Arithmetic Progression
Mathematics Class Tenth
Inroduction & NCERT Exercise 5.1
A series of numbers in which next term is obtained by adding a fixed number to the preceding term, is called an Arithmetic Progression. An Arithmetic Progression is written as `AP` in short
Example: `a, a+d, a+2d, a+3d,` `----, a+nd`
First Term: In the given, example of series, `a` is called the 1st term.
Last Term: Last number (term) of the series is called the last term.
In the given example of series, `(a+nd)` is the last term.
Common Difference: In the given example of series, difference between next term and preceding term is called the COMMON DIFFERENCE.
Example: Common difference `=((a+d)-a), ((a+2d)-(a+d))`, and so on
nth term of an Arithmetic Progression: nth term is generally denoted by `a_n`
∴ `a_n = a+(n-1)d`
Where, `a_n =` nth term`
`a` = first term
`n` = number of term
`d` = common difference
Sum of first n terms of an Arithmetic Progression (AP): Sum of first n term is generally denoted by letter `S`
`S=n/2[2a+(n-1)d]`
Where, `S` = Sum of first n terms
`n`= number of terms
`a`= First term
`d` = Common difference
Sum of all terms in an Arithmetic Progression (AP)
`S=n/2(a+l)`
Where, `S` = Sum of all terms
`n` = number of terms
`a` = First term
`l` = Last term
NCERT Exercise 5.1
Question (1) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
Solution:Given, Taxi Fare for 1st Km = Rs 15
And, fare for additional km = Rs 8
Thus, Fare for 1st km = Rs 15
∴ Fare for two (2) km = Rs 15 + Rs 8 = Rs 23
∴ Fare for three (3) km = Rs 23 + Rs 8 = Rs 31
∴ Fare for four (4) km = Rs 31 + Rs 8 = Rs 39
∴ Fare for five (5) km = Rs 39 + Rs 8 = Rs 47
................ And so on
Series created for km: 15, 23, 31, 39, 47, ......
Here, second term is obtained by addition of a fix number (8) to the preceding number, thus, list of numbers formed is in Arithmetic Progression (AP)
Question 1. (ii) The amount of air present in a cylinder when a vacuum pump removes `1/4` of the air remaining in the cylinder at a time
Solution:
Given, Total amount of air = 1
Vacuum pump removes `1/4` of air at a time
Air present, in cylinder = 1
Air left, in cylinder after first removal `=1-1/4=3/4`
Quantity of air removed in second step `=3/4xx1/4=3/16`
∴ Air left after seond removal `=3/4-3/16` `=(12-3)/16=9/16`
Quantity of air removed in third step `=9/16xx1/4=9/64`
∴ Air left after thrid removal `= 9/16-9/64` `=(36-9)/64=27/64`
......... And so on
List of numbers formed by the process of air removal is
`3/4, 9/16, 27/64` ....
Here difference between 2nd and 1st term
`=9/16-3/4` `=(9-12)/16=-3/16`
Difference between 3rd and 2nd term
`27/64-9/16 ` `=(27-36)/64=-9/64`
Since, differences between next and preceding terms are not common, i.e. this means next term of the list of numbers is not obtained by adding a fixed number, thus list of numbers formed after the process is not in Arithmetic Progression (AP)
Question 1 (iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.
Solution:
Cost of digging for first meter = Rs 150
Cost of digging for first 2 (two) meter = Rs 150 + 50 = 200
Cost of digging for first 3 (three) meter = Rs 200 + 50 = 250
Cost of digging for first 4 (four) meter = Rs 250 + 50 = 300
Cost of digging for first 5 (five) meter = Rs 300 + 50 = 350
........... And so on
List of number formed after above process
150, 200, 250, 300, 350, .....
Here common difference = 200 – 150 = 250 – 200 = 300 – 250 = .. = 50
Here, it is clear that next term is obtined after adding a fixed number (50) to the preceding term, thus, list formed in the given process is in Arithmetic Progression (AP)
Question 1 (iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Solution:
Given, Rs. deposited = 10000
Rate of interest per annum `=8%`
Amount in account in the begining of the year = Rs 10000
Interest @ 8 % at the end of 1st year
`=10000 xx 8/100=800`
∴ Total amount at the end of 1st year
= 10000 + 800 = 10800
Amount at the begining of 2nd year
= 10800 + 10000 = 20800
Interest on amount at the end of 2nd year
`20800 xx 8/100=1664`
Total amount at the end of 2nd year
` = 20800 + 1664 = 22464`
Amount at the begining of 3rd year = Rs 22464 + 10000= 32464
Interst at the end of 3rd year
`=32464 xx 8/100=2597.12`
∴ Total amount at the end of 3rd year
`=32464 + 2597.12 = 35061.12
......... and so on
List of numbers formed by the above process
=10000, 20800, 32464, 35061.12, ...
