NCERT Solutions for Exercise 5.3 Class 10 Maths Chapter 5 - Arithmetic Progressions

# NCERT Solutions for Exercise 5.3 Class 10 Maths Chapter 5 - Arithmetic Progressions

Edited By Ramraj Saini | Updated on Nov 17, 2023 04:55 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.3

NCERT Solutions for Exercise 5.3 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. The sum of n terms formula of the Arithmetic Progression is used to solve 20 problems in class 10 ex 5.3. There are also some word problems to help with comprehension. In exercise 5.3 Class 10 Maths, a number of standard questions are given, which help to enhance knowledge of the topic. It has 14 direct questions aiming to have good practice of the sum of n terms formula; then there are six-word problems in which the sum of n terms formula is used trickly

NCERT solutions for exercise 5.3 Class 10 Maths chapter 5 Arithmetic progression discusses the sum of n terms of the Arithmetic Progression , which provides a wonderful practice of this topic by presenting the standard set of problems. 10th class Maths exercise 5.3 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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## Assess NCERT Solutions for Class 10 Maths chapter 5 exercise 5.3

Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.3

$\small 2,7,12,...,$ to $\small 10$ terms.

Given AP is
$\small 2,7,12,...,$ to $\small 10$ terms
Here, $a = 2 \ and \ n = 10$
And
$d = a_2-a_1=7-2=5$
Now, we know that
$S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S = \frac{10}{2}\left \{ 2\times 2 +(10-1)5\right \}$
$\Rightarrow S = 5\left \{ 4 +45\right \}$
$\Rightarrow S = 5\left \{ 49\right \}$
$\Rightarrow S =245$
Therefore, the sum of AP $\small 2,7,12,...,$ to $\small 10$ terms is 245

$\small -37,-33,-29,...,$ to $\small 12$ terms.

Given AP is
$\small -37,-33,-29,...,$ to $\small 12$ terms.
Here, $a = -37 \ and \ n = 12$
And
$d = a_2-a_1=-33-(-37)=4$
Now, we know that
$S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S = \frac{12}{2}\left \{ 2\times (-37) +(12-1)4\right \}$
$\Rightarrow S = 6\left \{ -74 +44\right \}$
$\Rightarrow S = 5\left \{ -30\right \}$
$\Rightarrow S =-180$
Therefore, the sum of AP $\small -37,-33,-29,...,$ to $\small 12$ terms. is -180

$\small 0.6,1.7,2.8,...,$ to $\small 100$ terms.

Given AP is
$\small 0.6,1.7,2.8,...,$ to $\small 100$ terms..
Here, $a = 0.6 \ and \ n = 100$
And
$d = a_2-a_1=1.7-0.6=1.1$
Now, we know that
$S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S = \frac{100}{2}\left \{ 2\times (0.6) +(100-1)(1.1)\right \}$
$\Rightarrow S = 50\left \{ 1.2 +108.9\right \}$
$\Rightarrow S = 50\left \{ 110.1\right \}$
$\Rightarrow S =5505$
Therefore, the sum of AP $\small 0.6,1.7,2.8,...,$ to $\small 100$ terms. is 5505

$\small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $\small 11$ terms.

Given AP is
$\small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $\small 11$ terms.
Here, $a = \frac{1}{15} \ and \ n = 11$
And
$d = a_2-a_1=\frac{1}{12}-\frac{1}{15}= \frac{5-4}{60}= \frac{1}{60}$
Now, we know that
$S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S = \frac{11}{2}\left \{ 2\times \frac{1}{15} +(11-1)(\frac{1}{60})\right \}$
$\Rightarrow S = \frac{11}{2}\left \{ \frac{2}{15} +\frac{1}{6}\right \}$
$\Rightarrow S = \frac{11}{2}\left \{ \frac{9}{30}\right \}$
$\Rightarrow S =\frac{99}{60}= \frac{33}{20}$
Therefore, the sum of AP $\small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $\small 11$ terms. is $\frac{33}{20}$

