Question1
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 `15 for the first km and `8 for each additional km.
(ii) The amount of air present in a cylinder when a vacum pump removes of the air remaining in the cylinder at a time.
(iii)The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre.
(iv) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum.
Solution:
(i) The taxi fare after each km when the fare is `15 for the first km and `8 for each additional km.
The taxi fare will be 15, 15 + 8, 15 + 16, 15 + 24, 15 + 32, 15 + 40 etc.
(i.e.,) 15, 23, 31, 39, 47, 55 etc.
This is an AP as every succeeding term is obtained by adding 8 in its preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
Let amount of air present in the cylinder be "a".
The vacuum removes of air remaining in the cylinder.
âˆ´ The air present in the cylinder will be a, a  = , ==,
a, , , â€¦
This is not an AP because the difference between second term and first term is not the same as the difference between the third term and the second term.
(iii) The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre.
The cost of digging the well will be `150, `(150 + 50), `(150 + 100), `(150 + 150) etc.
(i.e.,) `150, `200, `250, `300, `350 etc
This is an AP as every succeeding term is obtained by adding `50 in its preceding term.
(iv) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum.
Initial amount of money = `10,000
Compound Interest = 8% p.a
Amount every year increase by 8% Compound Interest p.a.
In the second year amount increases to 10,000
In the 3^{rd} year amount increases to 10,000
In the 4^{th} year amount increases to 10,000
This is not an AP as the difference is not the same in this problem.
(i) The taxi fare after each km when the fare is `15 for the first km and `8 for each additional km.
(ii) The amount of air present in a cylinder when a vacum pump removes of the air remaining in the cylinder at a time.
(iii)The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre.
(iv) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum.
Solution:
(i) The taxi fare after each km when the fare is `15 for the first km and `8 for each additional km.
The taxi fare will be 15, 15 + 8, 15 + 16, 15 + 24, 15 + 32, 15 + 40 etc.
(i.e.,) 15, 23, 31, 39, 47, 55 etc.
This is an AP as every succeeding term is obtained by adding 8 in its preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
Let amount of air present in the cylinder be "a".
The vacuum removes of air remaining in the cylinder.
âˆ´ The air present in the cylinder will be a, a  = , ==,
a, , , â€¦
This is not an AP because the difference between second term and first term is not the same as the difference between the third term and the second term.
(iii) The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre.
The cost of digging the well will be `150, `(150 + 50), `(150 + 100), `(150 + 150) etc.
(i.e.,) `150, `200, `250, `300, `350 etc
This is an AP as every succeeding term is obtained by adding `50 in its preceding term.
(iv) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum.
Initial amount of money = `10,000
Compound Interest = 8% p.a
Amount every year increase by 8% Compound Interest p.a.
In the second year amount increases to 10,000
In the 3^{rd} year amount increases to 10,000
In the 4^{th} year amount increases to 10,000
This is not an AP as the difference is not the same in this problem.
Question2
Write first four terms of the A.P, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10 (ii) a = 2, d = 0 (iii) a = 4, d = 3 (iv) a = 1, d =
(v) a = 1.25, d = 0.25
Solution:
(i) a = 10, d = 10
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 10
a_{2 }= a_{1} + d = 10 + 10 = 20
a_{3} = a_{2} + d = 20 + 10 = 30
a_{4 }= a_{3} + d_{ }= 30 + 10 = 40
âˆ´ The first four terms of the A.P are 10, 20, 30, 40.
(ii) a = 2, d = 0
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 2
a_{2 }= a_{1} + d = 2 + 0 = 2
a_{3} = a_{2} + d = 2 + 0 = 2
a_{4 }= a_{3} + d_{ }= 2 + 0 = 2
Thus first four terms of the A.P are 2, 2, 2, 2
(iii) a = 4, d = 3
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 4
a_{2 }= a_{1} + d = 4 + (3) = 1
a_{3} = a_{2} + d = 1 + (3) = 2
a_{4 }= a_{3} + d_{ }= 2 + (3) = 5
The first 4 terms of the A.P are 4, 1, 2, 5
(iv) a = 1, d =
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 1
a_{2 }= a_{1} + d = 1 + = 
a_{3} = a_{2} + d =  + = 0
a_{4 }= a_{3} + d_{ }= 0 + =
The first 4 terms of the A.P are â€“1, , 0, .
(v) a = 1.25, d = 0.25
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 1.25
a_{2 }= a_{1} + d = 1.25 + (0.25) = 1.50
a_{3} = a_{2} + d = 1.50 + (0.25) = 1.75
a_{4 }= a_{3} + d_{ }= 1.75 + (0.25) = 2.0
The first 4 terms of the A.P are 1.25, 1.50, 1.75, 2.0.
(i) a = 10, d = 10 (ii) a = 2, d = 0 (iii) a = 4, d = 3 (iv) a = 1, d =
(v) a = 1.25, d = 0.25
Solution:
(i) a = 10, d = 10
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 10
a_{2 }= a_{1} + d = 10 + 10 = 20
a_{3} = a_{2} + d = 20 + 10 = 30
a_{4 }= a_{3} + d_{ }= 30 + 10 = 40
âˆ´ The first four terms of the A.P are 10, 20, 30, 40.
(ii) a = 2, d = 0
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 2
a_{2 }= a_{1} + d = 2 + 0 = 2
a_{3} = a_{2} + d = 2 + 0 = 2
a_{4 }= a_{3} + d_{ }= 2 + 0 = 2
Thus first four terms of the A.P are 2, 2, 2, 2
(iii) a = 4, d = 3
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 4
a_{2 }= a_{1} + d = 4 + (3) = 1
a_{3} = a_{2} + d = 1 + (3) = 2
a_{4 }= a_{3} + d_{ }= 2 + (3) = 5
The first 4 terms of the A.P are 4, 1, 2, 5
(iv) a = 1, d =
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 1
a_{2 }= a_{1} + d = 1 + = 
a_{3} = a_{2} + d =  + = 0
a_{4 }= a_{3} + d_{ }= 0 + =
The first 4 terms of the A.P are â€“1, , 0, .
(v) a = 1.25, d = 0.25
Let a_{1}, a_{2}, a_{3}, a_{4} be the first four terms of the A.P
Here a = a_{1 }= 1.25
a_{2 }= a_{1} + d = 1.25 + (0.25) = 1.50
a_{3} = a_{2} + d = 1.50 + (0.25) = 1.75
a_{4 }= a_{3} + d_{ }= 1.75 + (0.25) = 2.0
The first 4 terms of the A.P are 1.25, 1.50, 1.75, 2.0.
Question3
For the following A.Ps, write the first term and the common difference:
(i) 3, 1, 1, 3, â€¦ (ii) â€“5, 1, 3, 7, â€¦ (iii) â€¦ (iv) 0.6, 1.7, 2.8, 3.9, â€¦
Solution:
(i) The first term is 3 and common difference = a_{2} â€“ a_{1}= 1  3 = 2.
(ii) The first term is 5 and common difference = a_{2} â€“ a_{1}= (1) â€“ (5) = 4.
(iii) The first term is and common difference = a_{2} â€“ a_{1}=  = .
(iv) The first term is 0.6 and common difference = a_{2} â€“ a_{1}= (1.7) â€“ (0.6) = 1.1.
(i) 3, 1, 1, 3, â€¦ (ii) â€“5, 1, 3, 7, â€¦ (iii) â€¦ (iv) 0.6, 1.7, 2.8, 3.9, â€¦
Solution:
(i) The first term is 3 and common difference = a_{2} â€“ a_{1}= 1  3 = 2.
(ii) The first term is 5 and common difference = a_{2} â€“ a_{1}= (1) â€“ (5) = 4.
(iii) The first term is and common difference = a_{2} â€“ a_{1}=  = .
(iv) The first term is 0.6 and common difference = a_{2} â€“ a_{1}= (1.7) â€“ (0.6) = 1.1.
Question4
Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, â€¦ (ii) 2, , â€¦ (iii) â€“1.2, 3.2, 5.2, 7.2, â€¦
(iv) â€“10, 6, 2, 2, â€¦ (v) 3, 3 +, 3 + 2, 3 + 3, â€¦
Solution:
(i) 2, 4, 8, 16, â€¦
Here a_{2} â€“ a_{1} = 4 â€“ 2 = 2
a_{3} â€“ a_{2} = 8 â€“ 4 = 4
a_{4} â€“ a_{3} = 16 â€“ 8 = 8
Since a_{k+1}  a_{k} is not the same throughout, it is not an A.P.
(ii) 2, , â€¦
Here
a_{2} â€“ a_{1} = â€“ 2=
a_{3} â€“ a_{2} = 3 â€“ =
a_{4} â€“ a_{3} = â€“ 3 =
As a_{k+1}  a_{k} is the same throughout, it is an A.P with first term = 2 and common difference =
The next three terms are 4, 4 + = , + = = 5.
(iii) â€“1.2, 3.2, 5.2, 7.2, â€¦
Here
a_{2} â€“ a_{1} = 3.2 â€“ (1.2) = 3.2 + 1.2 = 2
a_{3} â€“ a_{2} = 5.2 â€“ (3.2)= 2
a_{4} â€“ a_{3} = 7.2 â€“ (5.2)= 2
Since a_{k+1} â€“ a_{k} is the same throughout it is an A.P with first term (a) = 1. 2 and common difference, (d) = 2
Thus the next three terms are
a_{5} = a_{4} + d = â€“7.2 â€“ 2 = 9.2
a_{6} = a_{5} + d = 9.2 â€“ 2 = 11.2
a_{7} = a_{6} + d =  11.2 â€“ 2 = 13.2
(iv) â€“10, 6, 2, 2, â€¦
Here
a_{2} â€“ a_{1} = 6 â€“ (10)= 4
a_{3} â€“ a_{2} = 2 â€“ (6) = 4
a_{4} â€“ a_{3} = 2 â€“ (2)= 4
As a_{k+1} â€“ a_{k} is the same throughout it is an A.P with 1^{st} term as 10 and common difference as 4.
Thus the next three terms are
a_{5} = a_{4 }+ d = 2 + 4 = 6
a_{6} = a_{5} + d = 6 + 4 = 10
a_{7} = a_{6} + d = 10 + 4 = 14
(v) 3, 3 +, 3 + 2, 3 + 3, â€¦
Here a_{2} â€“ a_{1} = 3 +  3 =
a_{3} â€“ a_{2} = 3 + 2 (3 +) =
a_{4} â€“ a_{3} = 3 + 3 (3 + 2) =
In this problem, a_{k+1} â€“a_{k} is the same throughout thus the given list is an A.P with 1^{st} term as 3 and common difference (d) as .
