NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5

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

Edited By Ramraj Saini | Updated on Nov 17, 2023 04:52 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2

NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5 Arithmetic Progressions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This class 10 ex 5.2 the consists of 20 questions solved using the formula of the sum of n terms of the Arithmetic Progression. It also has some word problems too to enhance the understanding of this concept.

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

NCERT solutions for Exercise 5.2 Class 10 Maths Chapter 5 Arithmetic Progressions focus on the Arithmetic Progression’s basic notion, i.e., how an Arithmetic Progression is formed? Also, it stresses to clear the understanding of the nth term of the Arithmetic Progression. 10th class Maths exercise 5.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Arithmetic Progressions Class 10 Chapter 5 Excercise: 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:

$\small a_n$

(i)

(ii)

(iii)

(iv)

(v)

$\small -18$

$\small ...$

$\small -18.9$

$\small 3.5$

$\small ...$

$\small -3$

$\small 2.5$

$\small ...$

105

$\small ...$

$\small -5$

$\small 3.6$

$\small ...$

Answer:

(i)
It is given that
$a=7, d = 3 , n = 8$
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) It is given that
$a=-18, n = 10, a_{10} = 0$
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) It is given that
$d=-3, n = 18, a_{18} = -5$
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) It is given that
$a=-18.9, d = 2.5, a_{n} = 3.6$
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) It is given that
$a=3.5, d = 0, n = 105$
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 is
$\small 10,7,4,...,$
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,
$a_{30} = 10+(30-1)(-3)$
$a_{30} = 10+(29)(-3)$
$a_{30} = 10-87 = -77$
Therefore, $\small 30$ th term of the AP: $\small 10,7,4,...,$ is -77
Hence, 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 is
$\small -3,-\frac{1}{2},2,...,$
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,
$a_{11} = -3+(11-1)\left ( \frac{5}{2} \right )$
$a_{11} = -3+(10)\left ( \frac{5}{2} \right )$
$a_{11} = -3+5\times 5 = -3+25 = 22$
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 AP series is
$\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26$
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,
$a_2= a_1+d$
$a_2= 2+12 = 14$
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,
$a_2= a_1+d$
$a_1= a = 13 - d= 13-(-5 ) = 18$
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,
$a_2= a_1+d$
$a_2 = 5+\frac{3}{2} = \frac{13}{2}$
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,
$a_2= a_1+d$
$a_2 = -4+2 = -2$
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 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,
$a_2= a_1+d$
$a_1 = 38-(-15) = 38+15 = 53$
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 is
$\small 3,8,13,18,...,$
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 is
$\small 7,13,19,...,205$
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 is
$\small 18,15\frac{1}{2},13,...,-47$
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 is
$\small 11,8,5,2...$
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:

It is given that
$\small 11$ 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:

It is given that
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:

It is given that
$\small 3$ 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:

It is given that
$\small 17$ 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 AP is
$\small 3,15,27,39,...$
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:

It is given that
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 saving will become Rs 20.75

Benefits of NCERT Solutions for Class 10 Maths Exercise 5.2

NCERT solutions for Class 10 Maths exercise 5.2 helps solve and revise all these exercises’ questions.
You would be able to gain more points if you go through Class 10 Maths chapter 5 NCERT solutions for exercise 5.2, and if you practice it thoroughly, it will help you score well in maths in examinations.
Exercise 5.2 in chapter 5 of maths for Class 10 is based on fundamental characteristics of Arithmetic Progression, i.e., any two consecutive terms of the Arithmetic Progression differ by the constant numerical value.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. How do we find a particular progression is an Arithmetic Progression?

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

2. How is the nth term of the Arithmetic Progression found?

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.

3. How can we find that a particular term belongs to a certain Arithmetic Progression?

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.

4. Can a whole Arithmetic Progression be determined by just any two non-consecutive terms of the Arithmetic Progression?

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.

5. How can we reverse the Arithmetic Progression?

For this, just do two things i.e.

Take the last term to be the first term.
Reverse the sign of the common difference(if it was +2, do it -2 and vice versa)

6. Any term or the Common Difference can be in fraction or not?

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

7. What is n in the Arithmetic Progression?

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.

8. How can we find the Common Difference of the Arithmetic Progression?

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

9. According to the NCERT answers for Class 10 Maths chapter 5 exercise 5.2, what is the nth term?

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.

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

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.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

sadhu ashram ramgart road aligarh

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

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vaibhavi bhalsod 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2

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