NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 - Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 - Arithmetic Progressions

Updated on 02 Jun 2025, 02:26 PM IST

The exercise demonstrates how to use the general formula of nth term to study arithmetic progressions better. The general formula helps identify any sequence term and demonstrates number growth patterns in a controlled way. Real-life scenarios displaying natural sequences form a significant part of the problems presented in this section. Through such problems we learn to think rationally as we use numerical methods in our daily life for savings calculations and measuring distances and following routines.

This Story also Contains

  1. NCERT Solutions Class 10 Maths Chapter 5: Exercise 5.2
  2. Access Solution of Arithmetic Progression Class 10 Chapter 5 Exercise: 5.2
  3. Topics Covered in Chapter 5, Arithmetic Progression: Exercise 5.2
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Class 10 Maths Chapter 5  Exercise 5.2 - Arithmetic Progressions
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 - Arithmetic Progressions

This section of the NCERT Solutions for Class 10 Maths focuses on solving questions based on the nth term of an AP using step-by-step reasoning from NCERT Books. The educational material guides students through three steps which involve recognizing unknown sequence components and sequence position identification together with algebra-based practical problem solving. This exercise contains problems which teach students to analyze details effectively to handle mathematical situations with precision and certainty.

NCERT Solutions Class 10 Maths Chapter 5: Exercise 5.2


Access Solution of Arithmetic Progression Class 10 Chapter 5 Exercise: 5.2

Q1 Fill in the blanks in the following table, given that a is the first term, d the common difference and $\small a_n$ the $\small n$ th term of the AP:


a

d

n

$\small a_n$

(i)

(ii)

(iii)

(iv)

(v)

7

$\small -18$

$\small ...$

$\small -18.9$

$\small 3.5$

3

$\small ...$

$\small -3$

$\small 2.5$

0

8

10

18

$\small ...$

105

$\small ...$

0

$\small -5$

$\small 3.6$

$\small ...$

Answer:
(i) Given $a=7, d = 3 , n = 8$
Now, we know that
$a_n = a+(n-1)d$

$a_8 = 7+(8-1)3= 7+7\times 3 = 7+21 = 28$
Therefore,
$a_8 = 28$

(ii) Given $a=-18, n = 10, a_{10} = 0$
Now, we know that
$a_n = a+(n-1)d$
$a_{10} = -18+(10-1)d$
$0 +18=9d$
$d = \frac{18}{9}=2$

(iii) Given $d=-3, n = 18, a_{18} = -5$
Now, we know that
$a_n = a+(n-1)d$
$a_{18} = a+(18-1)(-3)$
$-5=a+17\times (-3)$
$a = 51-5 = 46$
Therefore,
$a = 46$

(iv) Given $a=-18.9, d = 2.5, a_{n} = 3.6$
Now, we know that
$a_n = a+(n-1)d$
$a_{n} = -18.9+(n-1)2.5$
$3.6+18.9= 2.5n-2.5$
$n = \frac{22.5+2.5}{2.5}= \frac{25}{2.5}= 10$
Therefore,
$n = 10$

(v) Given $a=3.5, d = 0, n = 105$
Now, we know that
$a_n = a+(n-1)d$
$a_{105} = 3.5+(105-1)0$
$a_{105} = 3.5$
Therefore,
$a_{105} = 3.5$

Q2 (i) Choose the correct choice in the following and justify: $\small 30$ th term of the AP: $\small 10,7,4,...,$ is

(A) $\small 97$ (B) $\small 77$ (C) $\small -77$ (D) $\small -87$

Answer:
Given series $\small 10,7,4,...,$
Here, $a = 10$ and $d = 7 - 10 = -3$
Now, we know that
$a_n = a+(n-1)d$
It is given that $n = 30$
Therefore, after putting values, we get:
$a_{30} = 10+(30-1)(-3)$
$a_{30} = 10+(29)(-3)$
$a_{30} = 10-87 = -77$
Therefore, $\small 30$ th term of the AP: $\small 10,7,4,...,$ is -77
Hence, the correct answer is (C)

