NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2
NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions Exercise 5.2
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In an AP with first term a and common difference d, the nth term (or the general term) is given by
$a_n= a + (n - 1) d$
Question 2
Choose the correct choice in the following and justify
(i) 30^{th}term of the A.P: 10, 7, 4, …, is
(A)97
(B)77
(C)−77
(D)−87
(ii) 11^{th}term of the A.P. -3, -1/2, ,2 .... is
(A) 28
(B) 22
(C) - 38
(D)-97/2
Answer
(i) Given that
A.P. 10, 7, 4, …
First term,a= 10
Common difference,d=a_{2}−a_{1}= 7 − 10 = −3
We know that,a_{n}=a+ (n− 1)d
a_{30}= 10 + (30 − 1) (−3)
a_{30}= 10 − 87 = −77
Hence, the correct answer is optionC.
(ii) Given that A.P. is-3, -1/2, ,2 ...
First term ,a= - 3
Common difference,d=a_{2}−a_{1}= (-1/2) - (-3)
= (-1/2)+ 3 = 5/2
We know that,a_{n}=a+ (n− 1)d
a_{11}= 3+ (11 -1)(5/2)
a_{11}= -3+ 25
a_{11}= 22
Hence, the answer is option B.
Question 3
In the following APs find the missing term in the boxes.
Answer
Formula to consider for solving these questions
a_{n}=a+ (n− 1)d
Where d -> common difference
A -> first term
n-> term
a_{n} -> nth term
(ii)For this A.P.,
a_{2}= 13 and
a_{4}= 3
We know that,a_{n}=a+ (n− 1)d
a_{2}=a+ (2 - 1)d
13 =a+d...(i)
a_{4}=a+ (4 - 1)d
3 =a+ 3d...(ii)
Now solving these equations from the lesson of linear equations
On subtracting(i)from(ii), we get
- 10 = 2d
d= - 5
From equation(i), we get
13 =a+ (-5)
a= 18
a_{3}= 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For this A.P.,
a_{2}= 13 and
a_{4}= 3
We know that,a_{n}=a+ (n− 1)d
a_{2}=a+ (2 - 1)d
13 =a+ d ...(i)
a_{4}=a+ (4 - 1)d
3 =a+ 3d...(ii)
On subtracting(i)from(ii), we get,
- 10 = 2d
d = - 5
From equation(i), we get,
13 =a+ (-5)
a= 18
a_{3}= 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iv) For this A.P.,
a= −4 and
a_{6}= 6
We know that,
a_{n}=a+ (n− 1)d
a_{6}= a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d= 2
a_{2}=a+d= − 4 + 2 = −2
a_{3}=a+ 2d= − 4 + 2 (2) = 0
a_{4}=a+ 3d= − 4 + 3 (2) = 2
a_{5}=a+ 4d= − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v)
For this A.P.,
a_{2}= 38
a_{6}= −22
We know that
a_{n}=a+ (n− 1)d
a_{2}=a+ (2 − 1)d
38 =a+d...(i)
a_{6}=a+ (6 − 1)d
−22 =a+ 5d...(ii)
On subtracting equation(i)from(ii), we get
− 22 − 38 = 4d
−60 = 4d
d= −15
a=a_{2}−d= 38 − (−15) = 53
a_{3}=a+ 2d= 53 + 2 (−15) = 23
a_{4}=a+ 3d= 53 + 3 (−15) = 8
a_{5}=a+ 4d= 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
Question 4
Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer
We have
a= 3
d=a_{2}−a_{1}= 8 − 3 = 5
Let n^{th}term of this A.P. be 78.
a_{n}=a+ (n− 1)d
78 = 3 + (n− 1) 5
75 = (n− 1) 5
(n− 1) = 15
n= 16
Hence, 16^{th}term of this A.P. is 78.
Question 5
Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,31/2, 13,...., -47
Answer
(i) For this A.P.,
a= 7
d=a_{2}−a_{1}= 13 − 7 = 6
Let there are n terms in this A.P.
a_{n}= 205
We know that
a_{n}=a+ (n− 1)d
Therefore, 205 = 7 + (n− 1) 6
198 = (n− 1) 6
33 = (n− 1)
n= 34
Therefore, this given series has 34 terms in it.
(ii) For this A.P.,
a= 18
d=a_{2} – a_{1} = 31/2 - 18=-5/2
Let there are n terms in this A.P.
a_{n}= 205
a_{n}=a+ (n− 1)d
-47 = 18+ (n- 1) (-5/2)
-47 - 18 = (n- 1) (-5/2)
-65 = (n- 1)(-5/2)
(n- 1) = -130/-5
(n- 1) =26
n= 27
Therefore, this given A.P. has 27 terms in it.
