## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression (Ex 5.3) Exercise 5.3

Free PDF download of NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 (Ex 5.3) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise NCERT Solutions in your emails. Get access to Class 10 Science NCERT Solutions and solutions of other subjects for free.

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NCERT solutions for Class 10 Maths Arithmetic Progressions Chapter 5 allow students to analyse the various types of patterns and sequences that they are likely to encounter in their daily lives. The chapter delves into these patterns, which can be obtained by adding a certain number to the previous number or 'term.' This concept's application will help students solve day-to-day problems. The NCERT solutions class 10 maths chapter 5 introduces terminologies such as 'term,' common differences between the terms, and about the general form of an arithmetic progression or an AP. Using well-illustrated examples, how to find the 'n'-th term and the sum of 'n' consecutive terms is explained using the concept of arithmetic progression.

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## Access NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progression

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Exercise 5.3

1. Find the sum of the following APs.

I. \[\mathbf{2},\mathbf{7},\mathbf{12},....\] to \[\mathbf{10}\] terms.

Ans: Given, the first Term, $a=2$ ….. (1)

Given, the common Difference, \[d=7-2=5\] …..(2)

Given, the number of Terms, \[n=10\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{10}{2}\left[ 2\left( 2 \right)+\left( 10-1 \right)\left( 5 \right) \right]$

$\Rightarrow {{S}_{n}}=5\left[ 4+45 \right]$

$\therefore {{S}_{n}}=245$

II. \[-\mathbf{37},-\mathbf{33},-\mathbf{29},...\] to \[\mathbf{12}\] terms

Ans: Given, the first Term, $a=-37$ ….. (1)

Given, the common Difference, \[d=-33-\left( -37 \right)=4\] …..(2)

Given, the number of Terms, \[n=12\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{12}{2}\left[ 2\left( -37 \right)+\left( 12-1 \right)\left( 4 \right) \right]$

$\Rightarrow {{S}_{n}}=6\left[ -74+44 \right]$

$\therefore {{S}_{n}}=-180$

iii. \[\mathbf{0}.\mathbf{6},\mathbf{1}.\mathbf{7},\mathbf{2}.\mathbf{8},......\] to \[\mathbf{100}\] terms

Ans: Given, the first Term, $a=0.6$ ….. (1)

Given, the common Difference, \[d=1.7-0.6=1.1\] …..(2)

Given, the number of Terms, \[n=100\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{100}{2}\left[ 2\left( 0.6 \right)+\left( 100-1 \right)\left( 1.1 \right) \right]$

$\Rightarrow {{S}_{n}}=50\left[ 1.2+108.9 \right]$

$\therefore {{S}_{n}}=5505$

iv. $\dfrac{1}{15},\dfrac{1}{12},\dfrac{1}{10},.....$ to 11 terms

Ans: Given, the first Term, $a=\dfrac{1}{15}$ ….. (1)

Given, the common Difference, \[d=\dfrac{1}{12}-\dfrac{1}{15}=\dfrac{1}{60}\] …..(2)

Given, the number of Terms, \[n=11\] …..(3)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{11}{2}\left[ 2\left( \dfrac{1}{15} \right)+\left( 11-1 \right)\left( \dfrac{1}{60} \right) \right]$

$\Rightarrow {{S}_{n}}=\dfrac{11}{2}\left[ \dfrac{4+5}{30} \right]$

\[\therefore {{S}_{n}}=\dfrac{33}{20}\]

2. Find the sums given below

I. $7+10\dfrac{1}{2}+14+.....+84$

Ans: Given, the first Term, $a=7$ ….. (1)

Given, the common Difference, \[d=10\dfrac{1}{2}-7=\dfrac{7}{2}\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (3)

Substituting the values from (1) and (2) in (3) we get,

${{a}_{n}}=7+\dfrac{7}{2}\left( n-1 \right)=\dfrac{7}{2}\left( n+1 \right)$ ….. (4)

Given, last term of the series, \[{{a}_{n}}=84\] …..(5)

Substituting (5) in (4) we get, $84=\dfrac{7}{2}\left( n+1 \right)$

$\Rightarrow 24=\left( n+1 \right)$

\[\therefore n=23\] ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{23}{2}\left[ 7+84 \right]$

$\Rightarrow {{S}_{n}}=\dfrac{23}{2}\left( 91 \right)$

\[\therefore {{S}_{n}}=1046\dfrac{1}{2}\]

ii. \[\mathbf{34}+\mathbf{32}+\mathbf{30}+.....+\mathbf{10}\]

Ans: Given, the first Term, $a=34$ ….. (1)

Given, the common Difference, \[d=32-34=-2\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (3)

