NCERT Solutions for Exercise 5.3 Class 10 Maths Chapter 5

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

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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 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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Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.3

Q1 (i) Find the sum of the following APs: $\small 2,7,12,...,$ to $\small 10$ terms.

Answer:

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

Q1 (ii) Find the sum of the following APs: $\small -37,-33,-29,...,$ to $\small 12$ terms.

Answer:

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

Q1 (iii) Find the sum of the following APs: $\small 0.6,1.7,2.8,...,$ to $\small 100$ terms.

Answer:

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

Q1 (iv) Find the sum of the following APs: $\small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $\small 11$ terms.

Answer:

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 $\frac{33}{20}$

Q2 (i) Find the sums given below : $\small 7+10\frac{1}{2}+14+...+84$

Answer:

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 : $\small 34+32+30+...+10$

Answer:

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 : $\small -5+(-8)+(-11)+...+(-230)$

Answer:

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 $\small S_n$ .

Answer:

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 $\small S_1_3$ .

Answer:

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 $\small S_1_2$ .

Answer:

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 $\small S_1_0$
Answer:

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 $\small a_9$ .

Answer:

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 $\small a_n$ .

Answer:

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 $\small d$ .

Answer:

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 $\small a$ .

Answer:

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 $\small d$ .

Answer:

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 $\small a$ .

Answer:

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

Q4 How many terms of the AP: $\small 9,17,25,...$ must be taken to give a sum of $\small 636$ ?

Answer:

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 $\small 636$ .

Q5 The first term of an AP is $\small 5$ , the last term is $\small 45$ and the sum is $\small 400$ . Find the number of terms and the common difference.

Answer:

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

Q6 The first and the last terms of an AP are $\small 17$ and $\small 350$ respectively. If the common difference is $\small 9$ , how many terms are there and what is their sum?

Answer:

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

Q7 Find the sum of first $\small 22$ terms of an AP in which $\small d=7$ and $\small 22$ nd term is $\small 149$ .

Answer:

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

Q8 Find the sum of first $\small 51$ terms of an AP whose second and third terms are $\small 14$ and $\small 18$ respectively.

Answer:

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

Q9 If the sum of first $\small 7$ terms of an AP is $\small 49$ and that of $\small 17$ terms is $\small 289$ , find the sum of first $\small n$ terms.

Answer:

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 $n^2$

Q10 (i) Show that $\small a_1,a_2,...,a_n,...$ form an AP where an is defined as below : $\small a_n=3+4n$ Also find the sum of the first $\small 15$ terms.

Answer:

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

Q10 (ii) Show that $\small a_1,a _2,...,a_n,...$ form an AP where an is defined as below : $\small a_n=9-5n$ . Also find the sum of the first $\small 15$ terms in each case.

Answer:

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

Q11 If the sum of the first $\small n$ terms of an AP is $\small 4n-n^2$ , what is the first term (that is $\small S_1$ )? What is the sum of first two terms? What is the second term? Similarly, find the $\small 3$ rd, the $\small 10$ th and the $\small n$ th terms

Answer:

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,
$a_3= a+2d = 3+2(-2 )= 3-4=-1$
$a_{10}= a+9d = 3+9(-2 )= 3-18=-15$
$a_{n}= a+(n-1)d = 3+(n-1)(-2 )= 5-2n$

Q12 Find the sum of the first $\small 40$ positive integers divisible by $\small 6$ .

Answer:

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

Q13 Find the sum of the first $\small 15$ multiples of $\small 8$ .

Answer:

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

Q14 Find the sum of the odd numbers between $\small 0$ and $\small 50$ .

Answer:

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

Q15 A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs $\small 200$ for the first day, Rs $\small 250$ for the second day, Rs $\small 300$ for the third day, etc., the penalty for each succeeding day being Rs $\small 50$ more than for the preceding day. How much money the contractor has to pay a penalty, if he has delayed the work by $\small 30$ days?

Answer:

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

Q16 A sum of Rs $\small 700$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs $\small 20$ less than its preceding prize, find the value of each of the prizes.

Answer:

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

Q17 In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I???? will plant $\small 1$ tree, a section of Class II will plant $\small 2$ trees and so on till Class XII. There are three sections in each class. How many trees will be planted by the students?

Answer:

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

Q18 A spiral is made up of successive semicircles, with centres alternately at $\small A$ and $\small B$ ??????, starting with centre at $\small A$ , of radii $\small 0.5\hspace {1mm}cm,1.0\hspace {1mm}cm,1.5\hspace {1mm}cm,2.0\hspace {1mm}cm,...$ as shown in Fig. $\small 5.4$ . What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi =\frac{22}{7}$ )

1635921529757

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

Answer:

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

Q19 $\small 200$ logs are stacked in the following manner: $\small 20$ logs in the bottom row, $\small 19$ in the next row, $\small 18$ in the row next to it and so on (see Fig. $\small 5.5$ ). In how many rows are the $\small 200$ logs placed and how many logs are in the top row?

1635921556744

Answer:

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

Q20 In a potato race, a bucket is placed at the starting point, which is $\small 5$ m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. $\small 5.6$ ).

1635921559600

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 $\small 2\times5+2\times (5+3)$ ]

Answer:

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

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-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

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

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
- As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.
Ask for Directions:
- If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

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sai raja 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.

Read Complete Answer

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

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

NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.3

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