NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2:
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 focuses on Arithmetic Progressions (AP) and covers problems related to finding the nth term and the sum of n terms of an AP.
This exercise enhances understanding of APs through step-by-step solutions, enabling students to identify patterns, derive formulas, and solve real-life problems involving sequences. Key concepts include the general form of an AP, calculating terms,
and the sum formula
.
These solutions strengthen problem-solving skills and are essential for mastering the topic for exams.
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Overview
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 on Arithmetic Progressions (AP) are crucial for building a strong foundation in sequences and series. This exercise emphasizes deriving the nth term of an AP and calculating the sum of n terms.
Mastery of these concepts is vital as APs have practical applications in solving real-world problems, such as financial planning and pattern recognition. These solutions provide step-by-step explanations, fostering a deeper understanding, improving problem-solving skills, and preparing students for higher-level mathematics and competitive exams.
CBSE Class 10 Previous Year Question Papers PDF with Solutions
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 PDF
Below, we have provided the PDF for NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 on Arithmetic Progressions. This resource includes detailed solutions to all problems, covering the nth term formula and the sum of n terms in an AP. The step-by-step explanations aim to simplify complex concepts, making them easy to understand and practice. Download the PDF to enhance your preparation and excel in exams.
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 PDF
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Arithmetic Progressions
Below is the NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Arithmetic Progressions -
1. 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.
Solutions:
(i)
Given,
First term,
a
= 7
Common difference,
d
= 3
Number of terms,
n
= 8,
We have to find the nth term,
a
n
= ?
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values,
=> 7+(8 −1) 3
=> 7+(7) 3
=> 7+21 = 28
Hence,
a
n
= 28
(ii)
Given,
First term,
a
= -18
Common difference,
d
= ?
Number of terms,
n
= 10
Nth term,
a
n
= 0
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values,
0 = − 18 +(10−1)
d
18 = 9
d
d
= 18/9 = 2
Hence, common difference,
d
= 2
(iii)
Given,
First term,
a
= ?
Common difference,
d
= -3
Number of terms,
n
= 18
Nth term,
a
n
= -5
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values,
−5 =
a
+(18−1) (−3)
−5 =
a
+(17) (−3)
−5 =
a
−51
a
= 51−5 = 46
Hence,
a
= 46
(iv)
Given,
First term,
a
= -18.9
Common difference,
d
= 2.5
Number of terms,
n
= ?
Nth term,
a
n
= 3.6
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values,
3.6 = − 18.9+(
n
−1)2.5
3.6 + 18.9 = (
n
−1)2.5
22.5 = (
n
−1)2.5
(
n
– 1) = 22.5/2.5
n
– 1 = 9
n
= 10
Hence,
n
= 10
(v)
Given,
First term,
a
= 3.5
Common difference,
d
= 0
Number of terms,
n
= 105
Nth term,
a
n
= ?
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values,
a
n
= 3.5+(105−1) 0
a
n
= 3.5+104×0
a
n
= 3.5
Hence,
a
n
= 3.5
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)
Solutions:
(i) Given here,
A.P. = 10, 7, 4, …
Therefore, we can find,
First term,
a
= 10
Common difference,
d
=
a
2
−
a
1
= 7−10 = −3
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values;
a
30
= 10+(30−1)(−3)
a
30
= 10+(29)(−3)
a
30
= 10−87 = −77
Hence, the correct answer is option C.
(ii) Given here,
A.P. = -3, -1/2, ,2 …
Therefore, we can find,
First term
a
= – 3
Common difference,
d
=
a
2
−
a
1
= (-1/2) -(-3)
⇒(-1/2) + 3 = 5/2
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Putting the values;
a
11
= -3+(11-1)(5/2)
a
11
= -3+(10)(5/2)
a
11
= -3+25
a
11
= 22
Hence, the answer is option B.
3. In the following APs find the missing term in the boxes.
Solutions:
(i) For the given A.P., 2,2 , 26
The first and third term are;
a
= 2
a
3
= 26
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Therefore, putting the values here,
a
3
= 2+(3-1)
d
26 = 2+2
d
24 = 2
d
d
= 12
a
2
= 2+(2-1)12
= 14
Therefore, 14 is the missing term.