Question 6
Check whether -150 is a term of the A.P. 11, 8, 5, 2, … Answer
For this A.P.,
a= 11
d=a_{2}−a_{1}= 8 − 11 = −3
Let −150 be then^{th}term of this A.P.
We know that,
a_{n}=a+ (n− 1)d
-150 = 11+ (n- 1)(-3)
-150 = 11 - 3n+ 3
-164 = -3n
n= 164/3
Clearly,n is not an integer.
Therefore, - 150 is not a term of this A.P.
Question 7
Find the 31^{st}term of an A.P. whose 11^{th}term is 38 and the 16^{th}term is 73.
Answer
Given that,
a_{11}= 38
a_{16}= 73
We know that,
a_{n}=a+ (n− 1)d
a_{11}=a+ (11 − 1)d
38 =a+ 10d...(i)
Similarly,
a_{16}=a+ (16 − 1)d
73 =a+ 15d...(ii)
On subtracting(i)from(ii), we get
35 = 5d
d= 7
From equation(i),
38 =a+ 10 × (7)
38 − 70 =a
a= −32
Now that we know a and d of the A.P
a_{31}=a+ (31 − 1)d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^{st}term is 178.
Question 8
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.
Answer
Given that,
a_{3}= 12
a_{50}= 106
We know that,
a_{n}=a+ (n− 1)d
a_{3}=a+ (3 − 1)d
12 =a+ 2d...(i)
Similarly,a_{50}=a+ (50 − 1)d
106 =a+ 49d...(ii)
On subtracting(i)from(ii), we get
94 = 47d
d= 2
From equation(i), we get
12 =a+ 2 (2)
a= 12 − 4 = 8
a_{29}=a+ (29 − 1)d
a_{29}= 8 + (28)2
a_{29}= 8 + 56 = 64
Therefore, 29^{th}term is 64.
Question 9
If the 3^{rd}and the 9^{th}terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.
Answer
Given that,
a_{3}= 4
a_{9}= −8
We know that,
a_{n}=a+ (n− 1)d
a_{3}=a+ (3 − 1)d
4 =a+ 2d...(i)
a_{9}=a+ (9 − 1)d
−8 =a+ 8d...(ii)
On subtracting equation(i)from(ii), we get,
−12 = 6d
d= −2
From equation(i), we get,
4 =a+ 2 (−2)
4 =a− 4
a= 8
Let n^{th}term of this A.P. be zero.
a_{n}=a+ (n− 1)d
0 = 8 + (n− 1) (−2)
0 = 8 − 2n+ 2
2n= 10
n= 5
Hence, 5^{th}term of this A.P. is 0.
Question 10
If 17^{th}term of an A.P. exceeds its 10^{th}term by 7. Find the common difference.
Answer
We know that,
For an A.P.,a_{n}=a+ (n− 1)d
So 17^{th} term would be given as
a_{17}=a+ (17 − 1)d
a_{17}=a+ 16d
Similarly 10^{th} term would be given as
a_{10}=a+ 9d
It is given that in the question
a_{17}−a_{10}= 7
(a+ 16d) − (a+ 9d) = 7
7d= 7
d= 1
Therefore, the common difference is 1.
Question 11
Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^{th}term?
Answer
Given A.P. is 3, 15, 27, 39, …
a= 3
d=a_{2}−a_{1}= 15 − 3 = 12
a_{54}=a+ (54 − 1)d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n^{th}term be 771.
a_{n}=a+ (n− 1)d
771 = 3 + (n− 1) 12
768 = (n− 1) 12
(n− 1) = 64
n= 65
Therefore, 65^{th}term was 132 more than 54^{th}term.
Question 12
Two APs have the same common difference. The difference between their 100^{th}term is 100, what is the difference between their 1000^{th}terms? Answer
Let the first term of these A.P.s be a_{1}and a_{2}respectively and the common difference of these A.P.s be d
For first A.P.,
a_{100}=a_{1}+ (100 − 1)d
=a_{1}+ 99d
a_{1000}=a_{1}+ (1000 − 1)d
a_{1000}=a_{1}+ 999d
For second A.P.,
a_{100}=a_{2}+ (100 − 1)d
=a_{2}+ 99d
a_{1000}=a_{2}+ (1000 − 1)d
=a_{2}+ 999d
Given that, difference between
100^{th}term of these A.P.s = 100
Therefore, (a_{1}+ 99d) − (a_{2}+ 99d) = 100
a_{1}−a_{2}= 100 ...(i)
Difference between 1000^{th}terms of these A.P.s
(a_{1}+ 999d) − (a_{2}+ 999d) =a_{1}−a_{2}
From equation(i),
This difference,a_{1}−a_{2}= 100
Hence, the difference between 1000^{th}terms of these A.P. will be 100.
Question 13
How many three digit numbers are divisible by 7?