$\small 7+10\frac{1}{2}+14+...+84$

Given AP is
$\small 7+10\frac{1}{2}+14+...+84$
We first need to find the number of terms
Here, $a = 7 \ and \ a_n = 84$
And
$d = a_2-a_1=\frac{21}{2}-7= \frac{21-14}{2}= \frac{7}{2}$
Let suppose there are n terms in the AP
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow 84 = 7 + (n-1)\frac{7}{2}$
$\Rightarrow \frac{7n}{2}= 77+\frac{7}{2}$
$\Rightarrow n = 23$
Now, we know that
$S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S = \frac{23}{2}\left \{ 2\times7 +(23-1)(\frac{7}{2})\right \}$
$\Rightarrow S = \frac{23}{2}\left \{ 14 +77\right \}$
$\Rightarrow S = \frac{23}{2}\left \{ 91\right \}$
$\Rightarrow S =\frac{2093}{2}=1046\frac{1}{2}$
Therefore, the sum of AP $\small 7+10\frac{1}{2}+14+...+84$ is $1046\frac{1}{2}$

Q2 (ii) Find the sums given below : $\small 34+32+30+...+10$

Given AP is
$\small 34+32+30+...+10$
We first need to find the number of terms
Here, $a = 34 \ and \ a_n = 10$
And
$d = a_2-a_1=32-34=-2$
Let suppose there are n terms in the AP
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow 10 = 34 + (n-1)(-2)$
$\Rightarrow -26 = -2n$
$\Rightarrow n = 13$
Now, we know that
$S = \frac{n}{2}\left \{ a+a_n \right \}$

$\Rightarrow S = \frac{13}{2}\left \{ 44\right \}$
$\Rightarrow S =13\times 22 = 286$
Therefore, the sum of AP $\small 34+32+30+...+10$ is 286

Q2 (iii) Find the sums given below : $\small -5+(-8)+(-11)+...+(-230)$

Given AP is
$\small -5+(-8)+(-11)+...+(-230)$
We first need to find the number of terms
Here, $a = -5 \ and \ a_n = -230$
And
$d = a_2-a_1=-8-(-5)= -3$
Let suppose there are n terms in the AP
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow -230 = -5 + (n-1)(-3)$
$\Rightarrow -228 = -3n$
$\Rightarrow n = 76$
Now, we know that
$S = \frac{n}{2}\left \{ a+a_n \right \}$
$\Rightarrow S = \frac{76}{2}\left \{ (-5-230 )\right \}$

$\Rightarrow S = 38\left \{ -235\right \}$
$\Rightarrow S = -8930$
Therefore, the sum of AP $\small -5+(-8)+(-11)+...+(-230)$ is -8930

Q3 (i) In an AP: given $\small a=5$ , $\small d=3$ , $\small a_n=50$ , find $\small n$ and $\small S_n$ .

It is given that
$a = 5, d = 3 \ and \ a_n = 50$
Let suppose there are n terms in the AP
Now, we know that
$a_n = a+(n-1)d$
$\Rightarrow 50 = 5 + (n-1)3$
$\Rightarrow 48 = 3n$
$\Rightarrow n = 16$
Now, we know that
$S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S = \frac{16}{2}\left \{ 2\times(5) +(16-1)(3)\right \}$
$\Rightarrow S = 8\left \{ 10+45\right \}$
$\Rightarrow S = 8\left \{ 55\right \}$
$\Rightarrow S =440$
Therefore, the sum of the given AP is 440

Q3 (ii) In an AP: given $\small a=7$ , $\small a_1_3=35$ , find $\small d$ and $\small S_1_3$ .

It is given that
$a = 7 \ and \ a_{13} = 35$
$a_{13}= a+12d = 35$
$= 12d = 35-7 = 28$
$d = \frac{28}{12}= \frac{7}{3}$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{13} = \frac{13}{2}\left \{ 2\times(7) +(13-1)(\frac{7}{3})\right \}$
$\Rightarrow S_{13} = \frac{13}{2}\left \{14 +28\right \}$
$\Rightarrow S_{13} = \frac{13}{2}\left \{42\right \}$
$\Rightarrow S_{13} = 13 \times 21 = 273$
Therefore, the sum of given AP is 273

Q3 (iii) In an AP: given $\small a_1_2=37,d=3,$ find $\small a$ and $\small S_1_2$ .