Q2 (ii) Choose the correct choice in the following and justify : 11th term of the AP: $\small -3,-\frac{1}{2},2,...,$ is

(A) $\small 28$ (B) $\small 22$ (C) $\small -38$ (D) $\small -48\frac{1}{2}$

Answer:
Given series $\small -3,-\frac{1}{2},2,...,$
Here, $a = -3$ and $d =-\frac{1}{2} -(-3)= -\frac{1}{2} + 3 = \frac{-1+6}{2}= \frac{5}{2}$
Now, we know that
$a_n = a+(n-1)d$
It is given that $n = 11$
Therefore, after putting values, we get:
$a_{11} = -3+(11-1)\left ( \frac{5}{2} \right )$
$a_{11} = -3+(10)\left ( \frac{5}{2} \right )$
$a_{11} = -3+5\times 5 = -3+25 = 22$
Therefore, 11th term of the AP: $\small -3,-\frac{1}{2},2,...,$ is 22
Hence, the Correct answer is (B)

Q3 (i) In the following APs, find the missing terms in the boxes: $\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26$

Answer:
Given series $\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26$
Here, $a = 2 , n = 3 \ and \ a_3 = 26$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_3 =2+(3-1)d$
$\Rightarrow 26 -2=(2)d$
$\Rightarrow d = \frac{24}{2}= 12$
Now, after putting values we get:
$a_2= a_1+d$
$a_2= 2+12 = 14$
Therefore, the missing term is 14

Q3 (ii) In the following APs, find the missing terms in the boxes: $\small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3$

Answer:
Given AP series is
$\small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3$
Here, $a_2 = 13 , n = 4 \ and \ a_4 = 3$
Now,
$a_2= a_1+d$
$a_1= a = 13 - d$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_4 =13-d+(4-1)d$
$\Rightarrow 3-13=-d+3d$
$\Rightarrow d = -\frac{10}{2}= -5$
Now, after putting values we get:
$a_2= a_1+d$
$a_1= a = 13 - d= 13-(-5 ) = 18$
And
$a_3=a_2+d$
$a_3=13-5 = 8$
Therefore, missing terms are 18 and 8
AP series is 18,13,8,3

Q3 (iii) In the following APs, find the missing terms in the boxes: $\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}$

Answer:
Given AP series is $\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}$
Here, $a = 5 , n = 4 \ and \ a_4 = 9\frac{1}{2}= \frac{19}{2}$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_4 =5+(4-1)d$
$\Rightarrow \frac{19}{2} -5=3d$
$\Rightarrow d = \frac{19-10}{2\times 3} = \frac{9}{6} = \frac{3}{2}$
Now, after putting values we get:
$a_2= a_1+d$
$a_2 = 5+\frac{3}{2} = \frac{13}{2}$
And
$a_3=a_2+d$
$a_3=\frac{13}{2}+\frac{3}{2} = \frac{16}{2} = 8$
Therefore, missing terms are $\frac{13}{2}$ and 8
AP series is $5,\frac{13}{2}, 8 , \frac{19}{2}$

Q3 (iv) In the following APs, find the missing terms in the boxes: $\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6$
Answer:

Given AP series is $\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6$
Here, $a = -4 , n = 6 \ and \ a_6 = 6$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_6 =-4+(6-1)d$
$\Rightarrow 6+4 = 5d$
$\Rightarrow d = \frac{10}{5} = 2$
Now, after putting values we get:
$a_2= a_1+d$
$a_2 = -4+2 = -2$
And
$a_3=a_2+d$
$a_3=-2+2 = 0$
And
$a_4 = a_3+d$
$a_4 = 0+2 = 2$
And
$a_5 = a_4 + d$
$a_5 = 2+2 = 4$
Therefore, missing terms are -2, 0, 2, 4
AP series is -4, -2, 0, 2, 4, 6

Q3 (v) In the following APs, find the missing terms in the boxes: $\small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22$