Thus the next three terms are
a_{5} = a_{4 }+ d = (3 + 3) += 3 + 4
a_{6} = a_{5} + d = (3 + 4) + = 3 + 5
a_{7} = a_{6} + d = (3 + 5) += 3 + 6
(i) 2, 4, 8, 16, â€¦ (ii) 2, , â€¦ (iii) â€“1.2, 3.2, 5.2, 7.2, â€¦
(iv) â€“10, 6, 2, 2, â€¦ (v) 3, 3 +, 3 + 2, 3 + 3, â€¦
Solution:
(i) 2, 4, 8, 16, â€¦
Here a_{2} â€“ a_{1} = 4 â€“ 2 = 2
a_{3} â€“ a_{2} = 8 â€“ 4 = 4
a_{4} â€“ a_{3} = 16 â€“ 8 = 8
Since a_{k+1}  a_{k} is not the same throughout, it is not an A.P.
(ii) 2, , â€¦
Here
a_{2} â€“ a_{1} = â€“ 2=
a_{3} â€“ a_{2} = 3 â€“ =
a_{4} â€“ a_{3} = â€“ 3 =
As a_{k+1}  a_{k} is the same throughout, it is an A.P with first term = 2 and common difference =
The next three terms are 4, 4 + = , + = = 5.
(iii) â€“1.2, 3.2, 5.2, 7.2, â€¦
Here
a_{2} â€“ a_{1} = 3.2 â€“ (1.2) = 3.2 + 1.2 = 2
a_{3} â€“ a_{2} = 5.2 â€“ (3.2)= 2
a_{4} â€“ a_{3} = 7.2 â€“ (5.2)= 2
Since a_{k+1} â€“ a_{k} is the same throughout it is an A.P with first term (a) = 1. 2 and common difference, (d) = 2
Thus the next three terms are
a_{5} = a_{4} + d = â€“7.2 â€“ 2 = 9.2
a_{6} = a_{5} + d = 9.2 â€“ 2 = 11.2
a_{7} = a_{6} + d =  11.2 â€“ 2 = 13.2
(iv) â€“10, 6, 2, 2, â€¦
Here
a_{2} â€“ a_{1} = 6 â€“ (10)= 4
a_{3} â€“ a_{2} = 2 â€“ (6) = 4
a_{4} â€“ a_{3} = 2 â€“ (2)= 4
As a_{k+1} â€“ a_{k} is the same throughout it is an A.P with 1^{st} term as 10 and common difference as 4.
Thus the next three terms are
a_{5} = a_{4 }+ d = 2 + 4 = 6
a_{6} = a_{5} + d = 6 + 4 = 10
a_{7} = a_{6} + d = 10 + 4 = 14
(v) 3, 3 +, 3 + 2, 3 + 3, â€¦
Here a_{2} â€“ a_{1} = 3 +  3 =
a_{3} â€“ a_{2} = 3 + 2 (3 +) =
a_{4} â€“ a_{3} = 3 + 3 (3 + 2) =
In this problem, a_{k+1} â€“a_{k} is the same throughout thus the given list is an A.P with 1^{st} term as 3 and common difference (d) as .
Thus the next three terms are
a_{5} = a_{4 }+ d = (3 + 3) += 3 + 4
a_{6} = a_{5} + d = (3 + 4) + = 3 + 5
a_{7} = a_{6} + d = (3 + 5) += 3 + 6
Question5
Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms.
(i) 0.2, 0.22, 0.222, 0.222, â€¦ (ii) 0, 4, 8, 12, â€¦ (iii), â€¦ (iv) 1, 3, 9, 27, â€¦ (v) a, 2a, 3a, 4a, â€¦
Solution:
(i) 0.2, 0.22, 0.222, 0.2222, â€¦
Here,
a_{2} â€“ a_{1} = 0.22 â€“ 0.2 = 0.02
a_{3} â€“ a_{2} = 0.222 â€“ 0.22 = 0.002
a_{4} â€“ a_{3} = 0.2222 â€“ 0.222 = 0.0002
Hence a_{k+1} â€“ a_{k} is not the same throughout, hence it is not an A.P.
(ii) 0, 4, 8, 12
Here
a_{2} â€“ a_{1} = 4 â€“ 0 = 4
a_{3} â€“ a_{2} = 8 â€“ (4) = 4
a_{4} â€“ a_{3} = 12 â€“ (8) = 4
Here a_{k+1} â€“ a_{k} is the same throughout hence it is an A.P with 1^{st} term = 0
and common difference = 4
Thus the next three terms are
a_{5} = a_{4 }+ d = 12 + (4) = 16
a_{6} = a_{5} + d = 16 + (4) = 20
a_{7} = a_{6} + d = 20 + (4) = 24
(iii) , â€¦
Here
a_{2} â€“ a_{1} =  = 0
a_{3} â€“ a_{2} = â€“ = 0
a_{4} â€“ a_{3} = â€“ = 0
Here a_{k+1} â€“ a_{k} = 0 is the same throughout hence the list is an A.P with 1^{st} term = and common difference d = 0.
Thus the next three terms are
a_{5} = a_{4 }+ d = + 0 =
a_{6} = a_{5} + d = + 0 =
a_{7} = a_{6} + d = + 0 = .
(iv) 1, 3, 9, 27, â€¦
Here the 1^{st} term (a) = 1
a_{2} â€“ a_{1} = 3 â€“ 1 = 2
a_{3} â€“ a_{2} = 9  3 = 6
a_{4} â€“ a_{3} = 27 â€“ 9 = 18
Since a_{k+1} â€“ a_{k} is not the same throughout this is not an A.P.
(v) a, 2a, 3a, 4a
Here
a_{2} â€“ a_{1} = 2a  a = a
a_{3} â€“ a_{1} = 3a â€“ 2a = a
a_{4} â€“ a_{3} = 4a â€“ 3a = a
Since a_{k+1} â€“ a_{1} is the same throughout this is a A.P with first term = a and common difference = a
Thus the next three terms are
a_{5} = a_{4 }+ d = 4a + a = 5a
a_{6} = a_{5} + d = 5a + a = 6a
a_{7} = a_{6} + d = 6a + a = 7a
(i) 0.2, 0.22, 0.222, 0.222, â€¦ (ii) 0, 4, 8, 12, â€¦ (iii), â€¦ (iv) 1, 3, 9, 27, â€¦ (v) a, 2a, 3a, 4a, â€¦
Solution:
(i) 0.2, 0.22, 0.222, 0.2222, â€¦
Here,
a_{2} â€“ a_{1} = 0.22 â€“ 0.2 = 0.02
a_{3} â€“ a_{2} = 0.222 â€“ 0.22 = 0.002
a_{4} â€“ a_{3} = 0.2222 â€“ 0.222 = 0.0002
Hence a_{k+1} â€“ a_{k} is not the same throughout, hence it is not an A.P.
(ii) 0, 4, 8, 12
Here
a_{2} â€“ a_{1} = 4 â€“ 0 = 4
a_{3} â€“ a_{2} = 8 â€“ (4) = 4
a_{4} â€“ a_{3} = 12 â€“ (8) = 4
Here a_{k+1} â€“ a_{k} is the same throughout hence it is an A.P with 1^{st} term = 0
and common difference = 4
Thus the next three terms are
a_{5} = a_{4 }+ d = 12 + (4) = 16
a_{6} = a_{5} + d = 16 + (4) = 20
a_{7} = a_{6} + d = 20 + (4) = 24
(iii) , â€¦
Here
a_{2} â€“ a_{1} =  = 0
a_{3} â€“ a_{2} = â€“ = 0
a_{4} â€“ a_{3} = â€“ = 0
Here a_{k+1} â€“ a_{k} = 0 is the same throughout hence the list is an A.P with 1^{st} term = and common difference d = 0.
Thus the next three terms are
a_{5} = a_{4 }+ d = + 0 =
a_{6} = a_{5} + d = + 0 =
a_{7} = a_{6} + d = + 0 = .
(iv) 1, 3, 9, 27, â€¦
Here the 1^{st} term (a) = 1
a_{2} â€“ a_{1} = 3 â€“ 1 = 2
a_{3} â€“ a_{2} = 9  3 = 6
a_{4} â€“ a_{3} = 27 â€“ 9 = 18
Since a_{k+1} â€“ a_{k} is not the same throughout this is not an A.P.
(v) a, 2a, 3a, 4a
Here
a_{2} â€“ a_{1} = 2a  a = a
a_{3} â€“ a_{1} = 3a â€“ 2a = a
a_{4} â€“ a_{3} = 4a â€“ 3a = a
Since a_{k+1} â€“ a_{1} is the same throughout this is a A.P with first term = a and common difference = a
Thus the next three terms are
a_{5} = a_{4 }+ d = 4a + a = 5a
a_{6} = a_{5} + d = 5a + a = 6a
a_{7} = a_{6} + d = 6a + a = 7a
Question6
Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms.
(i) a, a^{2}, a^{3}, a^{4}, â€¦ (ii) ,,, , â€¦
(iii) , â€¦ (iv) 1^{2}, 3^{2}, 5^{2}, 7^{2}, â€¦ (v) 1^{2}, 5^{2}, 7^{2}, 73, â€¦
Solution:
(i) a, a^{2}, a^{3}, a^{4}, â€¦
Here
a_{2} â€“ a_{1} = a^{2}  a = a(a â€“ 1)
a_{3} â€“ a_{2} = a^{3} â€“ a^{2} = a^{2}(a â€“ 1)
a_{4} â€“ a_{3} = a^{4} â€“ a^{3} = a^{3}(a â€“ 1)
Since a_{k+1} â€“ a_{k} is not the same throughout this is not an A.P.
(ii) ,,,, â€¦
Here
a_{2} â€“ a_{1} =  = 2  =
a_{3} â€“ a_{2} =  = 3  2 =
a_{4} â€“ a_{3} =  = 4  3 =
Since a_{k+1} â€“ a_{k} is the same throughout it is an A.P with the first term as and common difference d =.
Thus the next three terms are
a_{5} = a_{4 }+ d = + = 4 + = 5
a_{6} = a_{5} + d = 5 + = 6
a_{7} = a_{6} + d = 6 + = 7.
(iii)
Here
a_{2} â€“ a_{1} = = (  1)
a_{3} â€“ a_{2} = = 3 = ( )
a_{4} â€“ a_{3} = = 2  3 = (2  ).
Here a_{k+1} â€“ a_{k} is not the same throughout hence the list is not an A.P.
(iv) 1^{2}, 3^{2}, 5^{2}, 7^{2}
Here
a_{2} â€“ a_{1} = 3^{2} â€“ 1^{2} = 9 â€“ 1 = 8
a_{3} â€“ a_{2} = 5^{2} â€“ 3^{2} = 25 â€“ 9 = 16
a_{4} â€“ a_{3} = 7^{2} â€“ 5^{2} = 49 â€“ 25 = 24
Since a_{k+1} â€“ a_{k} is not the same throughout hence the list not an A.P.
(v) 1^{2}, 5^{2}, 7^{2}, 73, â€¦
Here
a_{2} â€“ a_{1} = 5^{2} â€“ 1^{2} = 25 â€“ 1 = 24
a_{3} â€“ a_{2} = 7^{2} â€“ 5^{2} = 49 â€“ 25 = 24
a_{4} â€“ a_{3} = 73 â€“ 7^{2} = 73 â€“ 49 = 24
Since a_{k+1} â€“ a_{k} = 24 is the same throughout this list is an A.P with first term = 1^{2} and common difference = 24.