Answer
It is an interesting question which can be solved using arithmetic progression
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a= 105
d= 7
a_{n}= 994
n= ?
a_{n}=a+ (n− 1)d
994 = 105 + (n− 1) 7
(n− 1) = 127
n= 128
Therefore, 128 three-digit numbers are divisible by 7.
Question 14
How many multiples of 4 lie between 10 and 250?
Answer
It is an interesting question which can be solved using arithmetic progression.
First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be then^{th}term of this A.P.
a= 12
d= 4
a_{n}=248
a_{n}=a+ (n- 1)d
248 = 12+ (n- 1) × 4
59 =n- 1
n= 60
Therefore, there are 60 multiples of 4 between 10 and 250.
Question 15
For what value of n, are then^{th}terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Answer
First series
63, 65, 67, …
a= 63
d=a_{2}−a_{1}= 65 − 63 = 2
n^{th}term of this A.P. =a_{n}=a+ (n− 1)d
a_{n}= 63 + (n− 1) 2 = 63 + 2n− 2
a_{n}= 61 + 2n...(i)
Second Series
3, 10, 17, …
a= 3
d=a_{2}−a_{1}= 10 − 3 = 7
n^{th}term of this A.P. = 3 + (n− 1) 7
a_{n}= 3 + 7n− 7
a_{n}= 7n− 4 ...(ii)
It is given that,n^{th}term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n= 7n− 4
5n= 65
n= 13
Therefore, 13^{th}terms of both these A.P.s are equal to each other.
Question 16
Determine the A.P. whose third term is 16 and the 7^{th}term exceeds the 5^{th}term by 12.
Answer
We know that, nth term of the A.P is given by the expression
a_{n}=a+ (n− 1)d
Then 3th ,5^{th} and 7^{th} terms are
a_{3}= 16
a+ (3 − 1)d= 16
a+ 2d= 16 ...(i)
a_{7}= a+ (7 − 1)d= a+6d
a_{5}= a+ (5 − 1)d=a+4d
Then
a_{7}−a_{5}= 12
[a+ 6d] − [a+ 4d]= 12
2d= 12
d= 6
From equation(i), we get,
a+ 2 (6) = 16
a+ 12 = 16
a= 4
Therefore, A.P. will be
4, 10, 16, 22, …
Question 17
Find the 20^{th}term from the last term of the A.P. 3, 8, 13, …, 253.
Answer
Given A.P. is
3, 8, 13, …, 253
d=5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this reverse A.P.,
a= 253
d= 248 − 253 = −5
n= 20
a_{20}=a+ (20 − 1)d
a_{20}= 253 + (19) (−5)
a_{20}= 253 − 95
a= 158
Therefore, 20^{th}term from the last term is 158.
Question 18
The sum of 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.
Answer
We know that, nth term of the A.P is given by the expression
a_{n}=a+ (n− 1)d
Then 4^{th} term
a_{4}=a+ (4 − 1)d
a_{4}=a+ 3d
Similarly,8^{th},6^{th} and 10^{th} terms are
a_{8}=a+ 7d
a_{6}=a+ 5d
a_{10}=a+ 9d
Given that,
a_{4}+a_{8}= 24
a+ 3d+a+ 7d= 24
2a+ 10d= 24
a+ 5d= 12 ...(i)
a_{6}+a_{10}= 44
a+ 5d+a+ 9d= 44
2a+ 14d= 44
a+ 7d= 22 ...(ii)
Equation (i) and (ii) are linear equation in two variable
On subtracting equation(i)from(ii), we get,
2d= 22 − 12
2d= 10
d= 5
From equation(i), we get
a+ 5d= 12
a+ 5 (5) = 12
a= −13
a_{2}=a+d= − 13 + 5 = −8
a_{3}=a_{2}+d= − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.
Question 19
Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Answer
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here,a= 5000
d= 200
Let aftern^{th}year, his salary be Rs 7000.
Therefore,a_{n}=a+ (n− 1)d
7000 = 5000 + (n− 1) 200
200(n− 1) = 2000
(n− 1) = 10
n= 11
Therefore, in 11th year, his salary will be Rs 7000.
Question 20
Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in then^{th}week, her week, her weekly savings become Rs 20.75, find n.
Answer
Given that for the A.P
a= 5
d= 1.75
a_{n}= 20.75
We need to find out
n= ?
Now nth term is given by the expression
a_{n}=a+ (n− 1)d
20.75 = 5+ (n- 1) × 1.75
(n- 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n- 1 = 9
n= 10
Summary
NCERT Solutions for Class 10th Maths: Chapter 5 - Arithmetic Progressions Ex 5.2 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail.You can download this as pdf Download NCERT Solution Exercise 5.2 assignment as pdf
This chapter 5 has total 4 Exercise 5.1 ,5.2,5.3 and 5.4. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below