10th Maths 5.3

Chapter 5 Arithmetic Progressions

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NCERT Class 10th solution of Exercise 5.1

NCERT Class 10th solution of Exercise 5.2

NCERT Class 10th solution of Exercise 6.1

NCERT Class 10th Maths Projects

Exercise 5.3

Q1. Find the sum of the following APs:

i) $2, 7, 12, ...$ to $10$ terms.

ii) $-37, -33, -29, ...,$ to $12$ terms.

iii) $0.6, 1.7, 2.8, ...$ to $100$ terms.

iv) $1/15, 1/12, 1/10, ...$ to $11$ terms.

Sol. :

Given

$a=2, d=7-2=5,$ and

$n=10$

$S_n=n/2[2a+(n-1)d]$

$S_10=10/2[2(2)+(10-1)(5)]$

$S_10=5[4+(9)(5)]$

$S_10=5[4+45]$

$S_10=5[49]$

$S_10=245$

Answer :

$S_10=245.$

ii)

Given

$a=-37, d=(-33)-(-37)=-33+37=4,$ and

$n=12$

$S_n=n/2[2a+(n-1)d]$

$S_12=12/2[2(-37)+(12-1)(4)]$

$S_12=6[-74+(11)(4)]$

$S_12=6[-74+44]$

$S_12=6[-30]$

$S_12=-180$

Answer :

$S_12=-180.$

iii)

Given

$a=0.6, d=1.7-0.6=1.1,$ and

$n=100$

$S_n=n/2[2a+(n-1)d]$

$S_100=100/2[2(0.6)+(100-1)(1.1)]$

$S_100=50[1.2+(99)(1.1)]$

$S_100=50[1.2+108.9]$

$S_100=50[110.1]$

$S_100=5505$

Answer :

$S_100=5505.$

iv)

Given

$a=1/15, d=1/12-1/15=(5-4)/60=1/60,$ and

$n=11$

$S_n=n/2[2a+(n-1)d]$

$S_11=11/2[2(1/15)+(11-1)(1/60)]$

$S_11=11/2[2/15+(10)(1/60)]$

$S_11=11/2[2/15+1/6]$

$S_11=11/2[(4+5)/30]$

$S_11=11/2[9/30]$

$S_11=33/30$

Answer :

$S_11=33/30.$

Q2. Find the sums given below :

i) $7+10(1)/2+14+.....+84.$

ii) $34+32+30+.....+10.$

iii) $-5+(-8)+(-11)+.....+(-230)$

Sol. :

Given

$a=7, d=10(1)/2-7=3(2)/2=3.5,$ and

$l=a_n=84$

$a_n=a+(n-1)d$

$84=7+(n-1)(3.5)$

$84=7+3.5n-3.5$

$3.5n=84-7+3.5$

$n=80.5/3.5$

$n=23$

$S_n=n/2[2a+(n-1)d]$

we can write

$S_n=n/2[a+{a+(n-1)d}]$

$S_n=n/2[a+{l}]$

$S_23=23/2[7+84]$

$S_23=23/2[91]$

$S_23=2093/2$

$S_23=1042(1)/2$

$S_23=1042.5$

Answer :

$S_23=1046.5.$

ii)

Given

$a=34, d=32-34=-2,$ and

$l=a_n=10$

$a_n=a+(n-1)d$

$10=34+(n-1)(-2)$

$10=34-2n+2$

$2n=36-10$

$n=26/2$

$n=13$

$S_n=n/2[a+l]$

$S_13=13/2[34+10]$

$S_13=13/2[44]$

$S_13=13times22$

$S_13=286$

Answer :

$S_13=286.$

iii)

Given

$a=-5, d=(-8)-(-5)=-8+5=-3,$ and

$l=a_n=-230$

$a_n=a+(n-1)d$

$-230=-5+(n-1)(-3)$

$-230=-5+3n+3$

$3n=-2+230$

$n=228/3$

$n=76$

$S_n=n/2[a+l]$

$S_76=76/2[-5-230]$

$S_76=38[-235]$

$S_76=-8930$

Answer :

$S_76=-8930.$

Q3. In an AP:

i) given $a=5, d=3, a_n=50,$ find $n$ and $S_n.$

ii) given $a=7, a_13=35,$ find $d$ and $S_13.$

iii) given $a_12=37, d=3,$ find $a$ and $S_12.$

iv) given $a_3=15, S_10=125,$ find $d$ and $a_10.$

v) given $d=5, S_9=75,$ find $a$ and $a_9.$

vi) given $a=2, d=8, S_n=-14,$ find $n$ and $a_n.$

vii) given $a=8, a_n=62, S_n=210,$ find $n$ and $d.$

viii) given $a_n=4, d=2, S_n=-14,$ find $n$ and $a.$

ix) given $a=3, n=8, S=192,$ find $d.$

x) given $l=28, S=144,$ and there are total $19$ terms. Find $a.$

Sol. :