It is clear from the above list of numbers, that the next term can not be obtained by adding a fixed number to the preceding one, and hence list of numbers formed by the process given in the question is not in Arithmetic Progression (AP)
Question (2) Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) ` a= 10,\ d=10`
Solution:
We know that,
∴ `a_n = a+(n-1)d`
Where, `a_n =` nth term`
`a` = first term
`n` = number of term
`d` = common difference
First Term
`:. a_1 = 10`
Second Term
`a_2 = 10 + (2-1)xx10`
`= 10+10 = 20
`:. a_2 = 20`
Third Term
`a_3 = 10 + (3-1)xx10`
`= 10 + 2xx10 = 10+20`
`:. a_3 = 30`
Fourth Term
`a_4 = 10 + (4-1)xx10`
`=10+3xx10= 10+30=40`
`:.a_4 = 40`
∴ First four terms are
10, 20, 30, and 40 Answer
Question. 2 (ii) `a=-2,\ d=0`
Solution:
Here, since common difference, `d=0`
∴ `a_1 = -2, a_2 = -2, a_3 = -2` and `a_4 = -2`
All of the first four terms are equal to `-2`
Question 2. (iii) `a=4,\ d = -3`
Solution:
We know that, for an Arithmetic Progression
∴ `a_n = a+(n-1)d`
Where, `a_n =` nth term`
`a` = first term
`n` = number of term
`d` = common difference
First Term
`a_1 = 4`
Second Term
`a_2 = 4+(2-1)xx(-3)`
`=>a_2= 4-2 = 1`
Third Term
`a_3 = 4+(3-1)xx(-3)`
`= 4+(-6) = 4-6`
`=>a_3 = -2`
Fourth Term
`a_4 = 4 + (4-1)xx(-3)`
`=4+(-9)=4-9`
`a_4 = -5`
∴ `4, 1, -2` and `-5` are the first four terms Answer
Question. 2 (iv) `a=-1,\ d=1/2`
Solution:
We know that, for an Arithmetic Progression
∴ `a_n = a+(n-1)d`
Where, `a_n =` nth term`
`a` = first term
`n` = number of term
`d` = common difference
Here given, `a=-1,\ d=1/2`
∴ First Term
`a_1 = -1`
Second Term
`a_2 = -1 + (2-1)xx1/2`
`= -1 +1/2 = -1/2`
`:. a_2 = -1/2`
Third Term
`a_3 = -1 + (3-1)xx1/2`
`= -1 + 2xx1/2 = -1+1`
`=>a_3 = 0`
Fourth Term
`a_4 = -1+(4-1)xx1/2`
`=-1+3xx1/2=(-2+3)/2`
`a_4 = 1/2`
∴ `-1, -1/2, 0, 1/2` are the first four term of Arithmetic Progression for the given condition in question
Question 2 (v) `a=-1.25,\ d=-0.25`
Solution:
We know that, for an Arithmetic Progression
∴ `a_n = a+(n-1)d`
Where, `a_n =` nth term`
`a` = first term
`n` = number of term
`d` = common difference
Given, `a=-1.25,\ d=-0.25`
First Term `a_1 = -1.25`
Second Term
`a_2=-1.25+(2-1)xx(-0.25)`
`= -1.25 + (-0.25)`
`=-1.25-0.25= -1.50`
`:. a_2 = -1.50`
Third Term
`a_3 = -1.25+(3-1)xx(-0.25)`
`=-1.25+(2xx(-0.25))`
`=-1.25 - 0.5=-1.75`
`:. a_3 = -1.75`
Fourth Term
`a_4 = -1.25+(4-1)xx(-0.25)`
`=-1.25+[3xx(-0.25)]`
`=-1.25+(-0.75)=-2.0`
`:. a_4 = -2`
∴ `-1.25, -1.5, -1.75, 2` are the first four terms Answer
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