Thus the next three terms are
a_{5} = a_{4 }+ d = 73 + 24 = 97
a_{6} = a_{5} + d = 97 + 24 = 121
a_{7} = a_{6} + d = 121 + 24 = 145.
(i) a, a^{2}, a^{3}, a^{4}, â€¦ (ii) ,,, , â€¦
(iii) , â€¦ (iv) 1^{2}, 3^{2}, 5^{2}, 7^{2}, â€¦ (v) 1^{2}, 5^{2}, 7^{2}, 73, â€¦
Solution:
(i) a, a^{2}, a^{3}, a^{4}, â€¦
Here
a_{2} â€“ a_{1} = a^{2}  a = a(a â€“ 1)
a_{3} â€“ a_{2} = a^{3} â€“ a^{2} = a^{2}(a â€“ 1)
a_{4} â€“ a_{3} = a^{4} â€“ a^{3} = a^{3}(a â€“ 1)
Since a_{k+1} â€“ a_{k} is not the same throughout this is not an A.P.
(ii) ,,,, â€¦
Here
a_{2} â€“ a_{1} =  = 2  =
a_{3} â€“ a_{2} =  = 3  2 =
a_{4} â€“ a_{3} =  = 4  3 =
Since a_{k+1} â€“ a_{k} is the same throughout it is an A.P with the first term as and common difference d =.
Thus the next three terms are
a_{5} = a_{4 }+ d = + = 4 + = 5
a_{6} = a_{5} + d = 5 + = 6
a_{7} = a_{6} + d = 6 + = 7.
(iii)
Here
a_{2} â€“ a_{1} = = (  1)
a_{3} â€“ a_{2} = = 3 = ( )
a_{4} â€“ a_{3} = = 2  3 = (2  ).
Here a_{k+1} â€“ a_{k} is not the same throughout hence the list is not an A.P.
(iv) 1^{2}, 3^{2}, 5^{2}, 7^{2}
Here
a_{2} â€“ a_{1} = 3^{2} â€“ 1^{2} = 9 â€“ 1 = 8
a_{3} â€“ a_{2} = 5^{2} â€“ 3^{2} = 25 â€“ 9 = 16
a_{4} â€“ a_{3} = 7^{2} â€“ 5^{2} = 49 â€“ 25 = 24
Since a_{k+1} â€“ a_{k} is not the same throughout hence the list not an A.P.
(v) 1^{2}, 5^{2}, 7^{2}, 73, â€¦
Here
a_{2} â€“ a_{1} = 5^{2} â€“ 1^{2} = 25 â€“ 1 = 24
a_{3} â€“ a_{2} = 7^{2} â€“ 5^{2} = 49 â€“ 25 = 24
a_{4} â€“ a_{3} = 73 â€“ 7^{2} = 73 â€“ 49 = 24
Since a_{k+1} â€“ a_{k} = 24 is the same throughout this list is an A.P with first term = 1^{2} and common difference = 24.
Thus the next three terms are
a_{5} = a_{4 }+ d = 73 + 24 = 97
a_{6} = a_{5} + d = 97 + 24 = 121
a_{7} = a_{6} + d = 121 + 24 = 145.
Question7
Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n }the n^{th} term of the A.P:
a 
d 
n 
a_{n} 

(i) (ii) (iii) (iv) (v) 
7 18 â€¦. 18.9 3.5 
3 â€¦. 3 2.5 0 
8 10 18 â€¦ 105 
... 0 5 3.6 ... 
Solution:
(i) a = 7, d = 3, n = 8, a_{n} = ?
a_{n} = a + (n â€“ 1) Ã— d = 7 + (8  1) Ã— 3
= 7 + 7 Ã— 3 = 7 + 21 = 28.
(ii) a = 18
n = 10, a_{n} = 0, d = ?
a_{n} = a + (n â€“ 1) Ã— d
0 = 18 + (10 â€“ 1) Ã— d
0 = 18 + 9 Ã— d
0 = 18 + 9d
9d = 18
d = 2.
(iii) a = ?, d = 3, n = 18, a_{n} = 5
a_{n} = a + (n â€“ 1) Ã— d
5 = a + (18 â€“ 1) Ã— (3)
5 = a + 17 Ã— (3)
5 = a  51
a = 51 â€“ 5 = 46.
(iv) a = 18.9
d = 2.5, a_{n }= 3.6, n = ?
a_{n} = a + (n â€“ 1) Ã— d
3.6 = 18.9 +(n â€“ 1) Ã— 2.5
(n â€“ 1) Ã— 2.5 = 3.6 + 18.9
(n â€“ 1) = = ( To remove the decimal multiply and divide by 10)
â‡’ n â€“1 = 9
âˆ´ n = 10.
(v) a = 3.5, d = 0, n = 105
a_{n} = ?
a_{n} = a + (n â€“ 1) Ã— d
= 3.5 +(105 â€“ 1) Ã— 0
a_{n} = 3.5.
(i) a = 7, d = 3, n = 8, a_{n} = ?
a_{n} = a + (n â€“ 1) Ã— d = 7 + (8  1) Ã— 3
= 7 + 7 Ã— 3 = 7 + 21 = 28.
(ii) a = 18
n = 10, a_{n} = 0, d = ?
a_{n} = a + (n â€“ 1) Ã— d
0 = 18 + (10 â€“ 1) Ã— d
0 = 18 + 9 Ã— d
0 = 18 + 9d
9d = 18
d = 2.
(iii) a = ?, d = 3, n = 18, a_{n} = 5
a_{n} = a + (n â€“ 1) Ã— d
5 = a + (18 â€“ 1) Ã— (3)
5 = a + 17 Ã— (3)
5 = a  51
a = 51 â€“ 5 = 46.
(iv) a = 18.9
d = 2.5, a_{n }= 3.6, n = ?
a_{n} = a + (n â€“ 1) Ã— d
3.6 = 18.9 +(n â€“ 1) Ã— 2.5
(n â€“ 1) Ã— 2.5 = 3.6 + 18.9
(n â€“ 1) = = ( To remove the decimal multiply and divide by 10)
â‡’ n â€“1 = 9
âˆ´ n = 10.
(v) a = 3.5, d = 0, n = 105
a_{n} = ?
a_{n} = a + (n â€“ 1) Ã— d
= 3.5 +(105 â€“ 1) Ã— 0
a_{n} = 3.5.
Question8
Choose the correct choice in the following and justify:
(i) 30^{th} term of the A.P: 10, 7, 4, â€¦, is
(A) 97 (B) 77 (C) â€“77 (D) â€“ 87
(ii) 11^{th} term of the A.P: 3, â€“ , 2, â€¦, is
(A) 28 (B)22 (C) â€“38 (D) 48
Solution:
(i) 30^{th} term of the A.P: 10, 7, 4, â€¦, is
n^{th} term = a + (n â€“ 1)d, where n = 30
30^{th} term = 10 + (30 â€“ 1) Ã— (3)
= 10 + (29 Ã—  3)
= 10 â€“ 87 =  77
(C) is the correct answer.
(ii) 11^{th} term of the A.P: 3, â€“ , 2, â€¦, is
Here n = 11, a = 3
d = â€“â€“ (3)
= = = 2
a_{n} = a + (n â€“ 1)d
a_{n} = 3+ (n â€“ 1) Ã— (2)
= 3 +(11 â€“ 1) Ã— = 3 + 10 Ã—
=â€“ 3 + 25 = 22
a_{13} = 22.
(B) is the correct answer.
(i) 30^{th} term of the A.P: 10, 7, 4, â€¦, is
(A) 97 (B) 77 (C) â€“77 (D) â€“ 87
(ii) 11^{th} term of the A.P: 3, â€“ , 2, â€¦, is
(A) 28 (B)22 (C) â€“38 (D) 48
Solution:
(i) 30^{th} term of the A.P: 10, 7, 4, â€¦, is
n^{th} term = a + (n â€“ 1)d, where n = 30
30^{th} term = 10 + (30 â€“ 1) Ã— (3)
= 10 + (29 Ã—  3)
= 10 â€“ 87 =  77
(C) is the correct answer.
(ii) 11^{th} term of the A.P: 3, â€“ , 2, â€¦, is
Here n = 11, a = 3
d = â€“â€“ (3)
= = = 2
a_{n} = a + (n â€“ 1)d
a_{n} = 3+ (n â€“ 1) Ã— (2)
= 3 +(11 â€“ 1) Ã— = 3 + 10 Ã—
=â€“ 3 + 25 = 22
a_{13} = 22.
(B) is the correct answer.
Question9
In the following A.Ps, find the missing terms in the boxes:
(i) 2, ,26 (ii) , 13, , 3 (iii) 5, , , 9
(iv) 4,, , , , 6 (v) , 38, , , , 22
Solution:
(i) Let x be the missing term
x  2 = 26  x
2x = 28
x = 14
âˆ´ Missing term = 14
(ii) , 13, , 3
Let missing term in 1^{st} box be x, missing term in the 2^{nd} box be y,
13  x = y  13 = 3  y
Now considering the last two,
y  13 = 3  y
2y = 16y = 8.
The value of y = 8.
Here 13  x = y  13
â‡’13  x = 8  13
x = 26  8= 18.
(iii) 5, , , 9
Here a = 5, a_{4} = 9, n = 4, d = ?
a_{n} = a + (n 1) Ã— d
9 = 5 + (4  1) Ã— d
= 5 + 3d
3d =  5 = =
d = =
âˆ´ If a_{1} = 5
a_{2} = a_{1} + d = 5 + == = 6
a_{3} = 6 + = 8
(iv) 4, , , , , 6
a_{n} = 6
a = 4
d = ?
a_{n} = a + (n 1) Ã— d
6 = 4 + (6 1) Ã— d
6 = 4 + 5d
5d = 6 + 4 = 10
d = = 2
âˆ´ 4, 2, 0, 2, 4, 6 is the A.P series.
(v) , 38, , , , 22
a_{2} = 38, n = 6, a_{6} = 22
a_{n} = a + (n  1) Ã— d if n = 6,
22 = a + 5d â€¦â€¦. (1)
if n = 2,
38 = a + d â€¦â€¦â€¦. (2)
 (2) â‡’ 4d = 60
âˆ´ d = = 15.
Thus a_{2} = a + d = a + (15)
38 = a  15
a = 38 + 15 = 53.
a_{3} = 38 + (15) = 23
a_{4 }= 23 + (15) = 8
a_{5} = 8 + (15) = 7
âˆ´ 53, 38, 23, 8, 7, 22 is the A.P series
(i) 2, ,26 (ii) , 13, , 3 (iii) 5, , , 9
(iv) 4,, , , , 6 (v) , 38, , , , 22
Solution:
(i) Let x be the missing term
x  2 = 26  x
2x = 28
x = 14
âˆ´ Missing term = 14
(ii) , 13, , 3
Let missing term in 1^{st} box be x, missing term in the 2^{nd} box be y,
13  x = y  13 = 3  y
Now considering the last two,
y  13 = 3  y
2y = 16y = 8.