(ii) For the given A.P., , 13, ,3
a
2
= 13 and
a
4
= 3
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Therefore, putting the values here,
a
2
=
a
+(2-1)
d
13 =
a
+
d
……………….
(i)
a
4
=
a
+(4-1)
d
3 =
a
+3
d
…………..
(ii)
On subtracting equation
(i)
from
(ii)
, we get,
– 10 = 2
d
d
= – 5
From equation
(i)
, putting the value of d,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 the given A.P.,
a
= 5 and
a
4
= 19/2
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Therefore, putting the values here,
a
4
=
a
+(4-1)
d
19/2 =
5
+3d
(19/2) – 5 = 3d
3d = 9/2
d = 3/2
a
2
=
a
+(2-1)
d
a
2
=
5
+3/2
a
2
= 13/2
a
3
=
a
+(3-1)
d
a
3
=
5
+2×3/2
a
3
=
8
Therefore, the missing terms are 13/2 and 8, respectively.
(iv) For the given A.P.,
a
= −4 and
a
6
= 6
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Therefore, putting the values here,
a
6
= a+(6−1)d
6 = − 4+5
d
10 = 5
d
d
= 2
a
2
=
a
+
d
= − 4+2 = −2
a
3
=
a
+2
d
= − 4+2(2) = 0
a
4
=
a
+3
d
= − 4+ 3(2) = 2
a
5
=
a
+4
d
= − 4+4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v)
For the given A.P.,
a
2
= 38
a
6
= −22
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
Therefore, putting the values here,
a
2
=
a
+(2−1)
d
38 =
a
+
d
…………………….
(i)
a
6
=
a
+(6−1)
d
−22 =
a
+5
d
………………….
(ii)
On subtracting equation
(i)
from
(ii)
, we get
− 22 − 38 = 4
d
−60 = 4
d
d
= −15
a
=
a
2
−
d
= 38 − (−15) = 53
a
3
=
a
+ 2
d
= 53 + 2 (−15) = 23
a
4
=
a
+ 3
d
= 53 + 3 (−15) = 8
a
5
=
a
+ 4
d
= 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7, respectively.
4. Which term of the A.P. 3, 8, 13, 18, … is 78?
Solutions:
Given the A.P. series as3, 8, 13, 18, …
First term, a = 3
Common difference, d = a
2
− a
1
= 8 − 3 = 5
Let the
n
th
term of given A.P. be 78. Now as we know,
a
n
=
a
+(
n
−1)
d
Therefore,
78 = 3+(
n
−1)5
75 = (
n
−1)5
(
n
−1) = 15
n
= 16
Hence, 16
th
term of this A.P. is 78.
5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
Solutions:
(i) Given, 7, 13, 19, …, 205 is the A.P
Therefore
First term, a
= 7
Common difference, d
=
a
2
−
a
1
= 13 − 7 = 6
Let there are
n
terms in this A.P.
a
n
= 205
As we know, for an A.P.,
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.
First term, a = 18
Common difference, d = a
2
-a
1
=
d = (31-36)/2 = -5/2
Let there are n terms in this A.P.
a
n
= -47
As we know, for an A.P.,
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.
6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Solution:
For the given series, A.P. 11, 8, 5, 2..
First term, a
= 11
Common difference, d
=
a
2
−
a
1
= 8−11 = −3
Let −150 be the
n
th
term of this A.P.
As we know, for an A.P.,
a
n
=
a
+(
n
−1)
d
-150 = 11+(
n
-1)(-3)
-150 = 11-3
n
+3
-164 = -3
n
n
= 164/3
Clearly,
n
is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P.
7. Find the 31
st
term of an A.P. whose 11
th
term is 38 and the 16
th
term is 73.
Solution:
Given that,
11
th
term, a
11
= 38
and 16
th
term, a
16
= 73
We know that,
a
n
= a+(n−1)d
a
11
= a+(11−1)d
38 = a+10d ……………………………….
(i)
In the same way,
a
16
=
a
+(16−1)
d
73 =
a
+15
d
…………………………………………
(ii)
On subtracting equation
(i)
from
(ii)
, we get
35 = 5
d
d
= 7
From equation
(i)
, we can write,
38 =
a
+10×(7)
38 − 70 = a
a = −32
a
31
= a +(31−1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31
st
term is 178.