It is given that
$d = 3 \ and \ a_{12} = 37$
$a_{12}= a+11d = 37$
$= a= 37-11\times 3 = 37-33=4$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{12} = \frac{12}{2}\left \{ 2\times(4) +(12-1)3\right \}$
$\Rightarrow S_{12} = 6\left \{ 8+33\right \}$
$\Rightarrow S_{12} = 6\left \{41\right \}$
$\Rightarrow S_{12} =246$
Therefore, the sum of given AP is 246

Q3 (iv) In an AP: given $\small a_3=15, S_1_0=125,$ find $\small d$ and $\small S_1_0$

It is given that
$\small a_3=15, S_1_0=125$
$a_{3}= a+2d = 15 \ \ \ \ \ \ \ \ -(i)$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{10} = \frac{10}{2}\left \{ 2\times(a) +(10-1)d\right \}$
$\Rightarrow 125 = 5\left \{ 2a+9d\right \}$
$\Rightarrow 2a+9d = 25 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= 17 \ and \ d = -1$
Now,
$a_{10} = a+ 9d = 17 + 9(-1)= 17-9 = 8$
Therefore, the value of d and 10th terms is -1 and 8 respectively

Q3 (v) In an AP: given $\small d=5, S_9=75$ , find $\small a$ and $\small a_9$ .

It is given that
$\small d=5, S_9=75$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{9} = \frac{9}{2}\left \{ 2\times(a) +(9-1)5\right \}$
$\Rightarrow 75= \frac{9}{2}\left \{ 2a +40\right \}$
$\Rightarrow 150= 18a+360$
$\Rightarrow a = -\frac{210}{18}=-\frac{35}{3}$
Now,
$a_{9} = a+ 8d = -\frac{35}{3} + 8(5)= -\frac{35}{3}+40 = \frac{-35+120}{3}= \frac{85}{3}$

Q3 (vi) In an AP: given $\small a=2,d=8,S_n=90,$ find $\small n$ and $\small a_n$ .

It is given that
$\small a=2,d=8,S_n=90,$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 90 = \frac{n}{2}\left \{ 2\times(2) +(n-1)8\right \}$
$\Rightarrow 180 = n\left \{ 4+8n-8\right \}$
$\Rightarrow 8n^2-4n-180=0$
$\Rightarrow 4(2n^2-n-45)=0$
$\Rightarrow 2n^2-n-45=0$
$\Rightarrow 2n^2-10n+9n-45=0$
$\Rightarrow (n-5)(2n+9)=0$
$\Rightarrow n = 5 \ \ and \ \ n = - \frac{9}{2}$
n can not be negative so the only the value of n is 5
Now,
$a_{5} = a+ 4d = 2+4\times 8 = 2+32 = 34$
Therefore, value of n and nth term is 5 and 34 respectively

Q3 (vii) In an AP: given $\small a=8,a_n=62,S_n=210,$ find $\small n$ and $\small d$ .

It is given that
$\small a=8,a_n=62,S_n=210,$
Now, we know that
$a_n = a+(n-1)d$
$62 = 8+(n-1)d$
$(n-1)d= 54 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 210 = \frac{n}{2}\left \{ 2\times(8) +(n-1)d\right \}$
$\Rightarrow 420 = n\left \{ 16+54 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$\Rightarrow 420 = n\left \{ 70 \right \}$
$\Rightarrow n = 6$
Now, put this value in (i) we will get
$d = \frac{54}{5}$
Therefore, value of n and d are $6 \ and \ \frac{54}{5}$ respectively

Q3 (viii) In an AP: given $\small a_n=4,d=2,S_n=-14,$ find $\small n$ and $\small a$ .

It is given that
$\small a_n=4,d=2,S_n=-14,$
Now, we know that
$a_n = a+(n-1)d$
$4 = a+(n-1)2$
$a+2n = 6\Rightarrow a = 6-2n \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow -14 = \frac{n}{2}\left \{ 2\times(a) +(n-1)2\right \}$
$\Rightarrow -28 = n\left \{ 2(6-2n)+2n-2 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$\Rightarrow -28 = n\left \{ 10-2n \right \}$
$\Rightarrow 2n^2-10n-28=0$
$\Rightarrow 2(n^2-5n-14)=0$
$\Rightarrow n^2-7n+2n-14=0$
$\Rightarrow(n+2)(n-7)=0$
$\Rightarrow n = -2 \ \ and \ \ n = 7$
Value of n cannot be negative so the only the value of n is 7
Now, put this value in (i) we will get
a = -8
Therefore, the value of n and a are 7 and -8 respectively

Q3 (ix) In an AP: given $\small a=3,n=8,S=192,$ find $\small d$ .