Answer:
Given the AP series is
$\small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22$
Here, $a_2 = 38 , n = 6 \ and \ a_6 = -22$
Now,
$a_2=a_1+d$
$a_1=a =38-d \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow a_6 =38-d+(6-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$\Rightarrow -22-38-=-d+5d$
$\Rightarrow d = -\frac{60}{4} = - 15$
Now, after putting values we get:
$a_2= a_1+d$
$a_1 = 38-(-15) = 38+15 = 53$
And
$a_3=a_2+d$
$a_3=38-15 = 23$
And
$a_4 = a_3+d$
$a_4 = 23-15 = 8$
And
$a_5 = a_4 + d$
$a_5 =8-15 = -7$
Therefore, missing terms are 53, 23, 8, -7
AP series is 53, 38, 23, 8, -7, -22

Q4 Which term of the AP : $\small 3,8,13,18,...,$ is $\small 78$ ?

Answer:
Given AP series $\small 3,8,13,18,...,$
Let suppose that nth term of AP is 78
Here, $a = 3$ and $d = a_2-a_1 = 8 - 3 = 5$
Now, we know that that
$a_n = a + (n-1)d$
$\Rightarrow 78 = 3 + (n-1)5$
$\Rightarrow 78 -3 = 5n-5$
$\Rightarrow n = \frac{75 +5}{5}= \frac{80}{5} = 16$
Therefore, value of 16th term of given AP is 78

Q5 (i) Find the number of terms in each of the following APs: $\small 7,13,19,...,205$

Answer:
Given AP series $\small 7,13,19,...,205$
Let's suppose there are n terms in given AP
Then,
$a = 7 , a_n = 205$
And
$d= a_2-a_1 = 13-7 = 6$
Now, we know that
$a_n =a + (n-1)d$
$\Rightarrow 205=7 + (n-1)6$
$\Rightarrow 205-7 = 6n-6$
$\Rightarrow n = \frac{198+6}{6} = \frac{204}{6} = 34$
Therefore, there are 34 terms in given AP

Q5 (ii) Find the number of terms in each of the following APs: $\small 18,15\frac{1}{2},13,...,-47$

Answer:
Given AP series $\small 18,15\frac{1}{2},13,...,-47$
suppose there are n terms in given AP
Then,
$a = 18 , a_n = -47$
And
$d= a_2-a_1 = \frac{31}{2}-18 = \frac{31-36}{2} = -\frac{5}{2}$
Now, we know that
$a_n =a + (n-1)d$
$\Rightarrow -47=18 + (n-1)\left ( -\frac{5}{2} \right )$
$\Rightarrow -47-18= -\frac{5n}{2}+\frac{5}{2}$
$\Rightarrow -\frac{5n}{2}= -65-\frac{5}{2}$
$\Rightarrow -\frac{5n}{2}= -\frac{135}{2}$
$\Rightarrow n = 27$
Therefore, there are 27 terms in given AP

Q6 Check whether $\small -150$ is a term of the AP : $\small 11,8,5,2...$

Answer:
Given AP series $\small 11,8,5,2...$
Here, $a = 11$ and $d = a_2-a_1 = 8-11 = -3$
Now,
suppose -150 is nth term of the given AP
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow -150 = 11+(n-1)(-3)$
$\Rightarrow -150- 11=-3n+3$
$\Rightarrow =n = \frac{161+3}{3}= \frac{164}{3} = 54.66$
Value of n is not an integer
Therefore, -150 is not a term of AP $\small 11,8,5,2...$

Q7 Find the $\small 31$ st term of an AP whose $\small 11$ th term is $\small 38$ and the $\small 16$ th term is $\small 73$ .

Answer:
Given: $\small 11$ th term of an AP is $\small 38$ and the $\small 16$ th term is $\small 73$
Now,
$a_{11} =38= a+ 10d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_{16} =73= a+ 15d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= -32 \ \ \ and \ \ \ d = 7$
Now,
$a_{31} = a+30d = -32 + 30\times 7 = -32+210 = 178$
Therefore, 31st terms of given AP is 178

Q8 An AP consists of $\small 50$ terms of which $\small 3$ rd term is $\small 12$ and the last term is $\small 106$ . Find the $\small 29$ th term.
Answer:

