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

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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)

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, ${{S}_{n}}=\dfrac{11}{2}\left[ 2\left( \dfrac{1}{15} \right)+\left( 11-1 \right)\left( \dfrac{1}{60} \right) \right]$

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

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

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)

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), (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)

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), (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)

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) 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)

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) 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)

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) 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)

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) 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, $400=\dfrac{n}{2}\left[ 5+45 \right]$

$\Rightarrow 400=25n$ 

$\therefore n=16$ 

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, 

$45=5+\left( 16-1 \right)d$   

$\Rightarrow 40=15d$ 

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

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)

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), (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)

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) 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 $8$.  

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 $n=15$,

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

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

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

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)

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 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 $\pi \left( 0.5 \right)$. 

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,

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

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

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

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

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?

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 $-1$. 

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,

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

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

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

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

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

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

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

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.

Conclusion

Class 10 Ex 5.3 of Maths Chapter 5 - Arithmetic Progressions (AP), is crucial for a solid foundation in math. Understanding the concept of AP, identifying the common difference, and solving problems using the formula for nth term $a_n = a_1 + (n-1)d$ are key takeaways. Regular practice with NCERT solutions provided by platforms like Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward. Consistent practice and understanding will lead to proficiency in AP-related problems.

NCERT Solutions for Class 10 Maths Chapter 5 Exercises

CBSE Class 10 Maths Chapter 5 Other Study Materials

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.