It is given that
$\small a=3,n=8,S=192,$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 192 = \frac{8}{2}\left \{ 2\times(3) +(8-1)d\right \}$
$\Rightarrow 192=4\left \{6 +7d\right \}$
$\Rightarrow 7d = 48-6$
$\Rightarrow d = \frac{42}{7} = 6$
Therefore, the value of d is 6

Q3 (x) In an AP: given $\small l=28,S=144 \ and \ n = 9$ and there are total $\small 9$ terms. Find $\small a$ .

It is given that
$\small l=28,S=144 \ and \ n = 9$
Now, we know that
$l = a_n = a+(n-1)d$
$28 = a_n = a+(n-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 144 = \frac{9}{2}\left \{ a + a +(n-1)d\right \}$
$\Rightarrow 288 =9\left \{ a+ 28\right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(using \ (i))$
$\Rightarrow a+28= 32$
$\Rightarrow a=4$
Therefore, the value of a is 4

Given AP is
$\small 9,17,25,...$
Here, $a =9 \ and \ d = 8$
And $S_n = 636$
Now , we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 636 = \frac{n}{2}\left \{ 18+(n-1)8 \right \}$
$\Rightarrow 1272 = n\left \{ 10+8n \right \}$
$\Rightarrow 8n^2+10n-1272=0$
$\Rightarrow 2(4n^2+5n-636)=0$
$\Rightarrow 4n^2+53n-48n-636=0$
$\Rightarrow (n-12)(4n+53)=0$
$\Rightarrow n = 12 \ \ and \ \ n = - \frac{53}{4}$
Value of n can not be negative so the only the value of n is 12
Therefore, the sum of 12 terms of AP $\small 9,17,25,...$ must be taken to give a sum of $\small 636$ .

It is given that
$\small a=5,a_n=45,S_n=400,$
Now, we know that
$a_n = a+(n-1)d$
$45 = 5+(n-1)d$
$(n-1)d= 40 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 400 = \frac{n}{2}\left \{ 2\times(5) +(n-1)d\right \}$
$\Rightarrow 800 = n\left \{ 10+40 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$\Rightarrow 800 = n\left \{ 50 \right \}$
$\Rightarrow n = 16$
Now, put this value in (i) we will get
$d = \frac{40}{15}= \frac{8}{3}$
Therefore, value of n and d are $16 \ and \ \frac{8}{3}$ respectively

It is given that
$\small a=17,l=350,d=9,$
Now, we know that
$a_n = a+(n-1)d$
$350 = 17+(n-1)9$
$(n-1)9 = 333$
$(n-1)=37$
$n = 38$

Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{38}= \frac{38}{2}\left \{ 2\times(17) +(38-1)9\right \}$
$\Rightarrow S_{38}= 19\left \{ 34 +333\right \}$
$\Rightarrow S_{38}= 19\left \{367\right \}$
$\Rightarrow S_{38}= 6973$
Therefore, there are 38 terms and their sun is 6973

It is given that
$\small a_{22}=149,d=7,n = 22$
Now, we know that
$a_{22} = a+21d$
$149 = a+21\times 7$
$a = 149 - 147 = 2$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{22}= \frac{22}{2}\left \{ 2\times(2) +(22-1)7\right \}$
$\Rightarrow S_{22}= 11\left \{ 4 +147\right \}$
$\Rightarrow S_{22}= 11\left \{ 151\right \}$
$\Rightarrow S_{22}= 1661$
Therefore, there are 22 terms and their sum is 1661

It is given that
$\small a_{2}=14,a_3=18,n = 51$
And $d= a_3-a_2= 18-14=4$
Now,
$a_2 = a+d$
$a= 14-4 = 10$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{51}= \frac{51}{2}\left \{ 2\times(10) +(51-1)4\right \}$
$\Rightarrow S_{51}= \frac{51}{2}\left \{ 20 +200\right \}$
$\Rightarrow S_{51}= \frac{51}{2}\left \{ 220\right \}$
$\Rightarrow S_{51}= 51 \times 110$
$\Rightarrow S_{51}=5304$
Therefore, there are 51 terms and their sum is 5610

It is given that
$S_7 = 49 \ and \ S_{17}= 289$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{7}= \frac{7}{2}\left \{ 2\times(a) +(7-1)d\right \}$
$\Rightarrow 98= 7\left \{ 2a +6d\right \}$
$\Rightarrow a +3d = 7 \ \ \ \ \ \ \ -(i)$
Similarly,
$\Rightarrow S_{17}= \frac{17}{2}\left \{ 2\times(a) +(17-1)d\right \}$
$\Rightarrow 578= 17\left \{ 2a +16d\right \}$
$\Rightarrow a +8d = 17 \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
a = 1 and d = 2
Now, the sum of first n terms is
$S_n = \frac{n}{2}\left \{ 2\times 1 +(n-1)2 \right \}$
$S_n = \frac{n}{2}\left \{ 2 +2n-2 \right \}$
$S_n = n^2$
Therefore, the sum of n terms is $n^2$

$\small a_n=3+4n$ Also find the sum of the first $\small 15$ terms.

