Consider the following lists of numbers :
`(i) 1, 2, 3, 4, . . .`
`(ii) 100, 70, 40, 10, . . .`
`(iii) – 3, –2, –1, 0, . . .`
`(iv) 3, 3, 3, 3, . . .`
`(v) –1.0, –1.5, –2.0, –2.5, . . .`
● Each of the numbers in the list is called a term.
`=>`Let us observe and write the rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding
(or subtracting) 0 to the term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it.
`=>` In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an `"Arithmetic Progression ( AP )."`
`color{orange}{"So, An arithmetic progression is a list of numbers in which each term is obtained"}`
`color{orange}{" by adding a fixed number to the preceding term except the first term."}`
● This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero
`\color{green} ✍️` Let us denote the first term of an AP by `a_1`, second term by `a_2`, . . ., nth term by
`a_n` and the common difference by d. Then the AP becomes `a_1, a_2, a_3, . . ., a_n`.
So, `a_2 – a_1 = a_3 – a_2 = . . . = a_n – a_(n – 1) = d`.
Some more examples of AP are:
(a) The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149, . . ., 157.
(b) The minimum temperatures ( in degree celsius ) recorded for a week in the month of January in a city, arranged in ascending order are
`– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5`
(c) The balance money ( in Rs. ) after paying 5 % of the total loan of ` 1000` every month is `950, 900, 850, 800, . . ., 50.`
(d) The cash prizes ( in Rs. ) given by a school to the toppers of Classes I to XII are, respectively, `200, 250, 300, 350, . . ., 750.`
(e) The total savings (in Rs. ) after every month for 10 months when ` 50` are saved each month are `50, 100, 150, 200, 250, 300, 350, 400, 450, 500.` It is left as an exercise for you to explain why each of the lists above is an AP. You can see that
`a, a + d, a + 2d, a + 3d, . . .`
represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.
`=>` Note that in examples (a) to (e) above, there are only a finite number of terms. Such an AP is called a `"finite AP."` Also note that each of these Arithmetic Progressions (APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term.
`=>` Now, to know about an AP, what is the minimum information that you need? Is it enough to know the first term? Or, is it enough to know only the common difference? You will find that you will need to know both – the first term a and the common difference d.
● For instance if the first term a is 6 and the common difference d is 3, then the AP is
`6, 9,12, 15, . . .`
and if a is 6 and d is – 3, then the AP is
`6, 3, 0, –3, . . .`
Similarly, when
`a = – 7`, `d = – 2`, the AP is `– 7, – 9, – 11, – 13, . . .`
`a = 1.0, d = 0.1`, the AP is `1.0, 1.1, 1.2, 1.3, . . .`
`a = 0, d = 1 (1/2)` , the AP is `0, 1 (1/2 ) , 3, 4 (1/2) , 6, . . .`
`a = 2, d = 0`, the AP is 2, 2, 2, 2, . . .
`=>`We know that in an AP, every succeeding term is obtained by adding d to the preceding term. So, `d` found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP .
For example, for the list of numbers :
`6, 9, 12, 15, . . . ,`
We have `a_2 – a_1 = 9 – 6 = 3`,
`a_3 – a_2 = 12 – 9 = 3`,
`a_4 – a_3 = 15 – 12 = 3`
● Here the difference of any two consecutive terms in each case is 3. So, the given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers :` 6, 3, 0, – 3, . . .`,
`a_2 – a_1 = 3 – 6 = – 3`
`a_3 – a_2 = 0 – 3 = – 3`
`a_4 – a_3 = –3 – 0 = –3`
● Similarly this is also an AP whose first term is 6 and the common difference is –3.
In general, for an AP `a_1, a_2, . . ., a_n`, we have
`color{orange}{d = a_(k + 1) – a_k}`
`=>` where `a_(k + 1)` and `a_k` are the `( k + 1)`th and the `k^(th)` terms respectively.
● To obtain d in a given AP, we need not find all of `a_2 – a_1, a_3 – a_2, a_4 – a_3, . . . `
It is enough to find only one of them.
● Consider the list of numbers `1, 1, 2, 3, 5, . . . .` By looking at it, you can tell that the difference between any two consecutive terms is not the same. So, this is not an AP.
● Note that to find d in the `AP : 6, 3, 0, – 3, . . .`, we have subtracted 6 from 3 and not 3 from 6, i.e., we should subtract the `k^(th)` term from the `(k + 1)^( th)` term even if the `(k + 1)^( th)` term is smaller.