Substituting the values from (1) and (2) in (3) we get,

${{a}_{n}}=34-2\left( n-1 \right)=36-2n$ ….. (4)

Given, last term of the series, \[{{a}_{n}}=10\] …..(5)

Substituting (5) in (4) we get, $10=36-2n$

$\Rightarrow 2n=26$

\[\therefore n=13\] ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{13}{2}\left[ 34+10 \right]$

$\Rightarrow {{S}_{n}}=\dfrac{13}{2}\left( 44 \right)$

\[\therefore {{S}_{n}}=286\]

iii. \[-5+\left( -8 \right)+\left( -11 \right)+.....+\left( -230 \right)\]

Ans: Given, the first Term, $a=-5$ ….. (1)

Given, the common Difference, \[d=-8-\left( -5 \right)=-3\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (3)

Substituting the values from (1) and (2) in (3) we get,

${{a}_{n}}=-5-3\left( n-1 \right)=-2-3n$ ….. (4)

Given, last term of the series, \[{{a}_{n}}=-230\] …..(5)

Substituting (5) in (4) we get, $-230=-2-3n$

$\Rightarrow -228=-3n$

\[\therefore n=76\] ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{76}{2}\left[ -5+\left( -230 \right) \right]$

$\Rightarrow {{S}_{n}}=\dfrac{76}{2}\left( -235 \right)$

\[\therefore {{S}_{n}}=-8930\]

3. In an AP

i. Given \[a=5\], \[d=3\], \[{{\mathbf{a}}_{\mathbf{n}}}=\mathbf{50}\], find \[\mathbf{n}\] and \[{{\mathbf{S}}_{\mathbf{n}}}\].

Ans: Given, the first Term, $a=5$ ….. (1)

Given, the common Difference, \[d=3\] …..(2)

Given, ${{n}^{th}}$ term of the A.P., \[{{\mathbf{a}}_{\mathbf{n}}}=\mathbf{50}\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (4)

Substituting the values from (1), (2) and (3) in (4) we get,

$50=5+3\left( n-1 \right)=2+3n$

Simplifying it further we get,

$n=\dfrac{50-2}{3}$

$\therefore n=16$ …..(5)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(6)

Substituting the values from (1), (2) and (5) in (6) we get, ${{S}_{n}}=\dfrac{16}{2}\left[ 2\left( 5 \right)+\left( 16-1 \right)\left( 3 \right) \right]$

$\Rightarrow {{S}_{n}}=8\left[ 10+45 \right]$

\[\therefore {{S}_{n}}=440\]

ii. Given \[a=7\], \[{{a}_{13}}=35\], find \[d\] and \[{{\mathbf{S}}_{13}}\].

Ans: Given, the first Term, $a=7$ ….. (1)

Given, ${{13}^{th}}$ term of the A.P., \[{{a}_{13}}=35\] …..(2)

Substituting the values from (1), (2) in (3) we get,

$35=7+\left( 13-1 \right)d=7+12d$

Simplifying it further we get,

$d=\dfrac{28}{12}$

\[\therefore d=\dfrac{7}{3}\] …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(5)

Substituting the values from (1) and (4) in (5) we get, ${{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 7 \right)+\left( 13-1 \right)\left( \dfrac{7}{3} \right) \right]$

$\Rightarrow {{S}_{13}}=\dfrac{13}{2}\left[ 14+28 \right]$

\[\therefore {{S}_{13}}=273\]

iii. Given \[d=3\], \[{{\mathbf{a}}_{12}}=37\], find \[a\] and \[{{\mathbf{S}}_{12}}\].

Ans: Given, the common difference, $d=3$ ….. (1)

Given, ${{12}^{th}}$ term of the A.P., \[{{\mathbf{a}}_{12}}=37\] …..(2)

Substituting the values from (1), (2) in (3) we get,

$37=a+3\left( 12-1 \right)=a+33$

Simplifying it further we get,

\[\therefore a=4\] …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(5)

Substituting the values from (1) and (4) in (5) we get, ${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 4 \right)+\left( 12-1 \right)\left( 3 \right) \right]$

$\Rightarrow {{S}_{12}}=6\left[ 8+33 \right]$

\[\therefore {{S}_{12}}=246\]

iv. Given \[{{\mathbf{a}}_{3}}=1\mathbf{5}\], \[{{\mathbf{S}}_{10}}=125\] find \[{{a}_{10}}\] and \[d\].