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.
Solution:
Given that,
3
rd
term, a
3
= 12
50
th
term, a
50
= 106
We know that,
a
n
=
a
+(
n
−1)
d
a
3
=
a
+(3−1)
d
12 =
a
+2
d
…………………………….
(i)
In the same way,
a
50
=
a
+(50−1)
d
106 =
a
+49
d
………………………….
(ii)
On subtracting equation
(i)
from
(ii)
, we get
94 = 47
d
d
= 2 = common difference
From equation
(i)
, we can write now,
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.
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?
Solution:
Given that,
3
rd
term, a
3
= 4
and 9
th
term, a
9
= −8
We know that,
a
n
=
a
+(
n
−1)
d
Therefore,
a
3
=
a
+(3−1)
d
4 =
a
+2
d
………………………………………
(i)
a
9
=
a
+(9−1)
d
−8 =
a
+8
d
…………………………………………………
(ii)
On subtracting equation
(i)
from
(ii)
, we will get here,
−12 = 6
d
d
= −2
From equation
(i)
, we can write,
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 = 8−2
n
+2
2
n
= 10
n
= 5
Hence, 5
th
term of this A.P. is 0.
10. If 17
th
term of an A.P. exceeds its 10
th
term by 7. Find the common difference.
Solution:
We know that, for an A.P series;
a
n
=
a
+(
n
−1)
d
a
17
=
a
+(17−1)
d
a
17
=
a
+16
d
In the same way,
a
10
=
a
+9
d
As it is given in the question,
a
17
−
a
10
= 7
Therefore,
(
a
+16
d
)−(
a
+9
d
) = 7
7
d
= 7
d
= 1
Therefore, the common difference is 1.
11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54
th
term?
Solution:
Given A.P. is 3, 15, 27, 39, …
first term, a
= 3
common difference, d
=
a
2
−
a
1
= 15 − 3 = 12
We know that,
a
n
=
a
+(
n
−1)
d
Therefore,
a
54
=
a
+(54−1)
d
⇒3+(53)(12)
⇒3+636 = 639
a
54
= 639+132=771
We have to find the term of this A.P. which is 132 more than a
54,
i.e.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.
Or another method is;
Let
n
th
term be 132 more than 54
th
term.
n
= 54 + 132/2
= 54 + 11 = 65
th
term
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?
Solution:
Let, the first term of two APs be
a
1
and
a
2
respectively
And the common difference of these APs be
d
.
For the first A.P.,we know,
a
n
=
a
+(
n
−1)
d
Therefore,
a
100
=
a
1
+(100−1)
d
=
a
1
+ 99d
a
1000
=
a
1
+(1000−1)
d
a
1000
=
a
1
+999
d
For second A.P., we know,
a
n
=
a
+(
n
−1)
d
Therefore,
a
100
=
a
2
+(100−1)
d
=
a
2
+99
d
a
1000
=
a
2
+(1000−1)
d
=
a
2
+999
d
Given that, difference between 100
th
term of the two APs = 100
Therefore, (
a
1
+99
d
) − (
a
2
+99
d
) = 100
a
1
−
a
2
= 100………………………………………………………………..
(i)
Difference between 1000
th
terms of the two APs
(
a
1
+999
d
) − (
a
2
+999
d
) =
a
1
−
a
2
From equation
(i)
,
This difference,
a
1
−
a
2
= 100
Hence, the difference between 1000
th
terms of the two A.P. will be 100.
13. How many three digit numbers are divisible by 7?
Solution:
First three-digit number that is divisible by 7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, …
All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
As we know, the largest possible three-digit number is 999.
When we divide 999 by 7, the remainder will be 5.
Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
first term, a = 105
common difference, d = 7
a
n
= 994
n = ?
As we know,
a
n
= a+(n−1)d
994 = 105+(n−1)7
889 = (n−1)7
(n−1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 that is greater than 10 is 12.
Next multiple will be 16.
Therefore, the series formed as;
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, now;
12, 16, 20, 24, …, 248
Let 248 be the
n
th
term of this A.P.
first term, a
= 12
common difference, d
= 4
a
n
= 248
As we know,
a
n
= a+(n−1)d
248 = 12+(
n
-1)×4
236/4 = n-1
59 = n-1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
15. For what value of
n
, are the
n
th
terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Solution:
Given two APs as; 63, 65, 67,… and 3, 10, 17,….
