RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.2.
Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.2 solution. Solution is provided in this post. Here we have provide the solutions of RBSE Boards Books according to chapter wise.
 |
| Chapter 5 Arithmetic Progression Ex 5.2 |
RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.2.
Question 1.
Find
(i) 10th term of A.P. 2, 7, 12, ….
(ii) 18th term of A.P. √2, 3√2, 5√2, …..
(iii) 24th term of A.P. 9, 13, 17, 21, ….
Solution :
(i) Given A.P. 2, 7, 12, …..
First term a = 2
Common difference d = 7 – 2 = 5
nth term an = a + (n – 1)d
∴ a10 = 2 + (10 – 1) × 5
= 2 + 9 × 5
= 2 + 45
= 47
Hence, 10th term of given series 47.
Find
(i) 10th term of A.P. 2, 7, 12, ….
(ii) 18th term of A.P. √2, 3√2, 5√2, …..
(iii) 24th term of A.P. 9, 13, 17, 21, ….
Solution :
(i) Given A.P. 2, 7, 12, …..
First term a = 2
Common difference d = 7 – 2 = 5
nth term an = a + (n – 1)d
∴ a10 = 2 + (10 – 1) × 5
= 2 + 9 × 5
= 2 + 45
= 47
Hence, 10th term of given series 47.
(ii) Given A.P. √2, 3√2, 5√2, …..
First term a = √2
Common term d = 3√2 – √2
= √2(3 – 1) = 2√2
nth term an = a + (n – 1)d
∴ a18 = √2 + (18 – 1)2√2
= √2 + 17 × 2√2
= √2 + 34√2
= 35√2
Hence a18 = 35√2
(iii) Given A.P. 9, 13, 17, 21, …..
First term a = 9
common difference
d = 13 – 9 = 4
nth term an = a + (n – 1)d
∴ a24 = 9 + (24 – 1) × 4
= 9 + 23 × 4
= 9 + 92
= 101
Hence a24 = 101.
First term a = √2
Common term d = 3√2 – √2
= √2(3 – 1) = 2√2
nth term an = a + (n – 1)d
∴ a18 = √2 + (18 – 1)2√2
= √2 + 17 × 2√2
= √2 + 34√2
= 35√2
Hence a18 = 35√2
(iii) Given A.P. 9, 13, 17, 21, …..
First term a = 9
common difference
d = 13 – 9 = 4
nth term an = a + (n – 1)d
∴ a24 = 9 + (24 – 1) × 4
= 9 + 23 × 4
= 9 + 92
= 101
Hence a24 = 101.
|
| Chapter 5 Arithmetic Progression Ex 5.2 |
Question 2.
Solve
(i) Which term of A.P. 21, 18, 15, …… is – 81?
(ii) Which term of AP. 84, 80, 76, …. is zero ?
(iii) Is 301 any term of series 5, 11, 17, 23, …… ?
(iv) Is -150 is any term of A.P. 11, 8, 5, 2, …….
Solution :
(i) Given A.P. 21, 18, 15, ….
First term (a) = 21
Common difference
(d) = 18 – 21 = -3
∵ an = a + (n – 1)d
According to question
-81 = 21 + (n – 1)(-3)
⇒ -81-21 = (n – 1) × -3
⇒ -102 = (n – 1) × -3
⇒ (n – 1) =
⇒ n – 1 = 34
⇒ n = 34 + 1 = 35
Hence, 35th term of given series is -81.
(ii) Given A.P. 84, 80, 76…..
First term (a) = 84
Common difference (d) = 80 – 84 = -4
∵ an = a + (n – 1)d
According to question,
0 = 84 + (n – 1)(-4)
⇒ -84 = (n – 1) × -4
⇒ (n – 1) =
⇒ n – 1 = 21
⇒ n = 21 + 1 = 22
Hence. 22nd term of given A.P. is zero.
Solve
(i) Which term of A.P. 21, 18, 15, …… is – 81?
(ii) Which term of AP. 84, 80, 76, …. is zero ?
(iii) Is 301 any term of series 5, 11, 17, 23, …… ?
(iv) Is -150 is any term of A.P. 11, 8, 5, 2, …….
Solution :
(i) Given A.P. 21, 18, 15, ….
First term (a) = 21
Common difference
(d) = 18 – 21 = -3
∵ an = a + (n – 1)d
According to question
-81 = 21 + (n – 1)(-3)
⇒ -81-21 = (n – 1) × -3
⇒ -102 = (n – 1) × -3
⇒ (n – 1) =
⇒ n – 1 = 34
⇒ n = 34 + 1 = 35
Hence, 35th term of given series is -81.
(ii) Given A.P. 84, 80, 76…..