The value of y = 8.
Here 13  x = y  13
â‡’13  x = 8  13
x = 26  8= 18.
(iii) 5, , , 9
Here a = 5, a_{4} = 9, n = 4, d = ?
a_{n} = a + (n 1) Ã— d
9 = 5 + (4  1) Ã— d
= 5 + 3d
3d =  5 = =
d = =
âˆ´ If a_{1} = 5
a_{2} = a_{1} + d = 5 + == = 6
a_{3} = 6 + = 8
(iv) 4, , , , , 6
a_{n} = 6
a = 4
d = ?
a_{n} = a + (n 1) Ã— d
6 = 4 + (6 1) Ã— d
6 = 4 + 5d
5d = 6 + 4 = 10
d = = 2
âˆ´ 4, 2, 0, 2, 4, 6 is the A.P series.
(v) , 38, , , , 22
a_{2} = 38, n = 6, a_{6} = 22
a_{n} = a + (n  1) Ã— d if n = 6,
22 = a + 5d â€¦â€¦. (1)
if n = 2,
38 = a + d â€¦â€¦â€¦. (2)
 (2) â‡’ 4d = 60
âˆ´ d = = 15.
Thus a_{2} = a + d = a + (15)
38 = a  15
a = 38 + 15 = 53.
a_{3} = 38 + (15) = 23
a_{4 }= 23 + (15) = 8
a_{5} = 8 + (15) = 7
âˆ´ 53, 38, 23, 8, 7, 22 is the A.P series
Question10
Which term of the A.P: 3,8, 13, 18, â€¦,is 78
Solution:
Let 78 be the n^{th} term
Here a = 3
d = 5, a_{n} = 78
a_{n} = a + (n â€“1) Ã— d
78 = 3 + (n â€“ 1) Ã— d
78 â€“ 3 = (n â€“1) Ã— 5
75 = (n â€“ 1) 5
= (n â€“ 1)
n â€“ 1 = 15
n = 16
78 is the 16^{th} term.
Solution:
Let 78 be the n^{th} term
Here a = 3
d = 5, a_{n} = 78
a_{n} = a + (n â€“1) Ã— d
78 = 3 + (n â€“ 1) Ã— d
78 â€“ 3 = (n â€“1) Ã— 5
75 = (n â€“ 1) 5
= (n â€“ 1)
n â€“ 1 = 15
n = 16
78 is the 16^{th} term.
Question11
Find the number of terms in each of the following A.Ps:
(i) 7, 13, 19, â€¦, 205 (ii) 18, 15, 13, â€¦, 47
Solution:
(i) Let the number of terms be n
Here a = 7, d = 6, a_{n} = 205
a_{n} = a + (n â€“ 1) d
205 = 7 + (n â€“ 1) Ã— 6
(n â€“ 1)6 = 205 â€“ 7
n â€“ 1 =
n = 33 +1 = 34
âˆ´ Number of terms = 34.
(ii) 18, 15, 13, â€¦, 47
Here a = 18
d = 18 â€“ 15 = 2
a_{n} = 47
Let the number of terms be n
a_{n} = a + (n â€“ 1) d
47 = 18 + (n â€“1) (2 )
47 â€“ 18 = (n â€“ 1) ()
65 Ã— = (n â€“ 1)
26 + 1 = n
âˆ´ Number of terms = 27
(i) 7, 13, 19, â€¦, 205 (ii) 18, 15, 13, â€¦, 47
Solution:
(i) Let the number of terms be n
Here a = 7, d = 6, a_{n} = 205
a_{n} = a + (n â€“ 1) d
205 = 7 + (n â€“ 1) Ã— 6
(n â€“ 1)6 = 205 â€“ 7
n â€“ 1 =
n = 33 +1 = 34
âˆ´ Number of terms = 34.
(ii) 18, 15, 13, â€¦, 47
Here a = 18
d = 18 â€“ 15 = 2
a_{n} = 47
Let the number of terms be n
a_{n} = a + (n â€“ 1) d
47 = 18 + (n â€“1) (2 )
47 â€“ 18 = (n â€“ 1) ()
65 Ã— = (n â€“ 1)
26 + 1 = n
âˆ´ Number of terms = 27
Question12
Check whether â€“150 is a term of the A.P: 11, 8, 5, 2 â€¦â€¦â€¦
Solution:
Here a = 11, d = 3
If a_{n} = 150 be the n^{th} term of the A.P
a_{n} = a + (n â€“ 1) Ã— d
150 = 11 + (n â€“ 1) Ã— (3)
150  11 = (n â€“ 1) Ã— (3)
= (n â€“ 1)
n = = 53 + 1
which is not a whole number
âˆ´ 150 is not a term of the A.P.
Solution:
Here a = 11, d = 3
If a_{n} = 150 be the n^{th} term of the A.P
a_{n} = a + (n â€“ 1) Ã— d
150 = 11 + (n â€“ 1) Ã— (3)
150  11 = (n â€“ 1) Ã— (3)
= (n â€“ 1)
n = = 53 + 1
which is not a whole number
âˆ´ 150 is not a term of the A.P.
Question13
Find the 31^{st}_{ }term of an A.P whose 11^{th} term is 38 and the 16^{th} term is 73.
Solution:
11^{th} term is 38 and
16^{th} term is 73
a_{11} = 38 = a + 10d â€¦â€¦â€¦â€¦â€¦.(1)
a_{16} = 73 = a +15d â€¦â€¦â€¦â€¦â€¦â€¦..(2)
a + 10d = 38
a + 15d = 73
â€“5d = â€“35
d = 7
sub d = 7 in (1)
a + 10 d = 38
a + 10(7) = 38
a = 70 + 38
=  32
Thus 31^{st} term = a + (31 â€“ 1) Ã— d
= â€“32 + 30 Ã— 7
= â€“ 32 + 210
= 178.
Solution:
11^{th} term is 38 and
16^{th} term is 73
a_{11} = 38 = a + 10d â€¦â€¦â€¦â€¦â€¦.(1)
a_{16} = 73 = a +15d â€¦â€¦â€¦â€¦â€¦â€¦..(2)
a + 10d = 38
a + 15d = 73
â€“5d = â€“35
d = 7
sub d = 7 in (1)
a + 10 d = 38
a + 10(7) = 38
a = 70 + 38
=  32
Thus 31^{st} term = a + (31 â€“ 1) Ã— d
= â€“32 + 30 Ã— 7
= â€“ 32 + 210
= 178.
Question14
An A.P consists of 50 terms of which 3^{rd }term is 12 and the last term is 106. Find the 29^{th }term.
Solution:
3^{rd} term is a_{3} = a + (3 â€“ 1)d = 12 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
50^{th} term is a_{50} = a + (50 â€“ 1)d = 106 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)
(1) â‡’ a + 2d = 12 â€¦â€¦â€¦â€¦â€¦â€¦.. (3)
(2) â‡’ a + 49d = 106 â€¦â€¦â€¦â€¦â€¦â€¦.. (4)

(3) â€“ (4) Ãž 47d = 94
d = 2
sub d = 2 in (3)
a + 2(2) = 12
a = 12 â€“ 4 = 8
âˆ´ 29^{th} term of an A.P is a + (29 â€“ 1)d â€¦.(5)
substituting for a = 8 and d = 2 in (5),
a + 28d = 8 + 28 Ã— 2
= 8 + 56
= 64.
Solution:
3^{rd} term is a_{3} = a + (3 â€“ 1)d = 12 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
50^{th} term is a_{50} = a + (50 â€“ 1)d = 106 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)
(1) â‡’ a + 2d = 12 â€¦â€¦â€¦â€¦â€¦â€¦.. (3)
(2) â‡’ a + 49d = 106 â€¦â€¦â€¦â€¦â€¦â€¦.. (4)

(3) â€“ (4) Ãž 47d = 94
d = 2
sub d = 2 in (3)
a + 2(2) = 12
a = 12 â€“ 4 = 8
âˆ´ 29^{th} term of an A.P is a + (29 â€“ 1)d â€¦.(5)
substituting for a = 8 and d = 2 in (5),
a + 28d = 8 + 28 Ã— 2
= 8 + 56
= 64.
Question15
If the 3^{rd} and the 9^{th} terms of an A.P are 4 and â€“8 respectively, which term of this A.P is zero?
Solution:
3^{rd} term of the A.P is a + (3 â€“1)d = 4
(i.e.,) a + 2d = 4 â€¦â€¦â€¦â€¦. (1)
9^{th} term of the A.P is a + (9 â€“ 1)d = 8
a + 8d = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)
a + 2d = 4â€¦â€¦.(1)
a + 8d = 8â€¦..(2)
(1) â€“ (2)â‡’ 6d = 12
d = 2
substituting d = 2 in (1) â‡’ a â€“ 4 = 4
a = 8
Let the n^{th }term be 0 then a + (n â€“ 1) d = 0
(i.e.,) 8 + (n â€“ 1) (2) = 0
2n + 2 = 8
2n = 8 â€“2= 10
â‡’ n = 5
Thus n = 5. Hence the 5^{th }term is zero.
Solution:
3^{rd} term of the A.P is a + (3 â€“1)d = 4
(i.e.,) a + 2d = 4 â€¦â€¦â€¦â€¦. (1)
9^{th} term of the A.P is a + (9 â€“ 1)d = 8
a + 8d = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)
a + 2d = 4â€¦â€¦.(1)
a + 8d = 8â€¦..(2)
(1) â€“ (2)â‡’ 6d = 12
d = 2
substituting d = 2 in (1) â‡’ a â€“ 4 = 4
a = 8
Let the n^{th }term be 0 then a + (n â€“ 1) d = 0
(i.e.,) 8 + (n â€“ 1) (2) = 0
2n + 2 = 8
2n = 8 â€“2= 10
â‡’ n = 5
Thus n = 5. Hence the 5^{th }term is zero.
Question16
The 17^{th} term of an A.P exceeds its 10^{th} term by 7. Find the common difference.
Solution:
17^{th} term of the A.P = a +(17 â€“ 1)d
= a + 16 d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
10^{th} term of the A.P = a + (10 â€“ 1)d = a + 9d  (2)
Given (1) exceeds (2) by 7
[a + 16d] â€“ [a + 9d] = 7
16d â€“ 9d = 7
7d = 7
d = 1
Solution:
17^{th} term of the A.P = a +(17 â€“ 1)d
= a + 16 d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
10^{th} term of the A.P = a + (10 â€“ 1)d = a + 9d  (2)
Given (1) exceeds (2) by 7
[a + 16d] â€“ [a + 9d] = 7
16d â€“ 9d = 7
7d = 7
d = 1
Question17
Which term of the A.P: 3, 15, 27, 39, â€¦ will be 132 more than its 54^{th }term?