Ans: Given, ${{3}^{rd}}$ term of the A.P., \[{{\mathbf{a}}_{3}}=1\mathbf{5}\] …..(1)

Given, the sum of terms, \[{{\mathbf{S}}_{10}}=125\] ….. (2)

Substituting the values from (1) in (3) we get,

$15=a+\left( 3-1 \right)d=a+2d$ …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(5)

Substituting the values from (1) in (5) we get, $125=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right]$

$\Rightarrow 125=5\left[ 2a+9d \right]$

\[\therefore 25=2a+9d\] …..(5)

Let us solve equations (4) and (5) by subtracting twice of (4) from (5) we get,

\[25-30=\left( 2a+9d \right)-\left( 2a+4d \right)\]

$\Rightarrow -5=5d$

$\therefore d=-1$ …..(6)

From (4) and (6) we get, $a=17$ …..(7)

From (3), (6) and (7) for $n=10$ we get,

${{a}_{10}}=17-\left( 10-1 \right)$

$\therefore {{a}_{10}}=8$

v. Given \[{{\mathbf{S}}_{9}}=75\], \[d=5\] find \[a\] and \[{{a}_{9}}\].

Ans: Given, common difference, \[d=5\] …..(1)

Given, the sum of terms, \[{{\mathbf{S}}_{9}}=75\] ….. (2)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(3)

Substituting the values from (1), (2) in (3) we get, $75=\dfrac{9}{2}\left[ 2a+5\left( 9-1 \right) \right]$

$\Rightarrow 25=3\left[ a+20 \right]$

$\Rightarrow 3a=-35$

\[\therefore a=-\dfrac{35}{3}\] …..(4)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (5)

Substituting the values from (1), (4) in (5) we get,

${{a}_{9}}=-\dfrac{35}{3}+5\left( 9-1 \right)$

$\Rightarrow {{a}_{9}}=-\dfrac{35}{3}+40$

$\therefore {{a}_{9}}=\dfrac{85}{3}$

vi. Given \[a=2\], \[d=8\], \[{{S}_{\mathbf{n}}}=9\mathbf{0}\], find \[\mathbf{n}\] and \[{{a}_{\mathbf{n}}}\].

Ans: Given, common difference, \[d=8\] …..(1)

Given, first term, \[a=2\] …..(2)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=9\mathbf{0}\] ….. (3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(3)

Substituting the values from (1), (2), (3) in (4) we get, $90=\dfrac{n}{2}\left[ 2\left( 2 \right)+8\left( n-1 \right) \right]$

$\Rightarrow 45=n\left[ 2n-1 \right]$

$\Rightarrow 2{{n}^{2}}-n-45=0$

$\Rightarrow 2{{n}^{2}}-10n+9n-45=0$

$\Rightarrow 2n\left( n-5 \right)+9\left( n-5 \right)=0$

$\Rightarrow \left( n-5 \right)\left( 2n+9 \right)=0$

\[\therefore n=5\] …..(4)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (5)

Substituting the values from (1), (2), (4) in (5) we get,

${{a}_{5}}=2+8\left( 5-1 \right)$

$\Rightarrow {{a}_{5}}=2+32$

$\therefore {{a}_{5}}=34$

vii. Given \[a=8\], \[{{S}_{\mathbf{n}}}=21\mathbf{0}\], \[{{\mathbf{a}}_{\mathbf{n}}}=62\], find \[\mathbf{n}\] and \[d\].

Ans: Given, first term, \[a=8\] …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=21\mathbf{0}\] ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=62\] …..(3)

Substituting the values from (1), (2) in (4) we get, $210=\dfrac{n}{2}\left[ 2\left( 8 \right)+d\left( n-1 \right) \right]$

$\Rightarrow 420=n\left[ 16+\left( n-1 \right)d \right]$ …..(4)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (5)

Substituting the values from (1), (3) in (5) we get,

$62=8+\left( n-1 \right)d$ …..(6)

Let us solve equations (4) and (6) by subtracting $n$ times of (6) from (4) we get,

$420-62n=\left( 16n+n\left( n-1 \right)d \right)-\left( 8n+n\left( n-1 \right)d \right)$

$\Rightarrow 420-62n=8n$

$\Rightarrow 420=70n$

$\therefore n=6$ ……(7)

Substituting the values from (7) in (6) we get,

$62=8+\left( 6-1 \right)d$

$\Rightarrow 54=5d$

$\therefore d=\dfrac{54}{5}$

viii. Given \[{{S}_{\mathbf{n}}}=-14\], \[d=2\], \[{{\mathbf{a}}_{\mathbf{n}}}=4\], find \[\mathbf{n}\] and \[a\].