First term (a) = 84
Common difference (d) = 80 – 84 = -4
∵ an = a + (n – 1)d
According to question,
0 = 84 + (n – 1)(-4)
⇒ -84 = (n – 1) × -4
⇒ (n – 1) =
⇒ n – 1 = 21
⇒ n = 21 + 1 = 22
Hence. 22nd term of given A.P. is zero.
(iii) Given A.P. 5, 11, 17, 23 …..
First term (a) = 5
Common difference
(d) = 11 – 5 = 6
∵ an = a + (n – 1)d
According to question.,
⇒ 301 = 5 + (n – 1)(6)
⇒ 301 – 5 = 6(n—1)
⇒ 6(n – 1) = 296
⇒ (n – 1) =
⇒ n – 1 = 49.33
⇒ n = 49.33 + 1 = 50.33
∴ Value of n cannot be fraction, it means n is not a whole number. Thus no term can be 301 in given series A.P.
First term (a) = 5
Common difference
(d) = 11 – 5 = 6
∵ an = a + (n – 1)d
According to question.,
⇒ 301 = 5 + (n – 1)(6)
⇒ 301 – 5 = 6(n—1)
⇒ 6(n – 1) = 296
⇒ (n – 1) =
⇒ n – 1 = 49.33
⇒ n = 49.33 + 1 = 50.33
∴ Value of n cannot be fraction, it means n is not a whole number. Thus no term can be 301 in given series A.P.
|
| Chapter 5 Arithmetic Progression Ex 5.2 |
(iv) Given A.P. : 11, 8, 5, 2 …
First term (a) = 11 and common difference (d) = 8 – 11 = – 3
Let nth term, an = -150
⇒ a + (n – 1)d = -150
⇒ 11 + (n – 1) × (-3) = -150
⇒ -3(n – 1) = -150 – 11 = -161
⇒ (n – 1) =
= 53.6 (approx)
∴ n = 53.6 + 1 = 54.6
⇒ n is not a whole no.
Hence, -150 is no term is given A.P.
First term (a) = 11 and common difference (d) = 8 – 11 = – 3
Let nth term, an = -150
⇒ a + (n – 1)d = -150
⇒ 11 + (n – 1) × (-3) = -150
⇒ -3(n – 1) = -150 – 11 = -161
⇒ (n – 1) =
= 53.6 (approx)
∴ n = 53.6 + 1 = 54.6
⇒ n is not a whole no.
Hence, -150 is no term is given A.P.
|
| Chapter 5 Arithmetic Progression Ex 5.2 |
Question 3.
If 6th term and 17th turn of A.P. are 19 and 41 respectively, then find 40th term.
Solution :
Given
6th term a6 = 19
and 17th term a17 = 41
40th term a40 = ?
nth term, a = a + (n – 1)d
a6 = a + (6 – 1)d
⇒ 19 = a + 5d …..(i)
and a17 = a + (17 – 1)d
⇒ 41 = a + 16 d …(ii)
On subtracting (i) from (ii)
d =
d = 2
Putting d = 2 in equation (i)
a + 5 × 2 = 19
a = 19 – 10
a = 9
Thus, a40 = a + (40 – 1)d
= 9 + 39 × 2
= 9 + 78 = 87
Hence, 40th term of A.P. is 87.
If 6th term and 17th turn of A.P. are 19 and 41 respectively, then find 40th term.
Solution :
Given
6th term a6 = 19
and 17th term a17 = 41
40th term a40 = ?
nth term, a = a + (n – 1)d
a6 = a + (6 – 1)d
⇒ 19 = a + 5d …..(i)
and a17 = a + (17 – 1)d
⇒ 41 = a + 16 d …(ii)
On subtracting (i) from (ii)
d =
d = 2
Putting d = 2 in equation (i)
a + 5 × 2 = 19
a = 19 – 10
a = 9
Thus, a40 = a + (40 – 1)d
= 9 + 39 × 2
= 9 + 78 = 87
Hence, 40th term of A.P. is 87.
Question 4.
Third and ninth term of an A.P. are 4 and -8 respectively, then its which term will be zero?
Solution :
Let a is first term of A.P. and d is common difference.
Given, third term a3 = 4
a + (3 – 1)d = 4, [an = a + (n – 1)d]
⇒ a + 2d = 4
and ninth term a9 = -8
a + (9 – 1)d = -8
⇒ a + 8d = -8 …(ii)
Subtracting equation (i) from (ii)
∴ d =
= -2
Putting this value of d in equation (i)
a + 2(-2) = 4
or a – 4 = 4
∴ a = 4 + 4 = 8
Let nth term of series will be zero, than nth term
nth term an = 0
∴ a + (n – 1)d = 0
⇒ 8 + (n – 1) × (-2) = 0
⇒ -2(n – 1) = -8
⇒ (n – 1) = 4
∴ n = 5
Hence, 5th term of AP. will be zero.