Solution:
54^{th} term of the A.P is a + (54  1)d
Here a = 3
d = a_{2} â€“ a_{1} = 15 â€“ 3 = 12
âˆ´ 54^{th }term = 3 + (54 â€“ 1) Ã— 12
= 3 + 53 Ã— 12
= 3 + 636 = 639
132 more than 54^{th }term = 639 + 132 = 771
To find out which term of the A.P is 771, let us assume it is the n^{th} term
a + (n â€“ 1) Ã— d = 771
substituting for a = 3 and d = 12
3 + (n â€“ 1) Ã— 12 = 771
(n â€“ 1) Ã— 12 = 771  3
(n â€“ 1) = = 64
n = 65
Hence the 65^{th} term is 132 more than its 54^{th} term.
Solution:
54^{th} term of the A.P is a + (54  1)d
Here a = 3
d = a_{2} â€“ a_{1} = 15 â€“ 3 = 12
âˆ´ 54^{th }term = 3 + (54 â€“ 1) Ã— 12
= 3 + 53 Ã— 12
= 3 + 636 = 639
132 more than 54^{th }term = 639 + 132 = 771
To find out which term of the A.P is 771, let us assume it is the n^{th} term
a + (n â€“ 1) Ã— d = 771
substituting for a = 3 and d = 12
3 + (n â€“ 1) Ã— 12 = 771
(n â€“ 1) Ã— 12 = 771  3
(n â€“ 1) = = 64
n = 65
Hence the 65^{th} term is 132 more than its 54^{th} term.
Question18
Two A.Ps have the same common difference. The difference between their 100^{th} terms is 100, what is the difference between their 1000^{th} terms?
Solution:
Let a_{1} and a_{2} be the 1^{st }term of the 2 A.Ps and "d" be the common difference.
100^{th} term of 1^{st} A.P = a_{1} + (100 â€“ 1)d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)
100^{th} term of 2^{nd }A.P = a_{2} + (100 â€“ 1)d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)
Difference between (1) and (2) = 100
a_{1} + 99d â€“ (a_{2} + 99d) = 100
(i.e.,) a_{1} â€“ a_{2} = 100 â€¦â€¦â€¦â€¦â€¦.(3)
Difference between then 1000^{th} terms = (a_{1} + 999d) â€“ (a_{2} + 999d)
= a_{1} â€“ a_{2} = 100 (from (3))
Thus the difference between their 1000^{th }terms = 100.
Solution:
Let a_{1} and a_{2} be the 1^{st }term of the 2 A.Ps and "d" be the common difference.
100^{th} term of 1^{st} A.P = a_{1} + (100 â€“ 1)d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)
100^{th} term of 2^{nd }A.P = a_{2} + (100 â€“ 1)d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)
Difference between (1) and (2) = 100
a_{1} + 99d â€“ (a_{2} + 99d) = 100
(i.e.,) a_{1} â€“ a_{2} = 100 â€¦â€¦â€¦â€¦â€¦.(3)
Difference between then 1000^{th} terms = (a_{1} + 999d) â€“ (a_{2} + 999d)
= a_{1} â€“ a_{2} = 100 (from (3))
Thus the difference between their 1000^{th }terms = 100.
Question19
How many threedigit numbers are divisible by 7?
Solution:
The list of 3 digit numbers divisible by 7 are 105, 112, 119, â€¦, 994.
Solution:
The list of 3 digit numbers divisible by 7 are 105, 112, 119, â€¦, 994.
This is an A.P. with a = 105, d = 7
a_{n} = 994
âˆ´ 994 = a+(n â€“ 1) Ã— d
= 105+(n â€“ 1) Ã— 7
7(n â€“ 1) = 994 â€“ 105
= 889
n â€“ 1 = = 127
n = 128
Thus the number of threedigit numbers divisible by 7 are 128.
Question20
How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4 between 10 and 250 are 12, 16, 20, â€¦, 248.
Here
a = 12
d = 4
a_{n} = 248
âˆ´ 248 = 12 + (n â€“ 1) Ã— 4
(n â€“ 1) Ã— 4 = 248  12
(n â€“ 1) =
n = 59 + 1
= 60
âˆ´ Number of multiples of 4 between 10 and 250 = 60.
Solution:
Multiples of 4 between 10 and 250 are 12, 16, 20, â€¦, 248.
Here
a = 12
d = 4
a_{n} = 248
âˆ´ 248 = 12 + (n â€“ 1) Ã— 4
(n â€“ 1) Ã— 4 = 248  12
(n â€“ 1) =
n = 59 + 1
= 60
âˆ´ Number of multiples of 4 between 10 and 250 = 60.
Question21
For what value of n, are the n^{th} terms of two A.Ps: 63, 65, 67, â€¦, and 3, 10, 17, â€¦, equal?
Solution:
n^{th} term of the 1^{st} A.P = a + (n â€“1) Ã— d
Here a = 63, d = 2
â‡’ 63 + (n  1) Ã— 2 â€¦â€¦â€¦â€¦â€¦.(1)
n^{th} term of the 2^{nd} AP = a + (n â€“1) Ã— d
Here a = 3, d = 7
â‡’ 3 + (n â€“ 1) Ã— 7 â€¦â€¦â€¦â€¦â€¦â€¦.(2)
63 + ( n â€“ 1) Ã— 2 = 3 + (n â€“ 1) Ã— 7 (Given)
63 â€“ 3 = 7n â€“ 7 â€“ 2n + 2
60 + 5 = 5n
5n = 65
n = = 13.
13^{th} term of the 2 A.Ps are equal.
Solution:
n^{th} term of the 1^{st} A.P = a + (n â€“1) Ã— d
Here a = 63, d = 2
â‡’ 63 + (n  1) Ã— 2 â€¦â€¦â€¦â€¦â€¦.(1)
n^{th} term of the 2^{nd} AP = a + (n â€“1) Ã— d
Here a = 3, d = 7
â‡’ 3 + (n â€“ 1) Ã— 7 â€¦â€¦â€¦â€¦â€¦â€¦.(2)
63 + ( n â€“ 1) Ã— 2 = 3 + (n â€“ 1) Ã— 7 (Given)
63 â€“ 3 = 7n â€“ 7 â€“ 2n + 2
60 + 5 = 5n
5n = 65
n = = 13.
13^{th} term of the 2 A.Ps are equal.
Question22
Determine the A.P whose third term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.
Solution:
3^{rd} term = a + (3  1) Ã— d
= a + 2d = 16 â€¦â€¦â€¦â€¦â€¦. (1)
7^{th} term = a + 6d
5^{th} term = a + 4d
Given (a + 6d) â€“ (a + 4d) = 12
2d = 12 â‡’ d = 6
substitute d = 6 in (1)
a + 2(6) = 16
a + 12 = 16
a = 16 â€“ 12= 4.
a = 4, d = 6
âˆ´ The A.P is 4, 10, 16, â€¦
Solution:
3^{rd} term = a + (3  1) Ã— d
= a + 2d = 16 â€¦â€¦â€¦â€¦â€¦. (1)
7^{th} term = a + 6d
5^{th} term = a + 4d
Given (a + 6d) â€“ (a + 4d) = 12
2d = 12 â‡’ d = 6
substitute d = 6 in (1)
a + 2(6) = 16
a + 12 = 16
a = 16 â€“ 12= 4.
a = 4, d = 6
âˆ´ The A.P is 4, 10, 16, â€¦
Question23
Find the 20^{th} term from the last term of the A.P: 3, 8, 13, â€¦, 253.
Solution:
Here
a = 3, d = 5
Last term of the A.P is 253
Let 253 be the n^{th} term of A.P
n^{th} term = a + (n â€“ 1) Ã— d
= 3 + 5(n â€“ 1)
5n â€“ 5 = 250
5n = 255
n = 51
âˆ´ 253 is the 51^{st} term.
20^{th} term from the 51^{st} term is 32^{nd} term
32^{nd} term = a + (32 â€“ 1)d
= 3 + 31 Ã— 5
= 3 +155
= 158.
Solution:
Here
a = 3, d = 5
Last term of the A.P is 253
Let 253 be the n^{th} term of A.P
n^{th} term = a + (n â€“ 1) Ã— d
= 3 + 5(n â€“ 1)
5n â€“ 5 = 250
5n = 255
n = 51
âˆ´ 253 is the 51^{st} term.
20^{th} term from the 51^{st} term is 32^{nd} term
32^{nd} term = a + (32 â€“ 1)d
= 3 + 31 Ã— 5
= 3 +155
= 158.
Question24
The sum of the 4^{th} and 8^{th} terms of an A.P is 24 and the sum of the 6^{th} and 10^{th} terms is 44. Find the first three terms of the A.P.
Solution:
4^{th} term = a + 3d
8^{th }term = a + 7d
Sum of the 4^{th} term and 8^{th} term = a + 3d + a + 7d = 24
â‡’ 2d + 10d = 24
a + 5d = 12 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)
Sum of 6^{th} term and 10^{th} term = 44
â‡’ a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..â€¦â€¦â€¦(2)
a + 5d = 12 â€¦â€¦. (1)
a + 7d = 22 â€¦â€¦.. (2)

subtracting (1) â€“ (2) â‡’ 2d = 10
d = 5
substituting d = 5 in (1)
a + 5d = 12
a + 5(5) = 12
a = 12 â€“ 25 = 13.
âˆ´ The 1^{st} 3 terms are â€“13, 13 + 5, 13 + 10
(i.e.,) 13, 8, 3
Solution:
4^{th} term = a + 3d
8^{th }term = a + 7d
Sum of the 4^{th} term and 8^{th} term = a + 3d + a + 7d = 24
â‡’ 2d + 10d = 24
a + 5d = 12 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)
Sum of 6^{th} term and 10^{th} term = 44
â‡’ a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..â€¦â€¦â€¦(2)
a + 5d = 12 â€¦â€¦. (1)
a + 7d = 22 â€¦â€¦.. (2)

subtracting (1) â€“ (2) â‡’ 2d = 10
d = 5
substituting d = 5 in (1)
a + 5d = 12
a + 5(5) = 12
a = 12 â€“ 25 = 13.
âˆ´ The 1^{st} 3 terms are â€“13, 13 + 5, 13 + 10
(i.e.,) 13, 8, 3
Question25
Subba Rao started work in 1995 at an annual salary of `5000 and received an increment of `200 each year. In which year did his income reach `7000?
Solution:
Annual salary = `5000
Increment = `200
Total salary after n years = `7000
If the A.P. was of the form 5000, 5200, 5400, â€¦
When 1^{st} term a = 5000, d = 200
nth term = a + (n â€“ 1) d = 7000
= 5000 +(n â€“ 1) Ã— 200 = 7000
200(n â€“ 1) = 7000 â€“ 5000 = 2000
âˆ´ (n â€“ 1) =
n = 10 + 1
His income reached `7000 in the 11th year
Solution:
Annual salary = `5000
Increment = `200
Total salary after n years = `7000
If the A.P. was of the form 5000, 5200, 5400, â€¦
When 1^{st} term a = 5000, d = 200
nth term = a + (n â€“ 1) d = 7000
= 5000 +(n â€“ 1) Ã— 200 = 7000
200(n â€“ 1) = 7000 â€“ 5000 = 2000
âˆ´ (n â€“ 1) =
n = 10 + 1
His income reached `7000 in the 11th year
Question26
Ramkali saved `5 in the first week of a year and then increased her weekly savings by `1.75. If in the nth week, her weekly savings become `20.75, find n.