Ans: Given, common difference, \[d=2\] …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=-14\] ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=4\] …..(3)

Substituting the values from (1), (2) in (4) we get, $-14=\dfrac{n}{2}\left[ 2a+2\left( n-1 \right) \right]$

$\Rightarrow -14=n\left[ a+n-1 \right]$ …..(5)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (6)

Substituting the values from (1), (3) in (6) we get,

$4=a+2\left( n-1 \right)$ …..(7)

Let us solve equations (5) and (7) by substituting the value of $a$ from (7) in (5) we get,

$-14=n\left[ \left( 4-2\left( n-1 \right) \right)+n-1 \right]$

\[\Rightarrow -14=n\left[ 5-n \right]\]

$\Rightarrow {{n}^{2}}-5n-14=0$

$\Rightarrow {{n}^{2}}-7n+2n-14=0$

$\Rightarrow \left( n-7 \right)\left( n+2 \right)=0$

$\therefore n=7$ (Since $n$ cannot be negative) ……(8)

Substituting the values from (8) in (7) we get,

$4=a+2\left( 7-1 \right)$

$\Rightarrow 4=a+12$

$\therefore a=-8$

ix. Given \[a=3\], \[n=8\], \[S=192\], find \[d\].

Ans: Given, first term, \[a=3\] …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=192\] ….. (2)

Given, the number of terms, \[n=8\] …..(3)

Substituting the values from (1), (2) in (4) we get, $192=\dfrac{8}{2}\left[ 2\left( 3 \right)+d\left( 8-1 \right) \right]$

$\Rightarrow 192=4\left[ 6+7d \right]$

$\Rightarrow 48=6+7d$

$\Rightarrow 42=7d$

$\therefore d=6$

x. Given \[l=28\], \[S=144\] and there are total $9$ terms. Find \[a\].

Ans: Given, last term, \[l=28\] …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=144\] ….. (2)

Given, the number of terms, \[n=9\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(4)

Substituting the values from (1), (2) in (4) we get, $144=\dfrac{9}{2}\left[ a+28 \right]$

$\Rightarrow 32=a+28$

$\therefore a=4$

4. How many terms of the A.P. \[9,17,25...\] must be taken to give a sum of \[\mathbf{636}\]?

Ans: Given, common difference, \[d=17-9=8\] …..(1)

Given, first term, \[a=9\] …..(2)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=636\] ….. (3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(3)

Substituting the values from (1), (2), (3) in (4) we get, $636=\dfrac{n}{2}\left[ 2\left( 9 \right)+8\left( n-1 \right) \right]$

$\Rightarrow 636=n\left( 5+4n \right)$

$\Rightarrow 4{{n}^{2}}+5n-636=0$

$\Rightarrow 4{{n}^{2}}+53n-48n-636=0$

$\Rightarrow n\left( 4n+53 \right)-12\left( 4n+53 \right)=0$

$\Rightarrow \left( n-12 \right)\left( 4n+53 \right)=0$

$\Rightarrow n=12\text{ }or\text{ }-\dfrac{53}{4}$

Since $n$ can only be a natural number \[\therefore n=12\]

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

Ans: Given, first term, \[a=5\] …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=400\] ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=45\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(4)

Substituting the values from (1), (2), (3) in (4) we get, $400=\dfrac{n}{2}\left[ 5+45 \right]$

$\Rightarrow 400=25n$

$\therefore n=16$

Substituting the values from (1), (3) in (5) we get,

$45=5+\left( 16-1 \right)d$

$\Rightarrow 40=15d$

$\therefore d=\dfrac{8}{3}$

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

Ans: Given, first term, \[a=17\] …..(1)

Given, the common difference, \[d=9\] ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=350\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ….. (4)

Substituting the values from (1), (2), (3) in (4) we get,

$350=17+9\left( n-1 \right)$

$\Rightarrow 333=9\left( n-1 \right)$

$\Rightarrow 37=\left( n-1 \right)$

$\therefore n=38$ ……(5)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(6)

Substituting the values from (1), (5), (3) in (6) we get, ${{S}_{38}}=\dfrac{38}{2}\left[ 17+350 \right]$

$\Rightarrow {{S}_{38}}=19\left( 367 \right)$

$\therefore {{S}_{38}}=6973$

7. Find the sum of first \[22\] terms of an AP in which \[d=7\] and \[{{22}^{nd}}\] term is \[149\].

Ans: Given, the common difference, \[d=7\] ….. (1)

Given, the ${{22}^{nd}}$ term, \[{{\mathbf{a}}_{22}}=149\] …..(2)

Substituting the values from (1), (2) in (3) we get,

$149=a+7\left( 22-1 \right)$

$\Rightarrow 149=a+147$

$\therefore a=2$ ……(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$ …..(5)

Substituting the values from (1), (2), (4) in (5) we get, ${{S}_{22}}=\dfrac{22}{2}\left[ 2+149 \right]$

