specific term in a Geometric Progression
Question
Write down the 8th term in the Geometric Progression 1, 3, 9,
...
Answer
Finding the number of terms in a Geometric Progression
Question
Find the number of terms in the geometric progression 6, 12, 24,
..., 1536
Answer
Finding the sum of a Geometric Series
Question
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Find
the sum of each of the geometric series
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Answer
Finding the sum of a Geometric Series to Infinity
Question
Answer
Converting a Recurring Decimal to a Fraction
Decimals that occurs in repetition infinitely or are repeated in period are called recurring decimals.
For example, 0.22222222... is a recurring decimal because the
number 2 is repeated infinitely.
The recurring decimal 0.22222222... can be written as
Another example is 0.234523452345... is a recurring decimal
because the number 2345 is repeated periodically.
Thus, it can be written as
Question
Express 
as a fraction in their lowest terms.
as a fraction in their lowest terms.
Answer
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