Solution:
Here
Solution:
Here
a = 5
d = 1.75
In the nth week savingâ€™s = 5 + 1.75(n â€“ 1) = 20.75
1.75(n â€“ 1) = 20.75 â€“ 5
n  1 =
n 1 = 9
n = 10.
Question27
In an A.P:
(i) Given a = 5, d = 3, a_{n }= 50, find n and S_{n}.
(ii) Given a = 7, a_{13 }= 35, find d and S_{13}.
(iii) Given a_{12} = 37, d = 3, find a and S_{12}.
(iv) Given a_{3} = 15, S_{10 }= 125, find d and a_{10}.
(v) Given d = 5 , S_{9} = 75, find a and a_{9}.
Solution:
(i) Given a = 5, d = 3, a_{n }= 50, find n and S_{n}.
a = 5, d = 3, a_{n} = 50
n = = = = 16
S_{n} =
S_{16 } = = 8 Ã— 55 = 440.
Thus n = 16 and S_{16} = 440.
(ii) Given a = 7, a_{13 }= 35, find d and S_{13}.
a = 7, a_{13} = 35 find d and S_{13}
a = 7, l = 35, n = 13
We know that n =
13 =
13 â€“ 1 =
âˆ´ d = =
S_{n} =
S_{13} =
= = 273.
Thus d = and S_{13} = 273.
(iii) Given a_{12} = 37, d = 3, find a and S_{12}.
a_{12} = l = 37, d = 3,
n = 12, a = ?
n =
12 =
37 â€“ a = 11 Ã— 3
a = 33  37= 4
âˆ´ a = 4
S_{n} =
S_{12} =
= = 246.
Thus a = 4 and S_{12} = 246.
(iv) Given a_{3} = 15, S_{10 }= 125, find d and a_{10}.
a_{3} = 15, S_{10} = 125
a +2d = 15 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
S_{10 }= 125
â‡’(2a + 9d) = 125
â‡’ 5 (2a + 9d ) = 125
2a + 9d = = 25
2a + 9d = 25â€¦â€¦â€¦â€¦â€¦â€¦ (2)
(1) Ã—2 â‡’ 2a + 4d = 30 â€¦..(3)
Subtracting (2) from (3) weget,
5d = 5
âˆ´ d = 1
Substituting d = 1 in (1), we get,
a + 2d = 15
a + 2(1) = 15
a = 17
a_{10} = a + 9d = 17 + 9(1)
= 17 â€“ 9 = 8.
Thus d = 1 and a_{10} = 8.
(v) Given d = 5 , S_{9 }= 75, find a and a_{9}.
S_{9 }= 75 , n = 9
a_{n} = a +(n â€“ 1)5
a_{9} = a + 8 Ã— 5 â‡’ a_{9} = 40 + a â€¦â€¦â€¦â€¦â€¦.. (1)
S_{n} =
S_{9} = = 75
2a + 40 = 75 Ã— â€¦â€¦â€¦â€¦â€¦â€¦(2)
2a = â€“ 40
2a = =
a = =
Substituting a = in (1), we get,
a_{9} = 40 + = =
Thus a = and a_{9} = .
(i) Given a = 5, d = 3, a_{n }= 50, find n and S_{n}.
(ii) Given a = 7, a_{13 }= 35, find d and S_{13}.
(iii) Given a_{12} = 37, d = 3, find a and S_{12}.
(iv) Given a_{3} = 15, S_{10 }= 125, find d and a_{10}.
(v) Given d = 5 , S_{9} = 75, find a and a_{9}.
Solution:
(i) Given a = 5, d = 3, a_{n }= 50, find n and S_{n}.
a = 5, d = 3, a_{n} = 50
n = = = = 16
S_{n} =
S_{16 } = = 8 Ã— 55 = 440.
Thus n = 16 and S_{16} = 440.
(ii) Given a = 7, a_{13 }= 35, find d and S_{13}.
a = 7, a_{13} = 35 find d and S_{13}
a = 7, l = 35, n = 13
We know that n =
13 =
13 â€“ 1 =
âˆ´ d = =
S_{n} =
S_{13} =
= = 273.
Thus d = and S_{13} = 273.
(iii) Given a_{12} = 37, d = 3, find a and S_{12}.
a_{12} = l = 37, d = 3,
n = 12, a = ?
n =
12 =
37 â€“ a = 11 Ã— 3
a = 33  37= 4
âˆ´ a = 4
S_{n} =
S_{12} =
= = 246.
Thus a = 4 and S_{12} = 246.
(iv) Given a_{3} = 15, S_{10 }= 125, find d and a_{10}.
a_{3} = 15, S_{10} = 125
a +2d = 15 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
S_{10 }= 125
â‡’(2a + 9d) = 125
â‡’ 5 (2a + 9d ) = 125
2a + 9d = = 25
2a + 9d = 25â€¦â€¦â€¦â€¦â€¦â€¦ (2)
(1) Ã—2 â‡’ 2a + 4d = 30 â€¦..(3)
Subtracting (2) from (3) weget,
5d = 5
âˆ´ d = 1
Substituting d = 1 in (1), we get,
a + 2d = 15
a + 2(1) = 15
a = 17
a_{10} = a + 9d = 17 + 9(1)
= 17 â€“ 9 = 8.
Thus d = 1 and a_{10} = 8.
(v) Given d = 5 , S_{9 }= 75, find a and a_{9}.
S_{9 }= 75 , n = 9
a_{n} = a +(n â€“ 1)5
a_{9} = a + 8 Ã— 5 â‡’ a_{9} = 40 + a â€¦â€¦â€¦â€¦â€¦.. (1)
S_{n} =
S_{9} = = 75
2a + 40 = 75 Ã— â€¦â€¦â€¦â€¦â€¦â€¦(2)
2a = â€“ 40
2a = =
a = =
Substituting a = in (1), we get,
a_{9} = 40 + = =
Thus a = and a_{9} = .
Question28
In an A.P:
(i) Given a = 2 , d = 8, S_{n }= 90, find n and a_{n}.
(ii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
(iii) Given a_{n} = 4, d = 2, S_{n} = 14, find n and a.
(iv) Given a = 3, n = 8, S = 192, find d.
(v) Given l = 28, S = 144, and there are total 9 items. Find a.
Solution:
(i) Given a = 2 , d = 8, S_{n }= 90, find n and a_{n}.
a = 2, d = 8
S_{n} = 90, find n and a_{n}
S_{n} =
=
=
But S_{n} = 90,
=> n[8n â€“ 4] = 90 Ã— 2
8n^{2} â€“ 4n â€“ 180 = 0
Dividing throughout by 4, we get
2n^{2} â€“ n â€“ 45 = 0
2n^{2} â€“10n + 9n â€“ 45 = 0
2n(n  5) + 9(n  5) = 0
âˆ´ (2n + 9)(n â€“ 5) = 0
n = 5 (or) n =
âˆ´ Number of terms could be positive only, thus
n = 5 , and n = not possible
âˆ´ a_{n} = [a + (n â€“ 1) Ã— d]
a_{n} = [2 + (5 â€“ 1) Ã— 8] = 2 + 32 = 34.
Thus n = 5 and a_{n}= 34.
(ii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
Here, a = 8, a_{n} = 62, S_{n} = 210
Now we have to find n and d,
a_{n} = a + (n â€“ 1) Ã— d
= 8 + (n â€“ 1) Ã— d
(n â€“ 1) Ã— d = 62 â€“ 8 = 54 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)
S_{n} =
210 =
420 = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)
Substituting for (n â€“ 1) Ã— d = 54 from (1) in (2)
420 = n[16 +54]
âˆ´ n = = 6
Substitute n = 6 in (1), we get,
(n 1) Ã— d = 54
â‡’ (6  1) Ã— d = 54
â‡’ 5d = 54
â‡’ d =
Thus n = 6 and d = .
(iii) Given a_{n} = 4, d = 2, S_{n} = 14, find n and a.
Here, a_{n} = 4, d = 2, S_{n} = 14.
a_{n} = a +(n â€“ 1)d
4 = a +2(n â€“ 1)
â‡’ a = 4 â€“ [2(n 1)] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)
S_{n} =
14 =
=
Substitute a = 4 â€“ [2(n 1)] from (1)
14 = n[4 â€“ 2(n â€“ 1) + (n â€“ 1)]
14 = n[4 â€“ (n â€“ 1)]
= 4n â€“ n^{2} + n
n^{2} â€“ 5n â€“ 14 = 0
n^{2} â€“ 7n + 2n â€“ 14 = 0
n(n â€“ 7) + 2(n  7) = 0
(n  7)(n+2) = 0
n = 7, n = 2
Number of terms cannot be negative thus n = 7 9; 9;
Substitute n = 7 in (1), we get,
a = 4 â€“ [2 (7 â€“ 1)]
= 4 â€“ (2 Ã— 6)
= 4 â€“ 12 = 8
n = 7, a = 8
Thus n = 7 and a = 8.
(iv) Given a = 3, n = 8, S = 192, find d.
a = 3, n = 8, S = 192 find d
S_{n} =
192 =
192 = 4[(2 Ã— 3) + (8 â€“ 1)d]
192 = 4(6 + (8 â€“ 1)d)
6 + 7d =
6 + 7d = 48
7d = 42
d = = 6
Thus d = 6.
(v) Given l = 28, S = 144, and there are total 9 items. Find a.
l = 28, S = 144, n = 9, a = ?
S = (a+l)
144 =
144 Ã— = a + 28
32 = a + 28
â‡’ a = 4
Hence a = 4.
(i) Given a = 2 , d = 8, S_{n }= 90, find n and a_{n}.
(ii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
(iii) Given a_{n} = 4, d = 2, S_{n} = 14, find n and a.
(iv) Given a = 3, n = 8, S = 192, find d.
(v) Given l = 28, S = 144, and there are total 9 items. Find a.
Solution:
(i) Given a = 2 , d = 8, S_{n }= 90, find n and a_{n}.
a = 2, d = 8
S_{n} = 90, find n and a_{n}
S_{n} =
=
=
But S_{n} = 90,
=> n[8n â€“ 4] = 90 Ã— 2
8n^{2} â€“ 4n â€“ 180 = 0
Dividing throughout by 4, we get
2n^{2} â€“ n â€“ 45 = 0
2n^{2} â€“10n + 9n â€“ 45 = 0
2n(n  5) + 9(n  5) = 0
âˆ´ (2n + 9)(n â€“ 5) = 0
n = 5 (or) n =
âˆ´ Number of terms could be positive only, thus
n = 5 , and n = not possible
âˆ´ a_{n} = [a + (n â€“ 1) Ã— d]
a_{n} = [2 + (5 â€“ 1) Ã— 8] = 2 + 32 = 34.