Given

$a=5, d=3,$ and

$a_n=50$

$a_n=a+(n-1)d$

$50=5+(n-1)3$

$50=5+3n-3$

$3n=50-5+3$

$3n=48$

$n=48/3$

$n=16$

$S_n=n/2(a+a_n)$

$S_16=16/2(5+50)$

$S_16=8(55)$

$S_16=440$

Answer :

$n=16,$ and

$S_16=440$ .

ii)

Given

$a=7, a_13=35,$ and

$n=13$

$a_n=a+(n-1)d$

$35=7+(13-1)d$

$35=7+12d$

$12d=35-7$

$d=28/12=7/3$

$S_n=n/2(a+a_n)$

$S_13=13/2(7+35)$

$S_13=13/2(42)$

$S_13=13times21$

$S_13=273$

Answer :

$d=7/3$ and

$S_13=273$

iii)

Given

$a_12=37$ and

$d=3$

$a_n=a+(n-1)d$

$37=a+(12-1)3$

$37=a+11times3$

$37=a+33$

$a=37-33$

$a=4$

$S_n=n/2(a+a_n)$

$S_12=12/2(4+37)$

$S_12=6(41)$

$S_12=246$

Answer :

$a=4,$ and

$S_12=246.$

iv)

Given

$a_3=15, S_10=125,$ and

$n=10$

$a_n=a+(n-1)d$

$15=a+(3-1)d$

$15=a+2d$ ______________(1)

$S_n=n/2[2a+(n-1)d]$

$125=10/2[2a+(10-1)d]$

$125=5[2a+9d]$ [divide by

$5$ both sides]

$25=2a+9d$ _____________(2)

By equation (1)

$times2-(2)times1$

$30-25=2a+4d-(2a+9d)$

$5=2a+4d-2a+9d$

$5=-5d$

$d=-1$

Put in equation (1)

$15=a+2(-1)$

$15=a-2$

$a=15+2$

$a=17$

Now

$a_n=a+(n-1)d$

$a_10=17+(10-1)(-1)$

$a_10=17-9$

$a_10=8$

Answer :

$d=-1,$ and

$a_10=8.$

Given

$d=5, S_9=75,$ and

$n=9$

$S_n=n/2(a+a_n)$

$75=9/2(a+a_9)$

$75=(3times3)/2(a+a_9)$

$3a+3a_9=75times(2/3)$

$3a+3a_9=50$ ___________(1)

$S_9=9/2[2a+(9-1)5]$

$75=9a+180$

$9a=-180+75$

$a=-105/9=-35/3$

Put in equation (1)

$3(-35/3)+3a_9=50$

$-35+3a_9=50$

$3a_9=50+35$

$a_9=85/3$

Answer :

$a=-35/3,$ and

$a_9=85/3.$

vi)

Given

$a=2, d=8,$ and

$S_n=90$

$S_n=n/2[2a+(n-1)d]$

$90=n/2[2(2)+(n-1)8]$

$90=n/2[4+8n-8]$

$90=n/2[8n-4]$

$90=4n^2-2n$

$4n^2-2n-90=0$ [divide by

$2$ both sides]

$2n^2-n-45=0$

$2n^2-10n+9n-45=0$

$2n(n-5)+9(n-5)=0$

$(n-5)(2n+9)=0$

$n-5=0$

$n=5$

$2n+9=0$

$n=-9/2$ [ it is not possible ]

Now

$a_n=a+(n-1)d$

$a_n=2+(5-1)8$

$a_n=2+(4)8$

$a_n=34$

Answer :

$n=5$ and

$a_n=34.$

vii)

Given

$a=8, a_n=62,$ and

$S_n=210$

$S_n=n/2(a+a_n)$

$210=n/2(8+62)$

$210=n/2(70)$

$210=35n$

$n=210/35=6$

$a_n=a+(n-1)d$

$62=8+(6-1)d$

$62=8+5d$

$5d=62-8$

$d=54/5$

Answer :

$n=6,$ and

$d=54/5.$

viii)

Given

$a_n=4, d=2,$ and

$S_n=-14$

$a_n=a+(n-1)d$

$4=a+(n-1)2$

$4=a+2n-2$

$a=6-2n$ ________________(1)