$\Rightarrow {{S}_{22}}=11\left( 151 \right)$

$\therefore {{S}_{22}}=1661$

8. Find the sum of first \[51\] terms of an AP whose second and third terms are \[14\] and \[18\] respectively.

Ans: Given, the ${{2}^{nd}}$ term, \[{{\mathbf{a}}_{2}}=14\] ….. (1)

Given, the ${{3}^{rd}}$ term, \[{{\mathbf{a}}_{3}}=18\] …..(2)

Substituting the values from (1) in (3) we get,

$14=a+d$ …..(4)

Substituting the values from (2) in (3) we get,

$18=a+2d$ …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$18-14=\left( a+2d \right)-\left( a+d \right)$

$\therefore d=4$ ……(6)

Substituting the value from (6) in (4) we get $a=10$. …..(7)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ …..(8)

Substituting the values from (7), (6) in (8) we get for $n=51$ ,

${{S}_{51}}=\dfrac{51}{2}\left[ 2\left( 10 \right)+4\left( 51-1 \right) \right]$

$\Rightarrow {{S}_{51}}=\dfrac{51}{2}\left[ 20+200 \right]$

$\therefore {{S}_{51}}=5610$

9. If the sum of first \[7\] terms of an AP is \[49\] and that of \[17\] terms is \[289\], find the sum of first \[n\] terms.

Ans: Given, the sum of first $7$ terms, \[{{S}_{7}}=49\] ….. (1)

Given, the sum of first $17$ terms, \[{{S}_{17}}=289\] …..(2)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ ….. (3)

Substituting the values from (1) in (3) we get,

\[49=\dfrac{7}{2}\left[ 2a+\left( 7-1 \right)d \right]\]

\[\Rightarrow 7=a+3d\] …..(4)

Substituting the values from (2) in (3) we get,

\[289=\dfrac{17}{2}\left[ 2a+\left( 17-1 \right)d \right]\]

\[\Rightarrow 17=a+8d\] …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$17-7=\left( a+8d \right)-\left( a+3d \right)$

$\Rightarrow 10=5d$

$\therefore d=2$ ……(6)

Substituting the value from (6) in (4) we get $a=1$. …..(7)

Substituting the values from (7), (6) in (3) we get,

${{S}_{n}}=\dfrac{n}{2}\left[ 2+2\left( n-1 \right) \right]$

$\therefore {{S}_{n}}={{n}^{2}}$

10. Show that \[{{a}_{1}},{{a}_{2}}...,{{a}_{n}},...\] form an AP where ${{a}_{n}}$ is defined as below. Also find the sum of the first $15$ terms in each case.

i. \[{{a}_{n}}=3+4n\]

Ans: Consider two consecutive terms of the given sequence. Say ${{a}_{n}},{{a}_{n+1}}$. Difference between these terms will be

${{a}_{n+1}}-{{a}_{n}}=\left[ 3+4\left( n+1 \right) \right]-\left[ 3+4n \right]$

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=4\left( n+1 \right)-4n$

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=4$

Which is a constant $\forall n\in \mathbb{N}$.

For $n=1$, ${{a}_{1}}=3+4=7$

Therefore, it is an A.P. with first term $7$ and common difference $4$.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Therefore, ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 7 \right)+4\left( 15-1 \right) \right]$

$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left[ 14\left( 5 \right) \right]$

$\therefore {{S}_{15}}=525$

ii. \[{{a}_{n}}=9-5n\]

Ans: Consider two consecutive terms of the given sequence. Say ${{a}_{n}},{{a}_{n+1}}$. Difference between these terms will be

${{a}_{n+1}}-{{a}_{n}}=\left[ 9-5\left( n+1 \right) \right]-\left[ 9-5n \right]$

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=-5\left( n+1 \right)+5n$

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=-5$

Which is a constant $\forall n\in \mathbb{N}$.

For $n=1$, ${{a}_{1}}=9-5=4$

Therefore, it is an A.P. with first term $4$ and common difference $-5$.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Therefore, ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 4 \right)-5\left( 15-1 \right) \right]$

\[\Rightarrow {{S}_{15}}=15\left[ -31 \right]\]

$\therefore {{S}_{15}}=-465$

11. If the sum of the first $n$ terms of an AP is \[4n-{{n}^{2}}\], what is the first term (that is ${{S}_{1}}$)? What is the sum of first two terms? What is the second term? Similarly find the \[{{3}^{rd}}\], the \[{{10}^{th}}\] and the \[{{n}^{th}}\] terms.

Ans: Given, the sum of the first $n$ terms of an A.P. is \[4n-{{n}^{2}}\].

First term $={{S}_{1}}=4-1=3$. …..(1)

Sum of first two terms $={{S}_{2}}=8-{{\left( 2 \right)}^{2}}=4$ …..(2)

From (1) and (2), ${{2}^{nd}}$ term $={{S}_{2}}-{{S}_{1}}=4-3=1$.