Third and ninth term of an A.P. are 4 and -8 respectively, then its which term will be zero?
Solution :
Let a is first term of A.P. and d is common difference.
Given, third term a3 = 4
a + (3 – 1)d = 4, [an = a + (n – 1)d]
⇒ a + 2d = 4
and ninth term a9 = -8
a + (9 – 1)d = -8
⇒ a + 8d = -8 …(ii)
Subtracting equation (i) from (ii)
∴ d =
= -2Putting this value of d in equation (i)
a + 2(-2) = 4
or a – 4 = 4
∴ a = 4 + 4 = 8
Let nth term of series will be zero, than nth term
nth term an = 0
∴ a + (n – 1)d = 0
⇒ 8 + (n – 1) × (-2) = 0
⇒ -2(n – 1) = -8
⇒ (n – 1) = 4
∴ n = 5
Hence, 5th term of AP. will be zero.
|
| Chapter 5 Arithmetic Progression Ex 5.2 |
Question 5.
Third term of an A.P. Is 16 and 7th term is 12 more than 5th term, then find AP.
Solution :
Let first term of A.P. is a and common difference is d.
Given a3 = 16
a + (3 – 1)d = 16
⇒ a + 2d = 16 …(i)
According to question, a7 = 12 – a5
⇒ a7 – a5 = 12
[a + (7 – 1)d] – [a+(5 – 1)d] = 12
⇒ a7 + 6d – a – 4d = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
d =
= 6
Substituting value of d in equation (i)
a + 2(6) = 16
∴ a = 16 – 12 = 4
AP. a, a + d, a + 2d,…
= 4, 4 + 6, 4 + 2 × 6,…
= 4, 10, 16,…
Hence, required A.P. is 4, 10, 16, 22,…
Third term of an A.P. Is 16 and 7th term is 12 more than 5th term, then find AP.
Solution :
Let first term of A.P. is a and common difference is d.
Given a3 = 16
a + (3 – 1)d = 16
⇒ a + 2d = 16 …(i)
According to question, a7 = 12 – a5
⇒ a7 – a5 = 12
[a + (7 – 1)d] – [a+(5 – 1)d] = 12
⇒ a7 + 6d – a – 4d = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
d =
= 6Substituting value of d in equation (i)
a + 2(6) = 16
∴ a = 16 – 12 = 4
AP. a, a + d, a + 2d,…
= 4, 4 + 6, 4 + 2 × 6,…
= 4, 10, 16,…
Hence, required A.P. is 4, 10, 16, 22,…
Question 6.
How many three digit numbers are divisible by 7 ?
Solution :
Series of 3 digit numbers 100, 101, 102, …….. 999,
First three digit number divisible by 7 = 105 and last number = 994
Then series of three digit numbers divisible by 7
105, (105 + 7), (105 + 7+ 7), …. 994 = 105, 112, 119, …,994
Let total number of terms is n.
First term a = 105, common difference d = 7,
∴ nth term an = 994
⇒ a + (n – 1)d = 994
⇒ 105 + (n – 1) × 7 = 994
⇒ (n – 1) × 7=994 – 105 = 889
⇒ (n – 1) =
= 127
∴ n = 127 +1 = 128
Hence, three digit number divisible by 7 is 128.
How many three digit numbers are divisible by 7 ?
Solution :
Series of 3 digit numbers 100, 101, 102, …….. 999,
First three digit number divisible by 7 = 105 and last number = 994
Then series of three digit numbers divisible by 7
105, (105 + 7), (105 + 7+ 7), …. 994 = 105, 112, 119, …,994
Let total number of terms is n.
First term a = 105, common difference d = 7,
∴ nth term an = 994
⇒ a + (n – 1)d = 994
⇒ 105 + (n – 1) × 7 = 994
⇒ (n – 1) × 7=994 – 105 = 889
⇒ (n – 1) =
= 127∴ n = 127 +1 = 128
Hence, three digit number divisible by 7 is 128.
|
| Chapter 5 Arithmetic Progression Ex 5.2 |
Question 7.
Find 11th term from last of A.P. 10, 7, 4, …., -62
Solution :
Given A.P. 10, 7, 4….., -62
First term (a) = 10
Common difference (d) = 7 – 10 = – 3
Last term an = -62
Formula, rth term from last
= an – (r – 1)d
11th term from last = -62 – (11 – 1) × (-3)
= -62 – 10(-3)
= -62 + 30
= -32
Hence, 11th term from last of A.P. = 32.