Given: AP consists of $\small 50$ terms of which $\small 3$ rd term is $\small 12$ and the last term is $\small 106$
Now,
$a_3 = 12=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_{50} = 106=a+49d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= 8 \ \ \ and \ \ \ d = 2$
Now,
$a_{29} = a+28d=8+28\times 2 = 8 +56 = 64$
Therefore, 29th term of given AP is 64

Q9 If the $\small 3$ rd and the $\small 9$ th terms of an AP are $\small 4$ and $\small -8$ respectively, which term of this AP is zero?
Answer:

Given: $\small 3$ rd and the $\small 9$ th terms of an AP are $\small 4$ and $\small -8$ respectively
Now,
$a_3 = 4=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_{9} = -8=a+8d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= 8 \ \ \ and \ \ \ d = -2$
Now,
Let nth term of given AP is 0
Then,
$a_{n} = a+(n-1)d$
$0 = 8+(n-1)(-2)$
$2n = 8+2= 10$
$n = \frac{10}{2} = 5$
Therefore, 5th term of given AP is 0

Q10 The $\small 17$ th term of an AP exceeds its $\small 10$ th term by $\small 7$ . Find the common difference.
Answer:

Given: $\small 17$ th term of an AP exceeds its $\small 10$ th term by $\small 7$
i.e.
$a_{17}= a_{10}+7$
$\Rightarrow a+16d = a+9d+7$
$\Rightarrow a+16d - a-9d=7$
$\Rightarrow 7d=7$
$\Rightarrow d = 1$
Therefore, the common difference of AP is 1

Q11 Which term of the AP : $\small 3,15,27,39,...$ will be $\small 132$ more than its $\small 54$ th term?
Answer:

Given series $\small 3,15,27,39,...$
Here, $a= 3$ and $d= a_2-a_1 = 15 - 3 = 12$
Now, let's suppose nth term of given AP is $\small 132$ more than its $\small 54$ th term
Then,
$a_n= a_{54}+132$
$\Rightarrow a+(n-1)d = a+53d+132$
$\Rightarrow 3+(n-1)12 = 3+53\times 12+132$
$\Rightarrow 12n = 3+636+132+12$
$\Rightarrow 12n = 636+132+12$
$\Rightarrow n = \frac{780}{12}= 65$
Therefore, 65th term of given AP is $\small 132$ more than its $\small 54$ th term

Q12 Two APs have the same common difference. The difference between their $\small 100$ th terms is $\small 100$ , what is the difference between their $\small 1000$ th terms?
Answer:

Given: Two APs have the same common difference and difference between their $\small 100$ th terms is $\small 100$
i.e.
$a_{100}-a'_{100}= 100$
Let common difference of both the AP's is d
$\Rightarrow a+99d-a'-99d=100$
$\Rightarrow a-a'=100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - (i)$
Now, difference between 1000th term is
$a_{1000}-a'_{1000}$
$\Rightarrow a+999d -a'-999d$
$\Rightarrow a-a'$
$\Rightarrow 100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i) )$
Therefore, difference between 1000th term is 100

Q 13 How many three-digit numbers are divisible by $\small 7$ ?

Answer:

We know that the first three-digit number divisible by 7 is 105 and last three-digit number divisible by 7 is 994
Therefore,
$a = 105 , d = 7 \ and \ a_n = 994$
Let there are n three digit numbers divisible by 7
Now, we know that
$a_n = a+ (n-1)d$
$\Rightarrow 994 = 105 + (n-1)7$
$\Rightarrow 7n = 896$
$\Rightarrow n = \frac{896}{7} = 128$
Therefore, there are 128 three-digit numbers divisible by 7

Q14 How many multiples of $\small 4$ lie between $\small 10$ and $\small 250$ ?
Answer:

We know that the first number divisible by 4 between 10 to 250 is 12 and last number divisible by 4 is 248
Therefore,
$a = 12 , d = 4 \ and \ a_n = 248$
Let there are n numbers divisible by 4
Now, we know that
$a_n = a+ (n-1)d$
$\Rightarrow 248 = 12 + (n-1)4$
$\Rightarrow 4n = 240$
$\Rightarrow n = \frac{240}{4} = 60$
Therefore, there are 60 numbers between 10 to 250 that are divisible by 4

Q15 For what value of $\small n$ , are the $\small n$ th terms of two APs: $\small 63,65,67,...$ and $\small 3,10,17,...$ equal?