Taking first AP,
63, 65, 67, …
First term, a = 63
Common difference, d = a
2
−a
1
= 65−63 = 2
We know, n
th
term of this A.P. = a
n
= a+(n−1)d
a
n
= 63+(
n
−1)2 = 63+2
n
−2
a
n
= 61+2
n
……………………………………….
(i)
Taking second AP,
3, 10, 17, …
First term, a = 3
Common difference, d = a
2
− a
1
= 10 − 3 = 7
We know that,
n
th
term of this A.P. = 3+(
n
−1)7
a
n
= 3+7
n
−7
a
n
= 7
n
−4 ………………………………………………………..
(ii)
Given,
n
th
term of these A.P.s are equal to each other.
Equating both these equations, we get,
61+2
n
= 7
n
−4
61+4 = 5
n
5
n
= 65
n
= 13
Therefore, 13
th
terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7
th
term exceeds the 5
th
term by 12.
Solutions:
Given,
Third term, a
3
= 16
As we know,
a
+(3−1)
d
= 16
a
+2
d
= 16 ………………………………………. (i)
It is given that, 7
th
term exceeds the 5
th
term by 12.
a
7
−
a
5
= 12
[
a
+(7−1)
d
]−[
a
+(5−1)
d
]= 12
(
a
+6
d
)−(
a
+4
d
) = 12
2
d
= 12
d
= 6
From equation (i), we get,
a
+2(6) = 16
a
+12 = 16
a
= 4
Therefore, A.P. will be4, 10, 16, 22, …
17. Find the 20
th
term from the last term of the A.P. 3, 8, 13, …, 253.
Solution:
Given A.P. is3, 8, 13, …, 253
Common difference, d= 5.
Therefore, we can write the given AP in reverse order as;
253, 248, 243, …, 13, 8, 5
Now for the new AP,
first term, a = 253
and common difference, d = 248 − 253 = −5
n = 20
Therefore, using nth term formula, we get,
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 of the AP 3, 8, 13, …, 253
.
is 158.
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
.
Solution:
We know that, the nth term of the AP is;
a
n
=
a
+(
n
−1)
d
a
4
=
a
+(4−1)
d
a
4
=
a
+3
d
In the same way, we can write,
a
8
=
a
+7
d
a
6
=
a
+5
d
a
10
=
a
+9
d
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)
On subtracting equation
(i)
from
(ii)
, we get,
2d = 22 − 12
2d = 10
d
= 5
From equation
(i)
, we get,
a
+5
d
= 12
a
+5(5) = 12
a
+25 = 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.
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?
Solution:
It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.
Therefore, after 1995, the salaries of each year are;
5000, 5200, 5400, …
Here, first term,
a
= 5000
and common difference, d
= 200
Let after
n
th
year, his salary be Rs 7000.
Therefore, by the n
th
term formula of AP,
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.
20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the
n
th
week, her weekly savings become Rs 20.75, find
n.
Solution:
Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.
Hence,
First term, a = 5
and common difference, d = 1.75
Also given,
a
n
= 20.75
Find, n = ?
As we know, by the n
th
term formula,
a
n
=
a
+(
n
−1)
d
Therefore,
20.75 = 5+(
n
-1)×1.75
15.75 = (
n
-1)×1.75
(
n
-1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n
-1 = 9
n
= 10
Hence,
n
is 10.
Benefits of Using NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2
Clarity of Concepts
: Step-by-step solutions help students grasp the fundamentals of Arithmetic Progressions (AP), including the nth term and sum formulas.
Exam Preparation
: Solutions are aligned with the NCERT syllabus, ensuring comprehensive coverage of exam-relevant topics.
Improved Problem-Solving
: Detailed explanations enhance analytical thinking and accuracy in solving AP problems.
Real-Life Applications
: Understanding AP concepts aids in solving practical problems related to sequences and patterns.
Time Management
: Simplified solutions enable efficient practice and revision, helping students manage time effectively during exams.
Confidence Booster
: Practicing these solutions boosts confidence and reduces exam anxiety.