Sum of first three terms $={{S}_{3}}=12-{{\left( 3 \right)}^{2}}=3$ …..(3)

From (3) and (2), ${{3}^{rd}}$ term $={{S}_{3}}-{{S}_{2}}=3-4=-1$.

Similarly,

Sum of first $n$ terms $={{S}_{n}}=4n-{{n}^{2}}$ …..(4)

Sum of first $n-1$ terms $={{S}_{n-1}}=4\left( n-1 \right)-{{\left( n-1 \right)}^{2}}=-{{n}^{2}}+6n-5$ …..(5)

From (4) and (5), ${{n}^{th}}$ term $={{S}_{n}}-{{S}_{n-1}}=\left( 4n-{{n}^{2}} \right)-\left( -{{n}^{2}}+6n-5 \right)=5-2n$ …..(6)

From (6), ${{10}^{th}}$ term is $5-2\left( 10 \right)=-15$.

12. Find the sum of first $40$ positive integers divisible by $6$.

Ans: First positive integer that is divisible by $6$ is $6$ itself.

Second positive integer that is divisible by $6$ is \[6+6=12\].

Third positive integer that is divisible by $6$ is \[12+6=18\].

Hence, it is an A.P. with first term and common difference both as $6$.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Therefore, for $n=40$,

${{S}_{40}}=\dfrac{40}{2}\left[ 2\left( 6 \right)+6\left( 40-1 \right) \right]$

\[\Rightarrow {{S}_{40}}=120\left[ 41 \right]\]

$\therefore {{S}_{40}}=4920$

13. Find the sum of first $15$ multiples of $8$.

Ans: First positive integer that is divisible by $8$ is $8$ itself.

Second positive integer that is divisible by $8$ is \[8+8=16\].

Third positive integer that is divisible by $8$ is \[16+8=24\].

Hence, it is an A.P. with first term and common difference both as $8$.

Therefore, for $n=15$,

${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 8 \right)+8\left( 15-1 \right) \right]$

\[\Rightarrow {{S}_{15}}=60\left[ 16 \right]\]

$\therefore {{S}_{15}}=960$

14. Find the sum of the odd numbers between $0$ and $50$.

Ans: The odd numbers between $0$ and $50$ are $1,3,5,...,49$.

It is an A.P. with first term $1$ and common difference $2$. ….(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ … (2)

Substitute ${{a}_{n}}=49$ and values from (1) into (2)

$49=1+2\left( n-1 \right)$

$\Rightarrow 24=\left( n-1 \right)$

$\therefore n=25$ ……(3)

Substituting values from (1), (3) in (4) we get,

${{S}_{25}}=\dfrac{25}{2}\left[ 2+2\left( 25-1 \right) \right]$

\[\Rightarrow {{S}_{25}}=25\left[ 25 \right]\]

$\therefore {{S}_{25}}=625$

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

Ans: Penalty of delay for first day is Rs. $200$.

Penalty of delay for second day is Rs. $250$.

Penalty of delay for third day is Rs. $300$.

Hence it is an A.P. with first term $200$ and common difference $50$.

Money the contractor has to pay as penalty, if he has delayed the work by \[\mathbf{30}\] days is the sum of first $30$ terms of the A.P.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{30}}=\dfrac{30}{2}\left[ 2\left( 200 \right)+50\left( 30-1 \right) \right]$

\[\Rightarrow {{S}_{30}}=15\left[ 400+50\left( 29 \right) \right]\]

$\therefore {{S}_{30}}=27750$

Therefore, the contractor has to pay Rs \[27750\] as penalty.

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

Ans: Let the first prize be of Rs. $a$ then the second prize will be of Rs. $a-20$, the third prize will be of Rs. $a-40$.

Therefore, it is an A.P. with first term $a$ and common difference $-20$.

Given, ${{S}_{7}}=700$

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{7}}=\dfrac{7}{2}\left[ 2a-20\left( 7-1 \right) \right]$

\[\Rightarrow 700=7\left[ a-60 \right]\]

\[\Rightarrow 100=a-60\]

$\therefore a=160$

Therefore, the value of each of the prizes was \[Rs\text{ }160,\text{ }Rs\text{ }140,\text{ }Rs\text{ }120,Rs\text{ }100,\text{ }Rs\text{ }80,\text{ }Rs\text{ }60,\text{ }and\text{ }Rs\text{ }40.\]

17. 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 \[\mathbf{1}\] tree, a section of class II will plant \[\mathbf{2}\] trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Ans: Each section of class I will plant $1$ tree each. Therefore, total trees planted by class I are $3$.