It is given that
$\small a_n=3+4n$
We will check values of $a_n$ for different values of n
$a_1 = 3+4(1) =3+4= 7$
$a_2 = 3+4(2) =3+8= 11$
$a_3 = 3+4(3) =3+12= 15$
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 7 and common difference (d) equals to 4
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(7) +(15-1)4\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 14 +56\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 70\right \}$
$\Rightarrow S_{15}= 15 \times 35$
$\Rightarrow S_{15}= 525$
Therefore, the sum of 15 terms is 525

$\small a_n=9-5n$ . Also find the sum of the first $\small 15$ terms in each case.

It is given that
$\small a_n=9-5n$
We will check values of $a_n$ for different values of n
$a_1 = 9-5(1) =9-5= 4$
$a_2 = 9-5(2) =9-10= -1$
$a_3 = 9-5(3) =9-15= -6$
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 4 and common difference (d) equals to -5
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(4) +(15-1)(-5)\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 8 -70\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ -62\right \}$
$\Rightarrow S_{15}= 15 \times (-31)$
$\Rightarrow S_{15}= -465$
Therefore, the sum of 15 terms is -465

It is given that
the sum of the first $\small n$ terms of an AP is $\small 4n-n^2$
Now,
$\Rightarrow S_n = 4n-n^2$
Now, first term is
$\Rightarrow S_1 = 4(1)-1^2=4-1=3$
Therefore, first term is 3
Similarly,
$\Rightarrow S_2 = 4(2)-2^2=8-4=4$
Therefore, sum of first two terms is 4
Now, we know that
$\Rightarrow S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_2 = \frac{2}{2}\left \{ 2\times 3+(2-1)d \right \}$
$\Rightarrow 4 = \left \{6+d \right \}$
$\Rightarrow d = -2$
Now,
$a_2= a+d = 3+(-2 )= 1$
Similarly,
$a_3= a+2d = 3+2(-2 )= 3-4=-1$
$a_{10}= a+9d = 3+9(-2 )= 3-18=-15$
$a_{n}= a+(n-1)d = 3+(n-1)(-2 )= 5-2n$

Positive integers divisible by 6 are
6,12,18,...
This is an AP with
$here, \ a = 6 \ and \ d = 6$
Now, we know that
$S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{40}= \frac{40}{2}\left \{ 2\times 6+(40-1)6 \right \}$
$\Rightarrow S_{40}= 20\left \{12+234 \right \}$
$\Rightarrow S_{40}= 20\left \{246 \right \}$
$\Rightarrow S_{40}= 4920$
Therefore, sum of the first $\small 40$ positive integers divisible by $\small 6$ is 4920

First 15 multiples of 8 are
8,16,24,...
This is an AP with
$here, \ a = 8 \ and \ d = 8$
Now, we know that
$S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times 8+(15-1)8 \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 16+112 \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 128 \right \}$
$\Rightarrow S_{15}= 15 \times 64 = 960$
Therefore, sum of the first 15 multiple of 8 is 960

The odd number between 0 and 50 are
1,3,5,...49
This is an AP with
$here, \ a = 1 \ and \ d = 2$
There are total 25 odd number between 0 and 50
Now, we know that
$S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{25}= \frac{25}{2}\left \{ 2\times 1+(25-1)2 \right \}$
$\Rightarrow S_{25}= \frac{25}{2}\left \{ 2+48 \right \}$
$\Rightarrow S_{25}= \frac{25}{2}\times 50$
$\Rightarrow S_{25}= 25 \times 25 = 625$
Therefore, sum of the odd numbers between $\small 0$ and $\small 50$ 625

