Arithmetic Progression CBSE Class 10th

8/3/2019 Arithmetic Progression CBSE Class 10th

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Arithmetic

ProgressionBy Poojashri


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Definition

An arithmetic progression is a list of numbers

in which each term is obtained by adding afixed number to the preceding term exceptthe first term.


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General form of A.P

a, a + d, a + 2d, a + 3d, . . .

represents an arithmetic progression wherea is the first term and d the common

difference.

Types of A.P. Finite A.P : An arithmetic progression

sequence in which there are finitenumber of terms.


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Examples:

The heights ( in cm ) of some students of a schoolstanding in a queue in the morning assembly are147 , 148, 149, . . ., 157.

The minimum temperatures ( in degree Celsius )recorded for a week in the month of January in acity, arranged in ascending order are

3.1, 3.0, 2.9, 2.8, 2.7, 2.6, 2.5

The balance money ( in Rs ) after paying 5 % ofthe total loan of Rs 1000 every month is

950, 900, 850, 800, . . ., 50.

The total savings (in Rs) after every month for 10months when Rs 50 are saved each month are

50, 100, 150, 200, 250, 300, 350, 400, 450, 500.


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Infinite A.P: An arithmetic progression

sequence in which there are finite numberof terms.

Examples:

Real numbers1, 2, 3, 4, . . .

Imaginary numbers

3, 2, 1, 0, 1, 2, 3,. . .


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Terminology

} Term : Each of the numbers in the list orsequence is called a term.

}Common Difference(d) : Commondifference d is a fixed number by addingof which to any number gives the nextterm in arithmetic progression.

For A.P: 1, 1, 3, 5, . . .

d is -2.


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nth term of an APLet us consider an instance

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, . . . .What would be her monthly salary for

the fifth year?


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To answer this, let us first see what hermonthly salary for the second year

would be.

It would be Rs (8000 + 500) = Rs 8500. Inthe same way, we can find the monthly

salary for the 3rd, 4th and 5th year by

adding Rs 500 to the salary of the previousyear.

So, the salary for the 3rd year = Rs (8500 +500)

= Rs (8000 + 500 + 500)

= Rs (8000 + 2 500)

= Rs [8000 + (3 1) 500] (for the 3rd year)

= Rs 9000


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Salary for the 4th year = Rs (9000 + 500)

= Rs (8000 + 500 + 500 + 500)= Rs (8000 + 3 500)

= Rs [8000 + (4 1) 500] (for the 4th year)

= Rs 9500

Salary for the 5th year = Rs (9500 + 500)= Rs (8000+500+500+500 + 500)

= Rs (8000 + 4 500)

= Rs [8000 + (5 1) 500] (for the 5th year)

= Rs10000

Observe that we are getting a list of numbers

8000, 8500, 9000, 9500, 10000, . . .

These numbers are in AP. (Why?)


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Now, looking at the pattern formed above,

can you find her monthly salary for the 6thyear? The 15th year? And, assuming that

she will still be working in the job, whatabout the monthly salary for the 25th year?You would calculate this by adding Rs 500

each time to the salary of the previous yearto give the answer. Can we make this

process shorter? Let us see. You may havealready got some idea from the way

we have obtained the salaries above.


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Salary for the 15th year

= Salary for the 14th year + Rs 500= Rs [8000 + 14 500]

= Rs [8000 + (15 1) 500] = Rs 15000

i.e., First salary + (15 1) Annual

increment.

In the same way, her monthly salary for the25th year would be

Rs [8000 + (25 1) 500] = Rs 20000= First salary + (25 1) Annual increment


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Let a1, a2, a3, . . . be an AP whose first term

a1 is a and the common difference is d.Then,

the second term

a2 = a + d = a + (2 1) d

the third terma3 = a2 + d = (a + d) + d = a + 2d = a + (3

1) d

the fourth term

a4 = a3 + d = (a + 2d) + d = a + 3d = a + (4 1) d

. . . . . . . .