Each section of class II will plant $2$ trees each. Therefore, total trees planted by class II are $3\times 2=6$.

Each section of class III will plant $3$ trees each. Therefore, total trees planted by class III are $3\times 3=9$.

Therefore, it is an A.P series with first term and common difference both as $3$.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 3 \right)-3\left( 12-1 \right) \right]$

\[\Rightarrow {{S}_{12}}=6\left[ 39 \right]\]

$\therefore {{S}_{12}}=234$

Therefore, $234$ trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A of radii \[\mathbf{0}.\mathbf{5}\], \[\mathbf{1}.\mathbf{0}\] cm, \[\mathbf{1}.\mathbf{5}\] cm, \[\mathbf{2}.\mathbf{0}\] cm, ......... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

(image will be uploaded soon)

Ans: Length of first semi-circle ${{I}_{1}}=\pi \left( 0.5 \right)$ cm.

Length of second semi-circle ${{I}_{2}}=\pi \left( 1 \right)$ cm.

Length of third semi-circle ${{I}_{3}}=\pi \left( 1.5 \right)$ cm.

Therefore, it is an A.P series with first term and common difference both as $\pi \left( 0.5 \right)$.

${{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 0.5\pi \right)+\left( 0.5\pi \right)\left( 13-1 \right) \right]$

\[\Rightarrow {{S}_{13}}=7\times 13\times \left( 0.5\pi \right)\]

\[\Rightarrow {{S}_{13}}=7\times 13\times \dfrac{1}{2}\times \dfrac{22}{7}\]

$\therefore {{S}_{13}}=143$

Therefore, the length of such spiral of thirteen consecutive semi-circles

will be \[143\] cm.

19. The \[200\] logs are stacked in the following manner: \[20\] logs in the bottom row, \[19\] in the next row, \[18\] in the row next to it and so on. In how many rows are the \[200\] logs placed and how many logs are in the top row?

(Image will be uploaded soon)

Ans: Total logs in first row are $20$.

Total logs in second row are $19$.

Total logs in third row are $18$.

Therefore, it is an A.P series with first term $20$ and common difference $-1$.

$200=\dfrac{n}{2}\left[ 2\left( 20 \right)-\left( n-1 \right) \right]$

\[\Rightarrow 400=n\left[ 41-n \right]\]

\[\Rightarrow {{n}^{2}}-41n+400=0\]

\[\Rightarrow {{n}^{2}}-16n-25n+400=0\]

\[\Rightarrow n\left( n-16 \right)-25\left( n-16 \right)=0\]

\[\Rightarrow \left( n-16 \right)\left( n-25 \right)=0\]

For $n=25$, after ${{20}^{th}}$ term, all terms are negative, which is illogical as terms are representing the number of logs and number of logs being negative is illogical.

$\therefore n=16$

Total logs in ${{16}^{th}}$ row $=20-\left( 16-1 \right)=5$

Therefore, $200$ logs will be placed in $16$ rows and the total logs in ${{16}^{th}}$ row will be $5$.

20. In a potato race, a bucket is placed at the starting point, which is $5$m from the first potato and other potatoes are placed $3$m apart in a straight line. There are ten potatoes in the line.

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 \[\mathbf{2}\times \mathbf{5}+\mathbf{2}\times \left( \mathbf{5}+\mathbf{3} \right)\]]

(Image will be uploaded soon)

Ans: Total distance run by competitor to collect and drop first potato $=2\times 5=10$m.

Total distance run by competitor to collect and drop second potato $=2\times \left( 5+3 \right)=16$m.

Total distance run by competitor to collect and drop third potato $=2\times \left( 5+3+3 \right)=22$m.

Therefore, it is an A.P series with first term $10$ and common difference $6$.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore, to collect and drop $10$ potatoes total distance covered is

${{S}_{10}}=\dfrac{10}{2}\left[ 2\left( 10 \right)+6\left( 10-1 \right) \right]$

\[\Rightarrow {{S}_{10}}=5\left[ 74 \right]\]

$\therefore {{S}_{13}}=370$

Therefore, the competitor will run a total distance of \[370\]m.

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

Opting for the NCERT solutions for Ex 5.3 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 5.3 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Arithmetic Progression textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 5 Exercise 5.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 5 Exercise 5.3, there are plenty of exercises in this chapter that contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 5 Exercise 5.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

### NCERT Solutions for Class 10 Maths Chapter 5 Exercises

Chapter 5 - Arithmetic Progression All Exercises in PDF Format | |

4 Questions & Solutions | |

20 Questions & Solutions | |

5 Questions & Solutions |

### NCERT Solutions for Class 10 Maths

**1. What will be taught in this chapter?**

Arithmetic Progressions is one of the most important chapters of Class 10 NCERT Solutions book. This chapter will explain to you about the sequence of numbers and how each term is extracted by adding a fixed number to the preceding term by excluding the first term. This chapter introduces you to the basics of Arithmetic Progressions and their common difference, infinite and finite Arithmetic Progressions, nth term of an Arithmetic Progressions and sum of first n terms of an Arithmetic Progressions. An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

Example: 2,5,8,11,14…. is an arithmetic progression.