It is given that
Penalty for delay of completion beyond a certain date is Rs $\small 200$ for the first day, Rs $\small 250$ for the second day, Rs $\small 300$ for the third day and penalty for each succeeding day being Rs $\small 50$ more than for the preceding day
We can clearly see that
200,250,300,..... is an AP with
$a = 200 \ and \ d = 50$
Now, the penalty for 30 days is given by the expression
$S_{30}= \frac{30}{2}\left \{ 2\times 200+(30-1)50 \right \}$
$S_{30}= 15\left ( 400+1450 \right )$
$S_{30}= 15 \times 1850$
$S_{30}= 27750$
Therefore, the penalty for 30 days is 27750

It is given that
Each price is decreased by 20 rupees,
Therefore, d = -20 and there are total 7 prizes so n = 7 and sum of prize money is Rs 700 so $S_7 = 700$
Let a be the prize money given to the 1st student
Then,
$S_7 = \frac{7}{2}\left \{ 2a+(7-1)(-20) \right \}$
$700 = \frac{7}{2}\left \{ 2a-120 \right \}$
$2a - 120 = 200$
$a = \frac{320}{2}= 160$
Therefore, the prize given to the first student is Rs 160
Now,
Let $a_2,a_2,...,a_7$ is the prize money given to the next 6 students
then,
$a_2 = a+d = 160+(-20)=160-20=140$
$a_3 = a+2d = 160+2(-20)=160-40=120$
$a_4 = a+3d = 160+3(-20)=160-60=100$
$a_5 = a+4d = 160+4(-20)=160-80=80$
$a_6 = a+5d = 160+5(-20)=160-100=60$
$a_7 = a+6d = 160+6(-20)=160-120=40$
Therefore, prize money given to 1 to 7 student is 160,140,120,100,80,60.40

First there are 12 classes and each class has 3 sections
Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1. Thus every class will plant 3 times the number of their class
Similarly,

No. of trees planted by 3 sections of class 1 = 3

No. of trees planted by 3 sections of class 2 = 6

No. of trees planted by 3 sections of class 3 = 9

No. of trees planted by 3 sections of class 4 = 12
Its clearly an AP with first term (a) = 3 and common difference (d) = 3 and total number of classes (n) = 12

Now, number of trees planted by 12 classes is given by
$S_{12}= \frac{12}{2}\left \{ 2\times 3+(12-1)\times 3 \right \}$
$S_{12}= 6\left ( 6+33 \right )$
$S_{12}= 6 \times 39 = 234$
Therefore, number of trees planted by 12 classes is 234

[ Hint : Length of successive semicircles is $\small l_1,l_2,l_3,l_4,...$ with centres at $\small A,B,A,B,...,$ respectively.]

From the above-given figure

Circumference of 1st semicircle $l_1 = \pi r_1 = 0.5\pi$

Similarly,

Circumference of 2nd semicircle $l_2 = \pi r_2 = \pi$

Circumference of 3rd semicircle $l_3 = \pi r_3 = 1.5\pi$

It is clear that this is an AP with $a = 0.5\pi \ and \ d = 0.5\pi$

Now, sum of length of 13 such semicircles is given by

$S_{13} = \frac{13}{2}\left \{ 2\times 0.5\pi + (13-1)0.5\pi\right \}$
$S_{13} = \frac{13}{2}\left ( \pi+6\pi \right )$
$S_{13} = \frac{13}{2}\times 7\pi$
$S_{13} = \frac{91\pi}{2} = \frac{91}{2}\times \frac{22}{7}=143$
Therefore, sum of length of 13 such semicircles is 143 cm

As the rows are going up, the no of logs are decreasing,
We can clearly see that 20, 19, 18, ..., is an AP.
and here $a = 20 \ and \ d = -1$
Let suppose 200 logs are arranged in 'n' rows,
Then,
$S_n = \frac{n}{2}\left \{ 2\times 20 +(n-1)(-1) \right \}$
$200 = \frac{n}{2}\left \{ 41-n \right \}$
$\Rightarrow n^2-41n +400 = 0$
$\Rightarrow n^2-16n-25n +400 = 0$
$\Rightarrow (n-16)(n-25) = 0$
$\Rightarrow n = 16 \ \ and \ \ n = 25$
Now,
case (i) n = 25
$a_{25} =a+24d = 20+24\times (-1)= 20-24=-4$
But number of rows can not be in negative numbers
Therefore, we will reject the value n = 25

case (ii) n = 16

$a_{16} =a+15d = 20+15\times (-1)= 20-15=5$
Therefore, the number of rows in which 200 logs are arranged is equal to 5