Thus n = 5 and a_{n}= 34.
(ii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
Here, a = 8, a_{n} = 62, S_{n} = 210
Now we have to find n and d,
a_{n} = a + (n â€“ 1) Ã— d
= 8 + (n â€“ 1) Ã— d
(n â€“ 1) Ã— d = 62 â€“ 8 = 54 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)
S_{n} =
210 =
420 = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)
Substituting for (n â€“ 1) Ã— d = 54 from (1) in (2)
420 = n[16 +54]
âˆ´ n = = 6
Substitute n = 6 in (1), we get,
(n 1) Ã— d = 54
â‡’ (6  1) Ã— d = 54
â‡’ 5d = 54
â‡’ d =
Thus n = 6 and d = .
(iii) Given a_{n} = 4, d = 2, S_{n} = 14, find n and a.
Here, a_{n} = 4, d = 2, S_{n} = 14.
a_{n} = a +(n â€“ 1)d
4 = a +2(n â€“ 1)
â‡’ a = 4 â€“ [2(n 1)] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)
S_{n} =
14 =
=
Substitute a = 4 â€“ [2(n 1)] from (1)
14 = n[4 â€“ 2(n â€“ 1) + (n â€“ 1)]
14 = n[4 â€“ (n â€“ 1)]
= 4n â€“ n^{2} + n
n^{2} â€“ 5n â€“ 14 = 0
n^{2} â€“ 7n + 2n â€“ 14 = 0
n(n â€“ 7) + 2(n  7) = 0
(n  7)(n+2) = 0
n = 7, n = 2
Number of terms cannot be negative thus n = 7 9; 9;
Substitute n = 7 in (1), we get,
a = 4 â€“ [2 (7 â€“ 1)]
= 4 â€“ (2 Ã— 6)
= 4 â€“ 12 = 8
n = 7, a = 8
Thus n = 7 and a = 8.
(iv) Given a = 3, n = 8, S = 192, find d.
a = 3, n = 8, S = 192 find d
S_{n} =
192 =
192 = 4[(2 Ã— 3) + (8 â€“ 1)d]
192 = 4(6 + (8 â€“ 1)d)
6 + 7d =
6 + 7d = 48
7d = 42
d = = 6
Thus d = 6.
(v) Given l = 28, S = 144, and there are total 9 items. Find a.
l = 28, S = 144, n = 9, a = ?
S = (a+l)
144 =
144 Ã— = a + 28
32 = a + 28
â‡’ a = 4
Hence a = 4.
Question29
How many terms of the A.P: 9, 17, 25, â€¦ must be taken to give a sum of 636?
Solution:
The given A.P is 9, 17, 25, â€¦
S_{n} = 636
In this A.P, a = 9
d = a_{2} â€“ a_{1} = 17 â€“ 9 = 8
Let n be the number of terms
S_{n} =
636 =
1272 =
=
= 8n^{2} + 10n
8n^{2} + 10n â€“ 1272 = 0
Dividing throughout by 2, we get
4n^{2} + 5n â€“ 636 = 0
4n^{2} + 53n â€“ 48n  636 = 0
n(4n + 53) â€“ 12(4n + 53) = 0
(n  12)(4n + 53) = 0
â‡’ 4n = 53 â‡’ n = ; n = 12
Only one value n = 12 is admissible.
Thus the number of terms is 12.
Solution:
The given A.P is 9, 17, 25, â€¦
S_{n} = 636
In this A.P, a = 9
d = a_{2} â€“ a_{1} = 17 â€“ 9 = 8
Let n be the number of terms
S_{n} =
636 =
1272 =
=
= 8n^{2} + 10n
8n^{2} + 10n â€“ 1272 = 0
Dividing throughout by 2, we get
4n^{2} + 5n â€“ 636 = 0
4n^{2} + 53n â€“ 48n  636 = 0
n(4n + 53) â€“ 12(4n + 53) = 0
(n  12)(4n + 53) = 0
â‡’ 4n = 53 â‡’ n = ; n = 12
Only one value n = 12 is admissible.
Thus the number of terms is 12.
Question30
The first term of an A.P is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
a = 5, a_{n} = 45, S_{n} = 400
S_{n} = where l  the last term, a  first term and n  number of terms
400 =
=
400 = 25n
n = = 16
âˆ´ Number of terms = 16
a_{n} = a + (n1)d
45 = 5 +(16 â€“ 1) d
45 â€“ 5 = 15d
d =
Hence the number of terms and the common difference is 16 and .
Solution:
a = 5, a_{n} = 45, S_{n} = 400
S_{n} = where l  the last term, a  first term and n  number of terms
400 =
=
400 = 25n
n = = 16
âˆ´ Number of terms = 16
a_{n} = a + (n1)d
45 = 5 +(16 â€“ 1) d
45 â€“ 5 = 15d
d =
Hence the number of terms and the common difference is 16 and .
Question31
The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
In this problem,
a = 17, l = 350, d = 9
Let n be the no. of terms and S_{n} their sum
Let a_{n} = 350
a_{n} = a + (n â€“ 1)d
350 = 17 + (n â€“ 1) 9
350 â€“ 17 = (n â€“ 1) 9
= (n â€“ 1)
(n â€“ 1) = 37
n = 38
S_{n} =
=
= = 6973.
Thus the number of terms is 38 and its sum is 6973.
Solution:
In this problem,
a = 17, l = 350, d = 9
Let n be the no. of terms and S_{n} their sum
Let a_{n} = 350
a_{n} = a + (n â€“ 1)d
350 = 17 + (n â€“ 1) 9
350 â€“ 17 = (n â€“ 1) 9
= (n â€“ 1)
(n â€“ 1) = 37
n = 38
S_{n} =
=
= = 6973.
Thus the number of terms is 38 and its sum is 6973.
Question32
Find the sum of first 22 terms of an A.P in which d = 7 and 22^{nd} term is 149.
Solution:
d = 7, a_{22} = 149
n = 22
As a_{n} = a + (n â€“ 1)d
a_{22} = a + (n â€“ 1) Ã— 7
149 = a + 21 Ã— 7
= a + 147
a = 149  147
a = 2
âˆ´ S_{n} =
=
= 11 Ã— 151
âˆ´ The sum of first 22 terms = 1661
Solution:
d = 7, a_{22} = 149
n = 22
As a_{n} = a + (n â€“ 1)d
a_{22} = a + (n â€“ 1) Ã— 7
149 = a + 21 Ã— 7
= a + 147
a = 149  147
a = 2
âˆ´ S_{n} =
=
= 11 Ã— 151
âˆ´ The sum of first 22 terms = 1661
Question33
Find the sum of first 51 terms of an A.P whose second and third terms are 14 and 18 respectively.
Solution:
a_{2} = 14
a_{3 } = 18
a_{2} = 14 = a + (21)d
â‡’ a + d = 14 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)
a_{3} = 18
â‡’ a + 2d = 18 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)
a + d = 14
a + 2d = 18
(1) â€“ (2) â‡’ d = 4 â‡’ d = 4
Substituting d = 4 in (1), we get,
a + d = 14
a + 4 = 14
a = 14 â€“ 4 = 10
S_{n} =
S_{51} =
=
=
= = 5610.
âˆ´ The sum of first 51 terms = 5610.
Solution:
a_{2} = 14
a_{3 } = 18
a_{2} = 14 = a + (21)d
â‡’ a + d = 14 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)
a_{3} = 18
â‡’ a + 2d = 18 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)
a + d = 14
a + 2d = 18
(1) â€“ (2) â‡’ d = 4 â‡’ d = 4
Substituting d = 4 in (1), we get,
a + d = 14
a + 4 = 14
a = 14 â€“ 4 = 10
S_{n} =
S_{51} =
=
=
= = 5610.
âˆ´ The sum of first 51 terms = 5610.
Question34
If the sum of first 7 terms of an A.P is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
S_{7} = 49
S_{17} = 289
S_{n} =
S_{7} =
49 =
49 Ã— = (2a + 6d)
7= a + 3d  (1) (Dividing by 2)
S_{17} = [2a + (17 â€“ 1) d]
289 = [2a + (17 â€“ 1) d]
289 Ã— = (2a + 16d)
17 = a + 8d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2) (Dividing by 2)
(1)â‡’ 7= a + 3d
(2) (1) â‡’ 5d = 10
d = 2
Substituting d = 2 in (1), we get,
a + 3d = 7
a + 3(2) = 7
a = 7 â€“ 6
a = 1.
S_{n} =
= =
= n^{2}
Hence the sum of first n terms is n^{2}.
Solution:
S_{7} = 49
S_{17} = 289
S_{n} =
S_{7} =
49 =
49 Ã— = (2a + 6d)
7= a + 3d  (1) (Dividing by 2)
S_{17} = [2a + (17 â€“ 1) d]
289 = [2a + (17 â€“ 1) d]
289 Ã— = (2a + 16d)
17 = a + 8d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2) (Dividing by 2)
(1)â‡’ 7= a + 3d
(2) (1) â‡’ 5d = 10
d = 2
Substituting d = 2 in (1), we get,
a + 3d = 7
a + 3(2) = 7
a = 7 â€“ 6
a = 1.
S_{n} =
= =
= n^{2}
Hence the sum of first n terms is n^{2}.
Question35
Show that a_{1}, a_{2}, â€¦, a_{n}, â€¦ form an A.P where a_{n} is defined as below:
(i) a_{n} = 3 + 4n (ii) a = 9  5n
Solution:
(i) Given, a_{n} = 3 + 4n
When n = 1,
a_{1} = 3 + 4(1)= 3 + 4 = 7
When n = 2,
a_{2} = 3 + 4(2) = 3 + 8 = 11
When n = 3, a_{3} = 3 + 4(3) = 3 +12 = 15
Since a_{2} â€“ a_{1} = 11 â€“ 7 = 4 and a_{3} â€“ a_{2} = 15 â€“ 11 = 4.
i.e., a_{k+1 } a_{k} is the same every time, the given list forms an A.P with the common difference (d) = 4
S_{n} =
S_{15 }=
=
=
=
= = 525
This is an A.P with a = 7, d = 4, S_{15} = 525.
(ii) Given, a = 9  5n
When n = 1, a_{1} = 4
When n = 2, a_{2} = 9  10 = 1
When n = 3, a_{3} = 9 â€“ 15
= 6
Since a_{2} â€“ a_{1} = 1 â€“ 4 = 5 and a_{3} â€“ a_{2} = 6 + 1 = 5.
i.e., a_{k+1 } a_{k} is the same every time, the given list forms an A.P with the common difference (d) = 5
S_{n} =
S_{15 } =
=
=
=
=  465.