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Looking at the pattern, we can say that the

nth terman

=a

+ (n

1)d

.So, the nth term an of the AP with first terma and common difference d is given by

an= a + (n 1) d.

an

is also called the general term of the AP.Ifthere are m terms in the AP, then am

represents the last term which is sometimes

also denoted by l.


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Examples

} Example 1 : Find the 10th term of the AP :2, 7, 12, . . .

Solution : Here, a = 2, d = 7 2 = 5 and n =10.

We have an = a + (n1) d

So, a10 = 2 + (10 1) 5 = 2 + 45 = 47

Therefore, the 10th term of the given AP is47.


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} Example 2: Which term of the AP : 21, 18, 15, . . . is 81? Also, is any term 0? Give reason for your

answer.Solution : Here, a = 21, d = 18 21 = 3 and an = 81,and we have to find n.As an = a + ( n1) d,we have 81 = 21 + (n1)( 3)

81 = 24 3n105 = 3nSo, n = 35Therefore, the 35th term of the given AP is 81.Next, we want to know if there is any n for which an= 0. If such an n is there, then

21

+ (n1) (3) = 0,i.e., 3(n1) = 21

i.e., n = 8So, the eighth term is 0.


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}


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}


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}


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Sum of First n Terms of an AP

Let us consider an instance.

Shakila put Rs100 into her daughtersmoney box when she was one year old

and increased the amount by Rs 50 everyyear. The amounts of money (in Rs) in the

box on the 1st, 2nd, 3rd, 4th, . . . birthday

were

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


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How much money will be collected in the

money box by the time her daughter is 21

years old?

Here, the amount of money (in Rs) put inthe money box on her first, second, third, fourth

. . . birthday were respectively 100, 150, 200,

250, . . . till her 21st birthday. To find the totalamount in the money box on her 21st birthday,

we will have to write each of the 21 numbers in

the list above and then add them up. Dont youthink it would be a tedious and time consuming

process? Can we make the process shorter?This would be possible if we can find a method

for getting this sum. Let us see.


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S = 1 + 2 + 3 + . . . + 99 + 100

And then, reverse the numbers to writeS = 100 + 99 + . . . + 3 + 2 + 1

Adding these two, we get

2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 +99) + (1 + 100)

= 101 + 101 + . . . + 101 + 101 (100 times)

So, S =[(100101)/2] 5050,

i.e., the sum = 5050.

We will now use the same technique to find

the sum of the first n terms of an AP :a, a + d, a + 2d, . . .


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The nth term of this AP is a + (n 1) d. Let S

denote the sum of the first n terms of the AP.We have

S = a + (a + d ) + (a + 2d) + . . . + [a + (n 1)d ] ..(1)

Rewriting the terms in reverse order, wehave

S = [a + (n 1) d] + [a + (n 2) d ] + . . . + (a+ d) + a..(2)

On adding (1) and (2), term-wise. we get

2S = n [2a + (n 1) d ] (Since, there are n terms)


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}


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Now, if there are only n terms in an AP, then

an

= l,

the last term . From (3), we see that

This form of the result is useful when the firstand the last terms of an AP are

given and the common difference is notgiven.


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Now we return to the question that was

posed to us in the beginning. The amountof money (in Rs) in the money box ofShakilas daughter on 1st, 2nd, 3rd, 4thbirthday,

. . ., were 100, 150, 200, 250, . . ., respectivelyThis is an AP. We have to find the total

money collected on her 21st birthday, i.e.,the sum of the first 21 terms of this AP.


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}


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So, the amount of money collected on her

21st birthday is Rs 12600.


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}


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i.e., 910 = 91d

or, d = 10

Therefore, a20= 10 + (20 1) 10 = 200, i.e.

20th term is 200.

} Example 3 : Find the sum of :

(i) the first 1000 positive integers (ii) the first n

positive integers

Solution :(i) Let S = 1 + 2 + 3 + . . . + 1000


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}


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Arithmetic Progression CBSE Class 10th

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