Please Wait... While Loading Full Video#### Class 10 Chapter 5 - ARITHMETIC PROGRESSIONS

♦ Introduction

♦ Arithmetic Progressions

♦ Arithmetic Progressions

● In nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone etc.

● We now look for some patterns which occur in our day-to-day life. Some such examples are :

(i) Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of Rs. 8000, with an annual increment of Rs. 500 in her salary. Her salary (in Rs. ) for the 1st, 2nd, 3rd, . . . years will be, respectively

`8000, 8500, 9000, . . . .`

(ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see Fig. 5.1). The bottom rung is 45 cm in length. The lengths (in cm) of the 1st, 2nd, 3rd, . . ., 8th rung from the bottom to the top are, respectively

`45, 43, 41, 39, 37, 35, 33, 31`

(iii) In a savings scheme, the amount becomes 5/4 times of itself after every 3 years. The maturity amount (in Rs. ) of an investment of ` 8000` after 3, 6, 9 and 12 years will be, respectively : `10000, 12500, 15625, 19531.25`

(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2) are, respectively `12, 22, 32, . . . .`

(v) Shakila puts Rs. 100 into her daughter’s money box when she was one year old and increased the amount by Rs. 50 every year. The amounts of money (in Rs . ) in the box on the 1st, 2nd, 3rd, 4th, . . . birthday were

`100, 150, 200, 250, . . .`, respectively.

(vi) A pair of rabbits are too young to produce in their first month. In the second, and every subsequent month, they produce a new pair. Each new pair of rabbits produce a new pair in their second month and in every subsequent month (see Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd, . . ., 6th month, respectively are :

`1, 1, 2, 3, 5, 8`

`=>` In the examples above, we observe some patterns. In some, we find that the succeeding terms are obtained by adding a fixed number, in other by multiplying with a fixed number, in another we find that they are squares of consecutive numbers, and so on.

`=>` In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.

● We now look for some patterns which occur in our day-to-day life. Some such examples are :

(i) Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of Rs. 8000, with an annual increment of Rs. 500 in her salary. Her salary (in Rs. ) for the 1st, 2nd, 3rd, . . . years will be, respectively

`8000, 8500, 9000, . . . .`

(ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see Fig. 5.1). The bottom rung is 45 cm in length. The lengths (in cm) of the 1st, 2nd, 3rd, . . ., 8th rung from the bottom to the top are, respectively

`45, 43, 41, 39, 37, 35, 33, 31`

(iii) In a savings scheme, the amount becomes 5/4 times of itself after every 3 years. The maturity amount (in Rs. ) of an investment of ` 8000` after 3, 6, 9 and 12 years will be, respectively : `10000, 12500, 15625, 19531.25`

(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2) are, respectively `12, 22, 32, . . . .`

(v) Shakila puts Rs. 100 into her daughter’s money box when she was one year old and increased the amount by Rs. 50 every year. The amounts of money (in Rs . ) in the box on the 1st, 2nd, 3rd, 4th, . . . birthday were

`100, 150, 200, 250, . . .`, respectively.

(vi) A pair of rabbits are too young to produce in their first month. In the second, and every subsequent month, they produce a new pair. Each new pair of rabbits produce a new pair in their second month and in every subsequent month (see Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd, . . ., 6th month, respectively are :

`1, 1, 2, 3, 5, 8`

`=>` In the examples above, we observe some patterns. In some, we find that the succeeding terms are obtained by adding a fixed number, in other by multiplying with a fixed number, in another we find that they are squares of consecutive numbers, and so on.

`=>` In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.

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.

`(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.

Q 3129180011

For the `AP : 3/2 , 1/2 , - 1/2, -3/2 ,....,` write the first term a and the common difference d.

Class 10 Chapter 5 Example 1

Class 10 Chapter 5 Example 1

Here, `a = 3/2 , d = 1/2 -3/2 = -1` .

Remember that we can find d using any two consecutive terms, once we know that

the numbers are in AP.

Q 3139180012

Which of the following list of numbers form an AP? If they form an AP, write the next two terms :

(i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . .

(iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Class 10 Chapter 5 Example 2

(i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . .

(iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Class 10 Chapter 5 Example 2

(i) We have `a_2 – a_1 = 10 – 4 = 6`

`a_3 – a_2 = 16 – 10 = 6`

`a_4 – a_3 = 22 – 16 = 6`

i.e., `a_(k + 1) – a_k` is the same every time.

So, the given list of numbers forms an AP with the common difference d = 6.

The next two terms are: `22 + 6 = 28` and `28 + 6 = 34`.

(ii) `a_2 – a_1 = – 1 – 1 = – 2`

`a_3 – a_2 = – 3 – ( –1 ) = – 3 + 1 = – 2`

`a_4 – a_3 = – 5 – ( –3 ) = – 5 + 3 = – 2`

i.e., `a_(k + 1) – a_k` is the same every time.

So, the given list of numbers forms an AP with the common difference d = – 2.

The next two terms are:

– 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9

(iii) `a_2 – a_1 = 2 – (– 2) = 2 + 2 = 4`

`a_3 – a_2 = – 2 – 2 = – 4`

As `a_2 – a_1 ≠ a_3 – a_2` , the given list of numbers does not form an AP.

(iv) `a_2 – a_1 = 1 – 1 = 0`

`a_3 – a_2 = 1 – 1 = 0`

`a_4 – a_3 = 2 – 1 = 1`

Here, `a_2 – a_1 = a_3 – a_2 ≠ a_4 – a_3`.

So, the given list of numbers does not form an AP.