Answer:
Given two AP's are
$\small 63,65,67,...$ and $\small 3,10,17,...$
Let first term and the common difference of two AP's are a , a' and d , d'
$a = 63 \ , d = a_2-a_1 = 65-63 = 2$
And
$a' = 3 \ , d' = a'_2-a'_1 = 10-3 = 7$
Now,
Let nth term of both the AP's are equal
$a_n = a'_n$
$\Rightarrow a+(n-1)d=a'+(n-1)d'$
$\Rightarrow 63+(n-1)2=3+(n-1)7$
$\Rightarrow 5n=65$
$\Rightarrow n=\frac{65}{5} = 13$
Therefore, the 13th term of both the AP's are equal

Q16 Determine the AP whose third term is $\small 16$ and the $\small 7$ th term exceeds the $\small 5$ th term by $\small 12$ .
Answer:

It is given that
3rd term of AP is $\small 16$ and the $\small 7$ th term exceeds the $\small 5$ th term by $\small 12$
i.e.
$a_3=a+2d = 16 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_7=a_5+12$
$a+6d=a+4d+12$
$2d = 12$
$d = 6$
Put the value of d in equation (i) we will get
$a = 4$
Now, AP with first term = 4 and common difference = 6 is
4,10,16,22,.....

Q17 Find the $\small 20$ th term from the last term of the AP : $\small 3,8,13,...,253$ .

Answer:
Given AP is
$\small 3,8,13,...,253$
Here, $a = 3 \ and \ a_n = 253$
And
$d = a_2-a_1=8-3 = 5$
Let suppose there are n terms in the AP
Now, we know that
$a_n = a+(n-1)d$
$253= 3+(n-1)5$
$5n = 255$
$n = 51$
So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting
Therefore,
$a_{32} = a+31d$
$a_{32} = 3+31\times 5 = 3+155 = 158$
Therefore, 20th term from the of given AP is 158

Q18 The sum of the $\small 4$ th and $\small 8$ th terms of an AP is $\small 24$ and the sum of the $\small 6$ th and $\small 10$ th terms is $\small 44$ Find the first three terms of the AP.
Answer:

It is given that
sum of the < img alt="\small 4" class="fr-fic fr-dii" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%204"> th and $\small 8$ th terms of an AP is $\small 24$ and the sum of the $\small 6$ th and $\small 10$ th terms is $\small 44$
i.e.
$a_4+a_8=24$
$\Rightarrow a+3d+a+7d=24$
$\Rightarrow 2a+10d=24$
$\Rightarrow a+5d=12 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_6+a_{10}=44$
$\Rightarrow a+5d+a+9d=44$
$\Rightarrow 2a+14d=44$
$\Rightarrow a+7d=22 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= -13 \ and \ d= 5$
Therefore,first three of AP with a = -13 and d = 5 is
-13,-8,-3

Q19 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 is given that
Subba Rao started work at an annual salary of Rs 5000 and received an increment of Rs 200 each year
Therefore, $a = 5000 \ and \ d =200$
Let's suppose after n years his salary will be Rs 7000
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow 7000=5000+(n-1)200$
$\Rightarrow 2000=200n-200$
$\Rightarrow 200n=2200$
$\Rightarrow n = 11$
Therefore, after 11 years his salary will be Rs 7000
after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years
Thus , 1995+10 = 2005

Q20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs $\small 1.75$ . If in the $\small n$ th week, her weekly savings become Rs $\small 20.75$ , find $\small n$

Answer:
It is given that
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs $\small 1.75$
Therefore, $a = 5 \ and \ d = 1.75$
after $\small n$ th week, her weekly savings become Rs $\small 20.75$
Now, we know that
$a_n = a +(n-1)d$
$\Rightarrow 20.75= 5+(n-1)1.75$
$\Rightarrow 15.75= 1.75n-1.75$
$\Rightarrow 1.75n=17.5$
$\Rightarrow n=10$
Therefore, after 10 weeks, her savings will become Rs 20.75

Also Read-

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

Topics Covered in Chapter 5, Arithmetic Progression: Exercise 5.2

1. Application of the nth Term Formula: The formula helps identify any specific term present in an Arithmetic Progression (AP).