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[ Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $\small 2\times5+2\times (5+3)$ ]

Distance travelled by the competitor in picking and dropping 1st potato $= 2 \times 5 = 10 \ m$

Distance travelled by the competitor in picking and dropping 2nd potato $= 2 \times (5+3) =2\times 8 = 16 \ m$

Distance travelled by the competitor in picking and dropping 3rd potato $= 2 \times (5+3+3) =2\times 11 = 22 \ m$

and so on
we can clearly see that it is an AP with first term (a) = 10 and common difference (d) = 6
There are 10 potatoes in the line
Therefore, total distance travelled by the competitor in picking and dropping potatoes is
$S_{10}= \frac{10}{2}\left \{ 2\times 10+(10-1)6 \right \}$
$S_{10}= 5\left ( 20+54 \right )$
$S_{10}= 5\times 74 = 370$

Therefore, the total distance travelled by the competitor in picking and dropping potatoes is 370 m

## More About NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3:

It's about the sum of n terms of the Arithmetic Progression, it contains a variety of standard questions to give the depth knowledge about the relevant topic. Firstly, it includes the questions in which the formula for the sum of n terms is applied directly; then, it has some slightly tricky questions in which it requires a good understanding of the formula for the sum of n terms. Exercise 5.3 Class 10 Maths -Arithmetic Progression is the progression in which the difference between two consecutive terms is constant. Using this very concept, the sum of n terms of the Arithmetic Progression could be calculated. The NCERT solutions for Class 10 Maths exercise 5.3 mainly focuses on the sum of n terms of the Arithmetic Progression which could be computed efficiently by having clarity regarding the basic concepts of the chapter and which are cleared in preceding exercises; this exercise takes a step further to the topic Arithmetic Progression Exercise 5.3 Class 10 Maths. Students can quickly go through the Arithmetic Progressions Class 10 Notes to revise all concepts all together.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 5.3

• NCERT solutions for Class 10 Maths exercise 5.3 assist students in resolving and reviewing all of the exercises' questions.
• If you go over the NCERT solution for Class 10 Maths chapter 5 exercise 5.3 attentively, you will be able to acquire more marks and will be able to do really well in maths in examinations.
• Class 10 Maths chapter 5 exercise 5.3 is based on the sum of n terms of Arithmetic Progression.

Also see-

##### JEE Main high scoring chapters and topics

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##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## Subject Wise NCERT Exemplar Solutions

1. What criteria do we use to determine if a progression is an Arithmetic Progression?

Any two successive terms in Arithmetic Progression differ by a constant numerical value.

2. How do you find the Arithmetic Progression's nth term?

It might be simply determined by using the fact that any two successive phrases differ by a constant numerical value. Merely add the difference (n-1) times to the first term of the Arithmetic Progression to compute it.

3. What is meant by the sum of n terms of the Arithmetic Progression?

By sum of n terms of the Arithmetic Progression, it means the sum of all the terms of the Arithmetic Progression until the nth term.

4. Can the sum of n terms of the Arithmetic Progression be negative?

Yes, it could be. It happens only when it has negative terms in the progression.

5. Is it possible for any word or the Common Difference to be fractional or not?

Yes, it might be in a fraction, but the Arithmetic Progression's number of terms cannot be in fraction.

6. Is it possible to find the number of terms required to get the particular sum?

Yes, it could be done conveniently by using the sum of n terms formula, and since the sum would be given already, we all have to put the values in the formula and calculate the value of n in the formula.

7. How can we discover the Arithmetic Progression's Common Difference?

Simply differentiate the (n-1)th word from the nth term to discover the Arithmetic Progression's Common Difference.

8. According to NCERT solutions for Class 10 Maths chapter 5 exercise 5.3, what is the sum of n terms?

According to this exercise, the sum of n terms of the Arithmetic Progression is the mathematical sum of all the terms of the Arithmetic Progression till the nth term.

9. What kinds of questions do the NCERT solutions for Class 10 Mathematics chapter 5 exercise 5.3 cover?

The questions are based on the concept of the sum of n terms of ARithmetic Progression. To give a thorough practice on the topic, there are a set of standard questions available in the exercise; it contains direct formula-based questions and some word problems to enhance the understanding of the concept.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

• Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

• Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

• Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.

Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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