Consider the following lists of numbers :
`(i) 1, 2, 3, 4, . . .`
`(ii) 100, 70, 40, 10, . . .`
`(iii) – 3, –2, –1, 0, . . .`
`(iv) 3, 3, 3, 3, . . .`
`(v) –1.0, –1.5, –2.0, –2.5, . . .`
● Each of the numbers in the list is called a term.
`=>`Let us observe and write the rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding
(or subtracting) 0 to the term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it.
`=>` In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an `"Arithmetic Progression ( AP )."`
`color{orange}{"So, An arithmetic progression is a list of numbers in which each term is obtained"}`
`color{orange}{" by adding a fixed number to the preceding term except the first term."}`
● This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero
`\color{green} ✍️` Let us denote the first term of an AP by `a_1`, second term by `a_2`, . . ., nth term by
`a_n` and the common difference by d. Then the AP becomes `a_1, a_2, a_3, . . ., a_n`.
So, `a_2 – a_1 = a_3 – a_2 = . . . = a_n – a_(n – 1) = d`.
Some more examples of AP are:
(a) The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149, . . ., 157.
(b) The minimum temperatures ( in degree celsius ) recorded for a week in the month of January in a city, arranged in ascending order are
`– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5`
(c) The balance money ( in Rs. ) after paying 5 % of the total loan of ` 1000` every month is `950, 900, 850, 800, . . ., 50.`
(d) The cash prizes ( in Rs. ) given by a school to the toppers of Classes I to XII are, respectively, `200, 250, 300, 350, . . ., 750.`
(e) The total savings (in Rs. ) after every month for 10 months when ` 50` are saved each month are `50, 100, 150, 200, 250, 300, 350, 400, 450, 500.` It is left as an exercise for you to explain why each of the lists above is an AP. You can see that
`a, a + d, a + 2d, a + 3d, . . .`
represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.
`=>` Note that in examples (a) to (e) above, there are only a finite number of terms. Such an AP is called a `"finite AP."` Also note that each of these Arithmetic Progressions (APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term.
`=>` Now, to know about an AP, what is the minimum information that you need? Is it enough to know the first term? Or, is it enough to know only the common difference? You will find that you will need to know both – the first term a and the common difference d.
● For instance if the first term a is 6 and the common difference d is 3, then the AP is
`6, 9,12, 15, . . .`
and if a is 6 and d is – 3, then the AP is
`6, 3, 0, –3, . . .`
Similarly, when
`a = – 7`, `d = – 2`, the AP is `– 7, – 9, – 11, – 13, . . .`
`a = 1.0, d = 0.1`, the AP is `1.0, 1.1, 1.2, 1.3, . . .`
`a = 0, d = 1 (1/2)` , the AP is `0, 1 (1/2 ) , 3, 4 (1/2) , 6, . . .`
`a = 2, d = 0`, the AP is 2, 2, 2, 2, . . .
`=>`We know that in an AP, every succeeding term is obtained by adding d to the preceding term. So, `d` found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP .
For example, for the list of numbers :
`6, 9, 12, 15, . . . ,`
We have `a_2 – a_1 = 9 – 6 = 3`,
`a_3 – a_2 = 12 – 9 = 3`,
`a_4 – a_3 = 15 – 12 = 3`
● Here the difference of any two consecutive terms in each case is 3. So, the given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers :` 6, 3, 0, – 3, . . .`,
`a_2 – a_1 = 3 – 6 = – 3`
`a_3 – a_2 = 0 – 3 = – 3`
`a_4 – a_3 = –3 – 0 = –3`
● Similarly this is also an AP whose first term is 6 and the common difference is –3.
In general, for an AP `a_1, a_2, . . ., a_n`, we have
`color{orange}{d = a_(k + 1) – a_k}`
`=>` where `a_(k + 1)` and `a_k` are the `( k + 1)`th and the `k^(th)` terms respectively.
● To obtain d in a given AP, we need not find all of `a_2 – a_1, a_3 – a_2, a_4 – a_3, . . . `
It is enough to find only one of them.
● Consider the list of numbers `1, 1, 2, 3, 5, . . . .` By looking at it, you can tell that the difference between any two consecutive terms is not the same. So, this is not an AP.
● Note that to find d in the `AP : 6, 3, 0, – 3, . . .`, we have subtracted 6 from 3 and not 3 from 6, i.e., we should subtract the `k^(th)` term from the `(k + 1)^( th)` term even if the `(k + 1)^( th)` term is smaller.