2. Determining the Number of Terms: The problem requires determining the complete number of terms in an AP when particular terms and values are provided.

3. Identifying Specific Terms: The process of locating an AP term based on a particular provided value.

4. Solving Word Problems: Real-life sequences, such as distance computation or savings analysis, benefit from applying the AP practical methods.

5. Analysing Patterns: The process of analysing arithmetic sequences includes explaining patterns and structures within their system for making logical deductions.

Also see-

NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: According to the NCERT answers for Class 10 Maths chapter 5 exercise 5.2, what is the nth term?
A:

According to this exercise, the nth term is any term of the Arithmetic Progression that could be calculated by adding Common Difference (n-1) times to the first term of the Arithmetic Progression. This very concept is required to frame the whole Arithmetic Progression. 

Q: What sorts of questions are addressed in the NCERT solutions for Class 10 Maths chapter 5 exercise 5.2?
A:

The questions are based on the concept that the two consecutive terms of the Arithmetic Progression always differ by a constant numerical value. Based on this concept, there are word problems too available in this exercise.

Q: How do we find a particular progression is an Arithmetic Progression?
A:

In Arithmetic Progression, any two consecutive terms differ by a constant numerical value.

Q: How is the nth term of the Arithmetic Progression found?
A:

It could be calculated easily by using the prop[erty that any two consecutive terms differ by a constant numerical value. To calculate it, just add the difference (n-1) times to the first term of the Arithmetic Progression. 

Q: How can we find that a particular term belongs to a certain Arithmetic Progression?
A:

For this, just find the rank of that term in that Arithmetic Progression, and if the position comes out to be in fraction, then that term doesn’t belong to that Arithmetic Progression; otherwise, it is. 

Q: Can a whole Arithmetic Progression be determined by just any two non-consecutive terms of the Arithmetic Progression?
A:

 Yes, it could be done efficiently by using the nth term formula. There would be two unknowns, namely the first term and the common difference, and we would have two equations with us; solving them, we would get those. And once the first term and common difference are calculated then, Arithmetic Progression could be determined. 

Q: How can we reverse the Arithmetic Progression?
A:

For this, just do two things i.e.

  1. Take the last term to be the first term.

  2. Reverse the sign of the common difference(if it was +2, do it -2 and vice versa)

Q: Any term or the Common Difference can be in fraction or not?
A:

Yes, It could be in a fraction. Only the number of terms in the Arithmetic Progression can't be in a fraction.

Q: What is n in the Arithmetic Progression?
A:

It's just the generalized way to represent any term of the Arithmetic Progression. Based on the requirement, it could define the first term, last term, etc.

Q: How can we find the Common Difference of the Arithmetic Progression?
A:

To find the Common Difference of the Arithmetic Progression, just differentiate (n-1)th term from the nth term.

Articles
|
Next
Upcoming School Exams
Ongoing Dates
UP Board 12th Others

10 Aug'25 - 1 Sep'25 (Online)

Ongoing Dates
UP Board 10th Others

11 Aug'25 - 6 Sep'25 (Online)

Certifications By Top Providers
Explore Top Universities Across Globe

Questions related to CBSE Class 10th

On Question asked by student community

Have a question related to CBSE Class 10th ?

Hello,

Yes, you can give the CBSE board exam in 2027.

If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.

Hope it helps !

Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.

Hello Aspirant,

Here's how you can find it:

  • School ID Card: Your registration number is often printed on your school ID card.

  • Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.

  • School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.

  • Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.

Your school is the best place to get this information.

Hello,

It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.

The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.

For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.

Let me know if you need imformation about any exam eligibility criteria.

good wishes for your future!!

Hello Aspirant,

"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).

For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.