**2. Give a brief overview of the chapter.**

Arithmetic Progression is a significant chapter in your Class 10 Maths of CBSE curriculum. The chapter is a precursor to Geometric Progression. The chapter gives you an apparent idea about numeric sequences or progressions, where the difference between two consecutive numbers is consistent. Chapter 5 of Class 10 Maths presents objective problems where you have to deduce if a scenario follows an Arithmetic Progression or not, write AP sequences based on given first term and difference values, etc. Exercise 5.3 clubs further in-depth, with problems that require you to deduce the n-th term of an AP or the common difference of an Arithmetic Progression.

**3. How many questions are there in this exercise?**

There are a total of 20 questions in this particular exercise. Question 1 and question 2 consists of 4 and 3 sub-questions and in all the questions we have to find the sum of APs. In question 3, you’ll have to find the nth term or the mentioned terms. In question 4, you’ll have to check how many terms are there in the given sum, Question 5, 6, 7, 8 and 9 are similar types of questions. In which, you’ll have to find the first n terms.

Questions 10 and 11 ask you to find a1, a2... Till the first 15 terms. In question 12, 13 and 14 you have to find the sum of odd numbers between the given numbers. Question 15, 16, 17, 18, 19 and 20 are scenario-based questions. In which, you have to solve the questions based on the given scenario.

**4. Why should I choose Vedantu for preparation?**

NCERT Solutions for Class 10 Maths Chapter 5 are prepared in an easy to simple language to understand, and they are crisp and concise to the point. All the questions are accurately answered from the exercise given at the end of the NCERT Class 10 Maths of CBSE. Our NCERT Solutions for Class 10 Maths Chapter 5 have been drafted as per the latest CBSE Class 10 Maths Syllabus and NCERT CBSE Class 10 Maths Book.

Our NCERT Solutions gives you a clear idea and understanding of all the important concepts and help you develop a strong conceptual foundation. These solutions cover all possible questions and question types that can be asked in your Class 10 Maths exams.

**5. Mention the important concepts that you learn in NCERT Solutions for Chapter 5 Arithmetic Progression of Class 10 Maths.**

The important concepts that you learn in the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression are -

Arithmetic progression and its terms.

First term and common difference of an Arithmetic progression.

nth term of an Arithmetic progression.

Types of Arithmetic progressions.

Sum of all n terms of an Arithmetic progression.

Sum of n terms in an Arithmetic progression using its last term.

All these concepts are covered in the NCERT Solutions of Chapter 5 of Class 10 Maths and they are quite useful as they pave the way for a more comprehensive study pattern that leaves no room for error in examination.

**6. How many exercises are there in NCERT Solutions for Chapter 5 of Class 10 Maths?**

The NCERT Solutions for Chapter 5 of Class 10 Maths has four exercises:

The first exercise includes nine problems.

The second exercise includes two problems.

The third exercise includes ten problems.

The fourth exercise includes nine problems.

These questions, as well as their answers, are provided in the NCERT Solutions. Detailed solutions of the same are available free of cost on Vedantu website and also on the Vedantu Mobile app.

**7. Which questions are the most important questions of Exercise 5.3 of Chapter 5 of Class 10th Maths?**

The most important questions from Exercise 5.3 of Chapter 5 of Class 10 Maths are questions three, 18, 19 and 20. These questions are based on almost every concept covered till then. Regardless, all the questions must be attempted and practised thoroughly.

**8. What type of questions are there in Chapter 5 of Class 10 Maths?**

Multiple choice questions, descriptive questions, long answer type questions, short answer type questions, fill in the blanks, and everyday life examples are all included in Chapter 5 of Class 10 Maths . Students' problem-solving and time-management abilities should improve at the end of this chapter. This enables them to achieve excellent grades in their final exams.

**9. What are the most important definitions that I need to remember in Chapter 5 of Class 10 Maths ?**

The most important definitions that you need to remember in Chapter 5 of Class 10 Maths are the definitions of AP and Common Difference. Except for the initial term, an arithmetic progression (AP) is a list of values in which each term is produced by adding a fixed number d to the prior term. The common difference is denoted by the fixed number d. Numericals related to this topic are also very crucial to cover, both for knowledge purposes and as well as exam point of view.