This is an A.P with a = 4, d = 5, S_{15} = 465
(i) a_{n} = 3 + 4n (ii) a = 9  5n
Solution:
(i) Given, a_{n} = 3 + 4n
When n = 1,
a_{1} = 3 + 4(1)= 3 + 4 = 7
When n = 2,
a_{2} = 3 + 4(2) = 3 + 8 = 11
When n = 3, a_{3} = 3 + 4(3) = 3 +12 = 15
Since a_{2} â€“ a_{1} = 11 â€“ 7 = 4 and a_{3} â€“ a_{2} = 15 â€“ 11 = 4.
i.e., a_{k+1 } a_{k} is the same every time, the given list forms an A.P with the common difference (d) = 4
S_{n} =
S_{15 }=
=
=
=
= = 525
This is an A.P with a = 7, d = 4, S_{15} = 525.
(ii) Given, a = 9  5n
When n = 1, a_{1} = 4
When n = 2, a_{2} = 9  10 = 1
When n = 3, a_{3} = 9 â€“ 15
= 6
Since a_{2} â€“ a_{1} = 1 â€“ 4 = 5 and a_{3} â€“ a_{2} = 6 + 1 = 5.
i.e., a_{k+1 } a_{k} is the same every time, the given list forms an A.P with the common difference (d) = 5
S_{n} =
S_{15 } =
=
=
=
=  465.
This is an A.P with a = 4, d = 5, S_{15} = 465
Question36
If the sum of the first n terms of an A.P is 4n â€“ n^{2}, what is the first term (that is S_{1})? What is the sum of first two terms? What is the second term? Similarly, find the 3^{rd}, the 10^{th} and the n^{th} terms.
Solution:
S_{n} = (4n â€“ n^{2})
a_{1} = S_{1} = 4(1) â€“ (1)^{2}
= 4 â€“ 1 = 3
S_{2} = 4(2) â€“ 2^{2} = 4,
S_{3} = 4(3) â€“ 3^{2}
= 12 â€“ 9
= 3
a_{2} = S_{2 }â€“ S_{1} = 4 â€“ 3 = 1
a_{3} = S_{3 }â€“ S_{2} = 3 â€“ 4 = 1
a_{1}, a_{2}, a_{3} is given by 3, 1, 1
Here a = 3, d = 2, n = 10
âˆ´ a_{10} = a + (n â€“ 1)d
= 3 + 9 Ã— (2)
= 3 â€“ 18
= 15
a_{n} = 3 + (n â€“ 1) (â€“ 2)
a_{n} = 2n + 5 = 5 â€“ 2n
Thus a_{1}, a_{2}, a_{3} is given by 3, 1, 1, the 10^{th} term is â€“15, and the n^{th} term is 5 â€“ 2n.
Solution:
S_{n} = (4n â€“ n^{2})
a_{1} = S_{1} = 4(1) â€“ (1)^{2}
= 4 â€“ 1 = 3
S_{2} = 4(2) â€“ 2^{2} = 4,
S_{3} = 4(3) â€“ 3^{2}
= 12 â€“ 9
= 3
a_{2} = S_{2 }â€“ S_{1} = 4 â€“ 3 = 1
a_{3} = S_{3 }â€“ S_{2} = 3 â€“ 4 = 1
a_{1}, a_{2}, a_{3} is given by 3, 1, 1
Here a = 3, d = 2, n = 10
âˆ´ a_{10} = a + (n â€“ 1)d
= 3 + 9 Ã— (2)
= 3 â€“ 18
= 15
a_{n} = 3 + (n â€“ 1) (â€“ 2)
a_{n} = 2n + 5 = 5 â€“ 2n
Thus a_{1}, a_{2}, a_{3} is given by 3, 1, 1, the 10^{th} term is â€“15, and the n^{th} term is 5 â€“ 2n.
Question37
Find the sum of the first 40 positive integers divisible by 6.
Solution:
First term a = 6, d = 6, n = 40
Sum of the 40 positive integers divisible by 6
S_{n} =
S_{40} =
= 20[12 + (39 Ã— 6)] = 20 [12 + 234]
= 20 Ã— 246
âˆ´ S_{40}= 4920
Solution:
First term a = 6, d = 6, n = 40
Sum of the 40 positive integers divisible by 6
S_{n} =
S_{40} =
= 20[12 + (39 Ã— 6)] = 20 [12 + 234]
= 20 Ã— 246
âˆ´ S_{40}= 4920
Question38
Find the sum of the first 15 multiples of 8.
Solution:
First term a = 8, d = 8, n = 15
Sum of 1^{st} 15 multiples of 8:
S_{n} =
S_{15} =
=
=
= 960
Solution:
First term a = 8, d = 8, n = 15
Sum of 1^{st} 15 multiples of 8:
S_{n} =
S_{15} =
=
=
= 960
Question39
Find the sum of the odd numbers between 0 and 50.
Solution:
The list of odd numbers between 0 and 50: 1, 3, 5, â€¦, 49.
First term (a) = 1
Common difference (d) = 2
Last term (l) = 49
S_{n} = â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
Last term is given by
a_{n} = 1 + (n â€“ 1) Ã— d
= 1 + (n â€“ 1) Ã— 2
â‡’ 2n â€“ 2 = 49 â€“ 1
2n = 48 + 2
2n = 50
n = 25
Substituting n = 25 in (1), we get,
Sum = = 625.
âˆ´ The sum of odd numbers between 0 and 50 = 625
Solution:
The list of odd numbers between 0 and 50: 1, 3, 5, â€¦, 49.
First term (a) = 1
Common difference (d) = 2
Last term (l) = 49
S_{n} = â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
Last term is given by
a_{n} = 1 + (n â€“ 1) Ã— d
= 1 + (n â€“ 1) Ã— 2
â‡’ 2n â€“ 2 = 49 â€“ 1
2n = 48 + 2
2n = 50
n = 25
Substituting n = 25 in (1), we get,
Sum = = 625.
âˆ´ The sum of odd numbers between 0 and 50 = 625
Question40
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: `200 for the first day, `250 for the second day, `300 for the third day, etc., the penalty for each succeeding day being `50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
a_{1} = 200, a_{2} = 250, a_{3} = 300
(i.e.,) a = 200, d = 50
If n = 30,
S_{n} =
=
=
=
= 1850 Ã— 15 = `27,750
If the contractor delays the work by 30 days, he has to pay `27,750 as penalty.
Solution:
a_{1} = 200, a_{2} = 250, a_{3} = 300
(i.e.,) a = 200, d = 50
If n = 30,
S_{n} =
=
=
=
= 1850 Ã— 15 = `27,750
If the contractor delays the work by 30 days, he has to pay `27,750 as penalty.
Question41
A sum of `700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is `20 less than its preceding prize, find the value the work of each of the prizes.
Solution:
S_{7} = 700
d = `(20) less than preceding
= (20)
a_{1} = a + d = a â€“ 20
a_{2} = a  40
a_{3} = a â€“ 60
a_{7} = a â€“ 140
Now
S_{n} =
S_{7} =
=
700 =
a  60= 100
a = 160
Values of the prizes (in `) are 160, 140, 120, 100, 80, 60, 40.
Solution:
S_{7} = 700
d = `(20) less than preceding
= (20)
a_{1} = a + d = a â€“ 20
a_{2} = a  40
a_{3} = a â€“ 60
a_{7} = a â€“ 140
Now
S_{n} =
S_{7} =
=
700 =
a  60= 100
a = 160
Values of the prizes (in `) are 160, 140, 120, 100, 80, 60, 40.
Question42
In a School, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Hence the A.P will be of the form 1, 2, 3, 4, â€¦, 12
1st term (a) = 1
Last term (l) = 12
Common difference (d) = 1
Sum of trees planted by one section of each class =
=
= 13 Ã— 6
= 78 .
Since there are 3 sections in each class the total number of trees planted = 78 Ã— 3 = 234.
âˆ´ Total number of trees planted by the students = 234.
Solution:
Hence the A.P will be of the form 1, 2, 3, 4, â€¦, 12
1st term (a) = 1
Last term (l) = 12
Common difference (d) = 1
Sum of trees planted by one section of each class =
=
= 13 Ã— 6
= 78 .
Since there are 3 sections in each class the total number of trees planted = 78 Ã— 3 = 234.
âˆ´ Total number of trees planted by the students = 234.
Question43
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, â€¦ as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take p = 22/7)
Solution:
Length of semicircle with the centre at A = l_{1}
Length of semicircle with the centre at B = l_{2}
Length of semicircle with the centre at A = l_{3} and so onâ€¦
Applying the formula, length of semicircle =
As a spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, â€¦ as shown in the figure
The total length of all the semicircles = Ï€ Ã— (0.5) + Ï€ Ã— (1.0) + Ï€ Ã— (1.5) +â€¦
= Ï€ [0.5 + 1 + 1.5 +â€¦]
This is an AP with first term = 0.5 and d = 0.5
If n = 13,
S_{n} =
& S_{13} =
= = = =
Total length of the thirteen consecutive semicircles = Ï€ Ã—
= = 143 cm.
Question44
200 logs are stacked in the following manner: 20 logs in the bottom now, 19 in the next row, 18 in the row next to it and so on. In how may rows are the 200 logs placed and how many logs are in the top row?
Solution:
Number of logs in 1^{st} row (a) = 20
Total number of logs =_{ }S_{n} = 200
d = (1)
S_{n} =
200 =
=
400 = n(41 â€“ n)
(41 â€“ n)n = 400
41n  n^{2} â€“ 400 = 0
41n â€“ n^{2} â€“ 400 = 0
n^{2} â€“ 41n + 400 = 0
n^{2} â€“ 16n â€“ 25n + 400 = 0
n(n â€“ 16) â€“ 25(n â€“ 16) = 0
(n â€“ 25)(n â€“ 16) = 0
n = 16 (or) n = 25
Number of rows is either 16 or 25 (substituting for a and d)
If n = 16,
a_{n} = a + (n â€“ 1) d
a_{16 }= 20 + (16 â€“ 1)(1)
a_{16} = 20  15 = 5
Substituting for a and d
If n = 25, a_{25} = 20 +(25 â€“ 1) (1)
= 20 â€“ 24 = ve
Since the number of logs in the row cannot be negative (n â‰ 25)
Hence the number of rows = 16, and number of logs in the top row = 5
[(i.e.,) n = 16, a_{16} = 5]
Question45
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the lineA competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Solution:
To pick up the 1^{st} potato the competitor has to run 2 Ã— 5 m. Similarly for the 2^{nd} potato
2 Ã— 5 + 2 Ã— 3 = 2 Ã— (5 + 3)
Similarly for the 3^{rd} potato the competitor has to run = 2 Ã— (5 + 3 + 3)
If we write down the distance traveled by the competitor for each potato it can be given by
2 Ã— 5, 2 Ã— (5 + 3), 2 Ã— (5 + 3 + 3) etc
(i.e.,) 2 times 5, 8, 11, â€¦
a = 5, d = 3, n =10
S_{n} =
S_{10} =
=
= 5 (10 + 27)
= 5 (37)
= 185
âˆ´ Total distance traveled = 2 times (5 + 8 + 11 +â€¦)
= 2 Ã— 185
= 370 m.