Find 11th term from last of A.P. 10, 7, 4, …., -62
Solution :
Given A.P. 10, 7, 4….., -62
First term (a) = 10
Common difference (d) = 7 – 10 = – 3
Last term an = -62
Formula, rth term from last
= an – (r – 1)d
11th term from last = -62 – (11 – 1) × (-3)
= -62 – 10(-3)
= -62 + 30
= -32
Hence, 11th term from last of A.P. = 32.
Question 8.
Find the 12th term from last of A.P. 1, 4, 7, 10, …., 88
Solution :
Given A.P. 1, 4, 7, 10……, 88
First term (a) = 1
Common difference (d) = 4 – 1 = 3
Last term an = 88
Formula, rth term from last
= an – (r – 1)d
12th term from last = 88 – (12 – 1) × 3
= 88 – 11 × 3
= 88 – 33 = 55
Hence, 12th term from last term of A.P will be 55.
Find the 12th term from last of A.P. 1, 4, 7, 10, …., 88
Solution :
Given A.P. 1, 4, 7, 10……, 88
First term (a) = 1
Common difference (d) = 4 – 1 = 3
Last term an = 88
Formula, rth term from last
= an – (r – 1)d
12th term from last = 88 – (12 – 1) × 3
= 88 – 11 × 3
= 88 – 33 = 55
Hence, 12th term from last term of A.P will be 55.
|
| Chapter 5 Arithmetic Progression Ex 5.2 |
Question 9.
There are 60 terms in an AP. If its first and last term are 7 and 125 respectively, then find its 32nd term.
Solution :
Number of terms (n) = 60
First term (a) = 7
Last term (an) = 125
Formula, an = a + (n – 1)d
⇒ 125 = 7 + (60 – 1)d
⇒ 125 – 7 = 59 d
⇒ 118 = 59 d
⇒ d =
⇒ d = 2
Thus, 32nd term
a32 = a + (32 – 1)d
= 7 + 31 × 2
= 7 + 62 = 69
Hence, 32nd term of A.P. is 69.
There are 60 terms in an AP. If its first and last term are 7 and 125 respectively, then find its 32nd term.
Solution :
Number of terms (n) = 60
First term (a) = 7
Last term (an) = 125
Formula, an = a + (n – 1)d
⇒ 125 = 7 + (60 – 1)d
⇒ 125 – 7 = 59 d
⇒ 118 = 59 d
⇒ d =
⇒ d = 2
Thus, 32nd term
a32 = a + (32 – 1)d
= 7 + 31 × 2
= 7 + 62 = 69
Hence, 32nd term of A.P. is 69.
Question 10.
Four numbers are ¡n A.P. If sum of numbers is 50 and larger number is 4 times the smaller number, then find the number.
Solution :
Let four numbers in A.P. are
a, a + d, a + 2d, a + 3d
According to question
a + (a + d) + (a + 2d) + (a + 3d) = 50
⇒ 4a + 6d = 50
⇒ 2(2a + 3d) = 50
⇒ 2a + 3d = 25 …..(i)
If larger number is 4 times the smaller number, then equation will be as follow,
a + 3d = 4 x a
⇒ a + 3d = 4a
⇒ 3d = 3a
⇒ d = a …(ii)
from equation (i) and (ii)
2a + 3a = 25 [∴ d = a]
a = 25
a = 5
∴ d = 5 [∵ eqn(ii)]
∴ Numbers a = 5
a + d = 5 + 5 = 10
a + 2d = 5 + 2 × 5 = 15
a + 3d = 5 + 3 × 5 = 20
Hence, four numbers are 5, 10, 15, and 20.
Four numbers are ¡n A.P. If sum of numbers is 50 and larger number is 4 times the smaller number, then find the number.
Solution :
Let four numbers in A.P. are
a, a + d, a + 2d, a + 3d
According to question
a + (a + d) + (a + 2d) + (a + 3d) = 50
⇒ 4a + 6d = 50
⇒ 2(2a + 3d) = 50
⇒ 2a + 3d = 25 …..(i)
If larger number is 4 times the smaller number, then equation will be as follow,
a + 3d = 4 x a
⇒ a + 3d = 4a
⇒ 3d = 3a
⇒ d = a …(ii)
from equation (i) and (ii)
2a + 3a = 25 [∴ d = a]
a = 25
a = 5
∴ d = 5 [∵ eqn(ii)]
∴ Numbers a = 5
a + d = 5 + 5 = 10
a + 2d = 5 + 2 × 5 = 15
a + 3d = 5 + 3 × 5 = 20
Hence, four numbers are 5, 10, 15, and 20.
Hope that the Solutions provided here for Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.1 solution drop a comment below and like and share the post.
Thank you.
0 Comments