$S_n=n/2(a+a_n)$

$-14=n/2(a+4)$

$-28=n(a+4)$ _____________(2)

from equation (1) and (2)

$-28=n(6-2n+4)$

$-28=n(10-2n)$

$-28=10n-2n^2$

$2n^2-10n-28=0$ [ divide by

$2$ both sides]

$n^2-5n-14=0$

$n^2-7n+2n-14=0$

$n(n-7)+2(n-7)=0$

$(n-7)(n+2)=0$

$n-7=0$

$n=7$

$n+2=0$

$n=-2$ [it is not possible ]

Put in equation (1)

$a=6-2times7$

$a=6-14$

$a=-8$

Answer :

$a=-8,$ and

$n=7.$

ix)

Given

$a=3, n=8,$ and

$S_9=192$

$S_n=n/2[2a+(n-1)d]$

$192=8/2[2(3)+(8-1)d]$

$192=4[6-7d]$

$48=6-7d$

$7d=6-48$

$d=42/7=6$

Answer :

$d=6.$

Given

$l=a_9=28, S_9=144,$ and

$n=9$

$S_9=n/2(a+l)$

$144=9/2(a+28)$

$288=9a+252$

$9a=252-288$

$a=36/9=4$

Answer :

$a=4.$

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

Sol. :

Given

$a=9, d=17-9=8,$ and

$S_n=636$

$S_n=n/2[2a+(n-1)d]$

$636=n/2[2(9)+(n-1)8]$

$636=n/2[18+8n-8]$

$636=9n+4n_2-4n$

$4n_2+5n-636=0$

$4n_2+53n-48n-636=0$

$n(4n+53)-12(4n+53)=0$

$(4n+53)(n-12)=0$

$4n+53=0$

$n=-53/4$ [ it is not possible. ]

$n-12=0$

$n=12$

Answer :

$n=12.$

Q5. 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.

Sol.

Given

$a=5, a_n=45,$ and

$S_n=400$

$S_n=n/2(a+a_n)$

$400=n/2(5+45)$

$400=25n$

$n=400/25$

$n=16$

$a_n=a(n-1)d$

$45=5+(16-1)d$

$40=15d$

$d=40/15$

$d=8/3$

Answer :

$n=16,$ and

$d=8/3.$

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

Sol. :

Given

$a=17, a_n=350,$ and

$d=9$

$a_n=a+(n-1)d$

$350=17+(n-1)9$

$350=17+9n-9$

$9n=350-8$

$n=342/9$

$n=38$

$S_n=n/2(a+a_n)$

$S_n=38/2(17+350)$

$S_n=19times367$

$S_n=6973$

Answer :

$n=38,$ and

$S_n=6973.$

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

Sol. :

Given

$d=7, a_22=149,$ and

$n=22$

$a_n=a+(n-1)d$

$149=a+(22-1)7$

$149=a+21times7$

$a=149-147$

$a=2$

$S_n=n/2(a+a_n)$

$S_22=22/2(2+149)$

$S_22=11(151)$

$S_22=1661$

Answer :

$a=2,$ and

$S_n=1661.$

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

Sol. :

Given

$a_2=14, a_3=18$

$a+d=14$ _____________(1)

$a+2d=18$ ____________(2)

By equation (2)-(1)

$a+d-(a+2d)=18-14$

$a+d-a-2d=4$

$d=4$

Put in equation (1)

$a+4=14$

$a=14-4$

$a=10$

$S_n=n/2[2a+(n-1)d]$

$S_51=51/2[2(10)+(51-1)4]$

$S_51=51/2[20+(50)4]$

$S_51=51/2[20+200]$

$S_51=51/2[220]$

$S_51=51times110$

$S_51=5610$

Answer :

$d=4, a=10,$ and

$S_51=5610.$

Q9. 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.

Sol. :

Given

$S_7=49,$ and

$S_17=289$

$S_n=n/2[2a+(n-1)d]$

$49=7/2[2a+(7-1)d]$

$49=7/2[2a+6d]$

$49/7=a+3d$

$7=a+3d$ _____________(1)

$S_17=17/2[2a+(17-1)d$

$289=17/2[2a+16d]$

$289/17=a+8d$

$17=a+8d$ ____________(2)

By equation (2)-(1)

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

$10=a+8d-a-3d$

$10=5d$

$d=10/5$

$d=2$

Put in equation (1)

$7=a+3times2$

$7=a+6$

$a=7-6$

$a=1$

$S_n=n/2[2a+(n-1)d]$

$S_n=n/2[2(1)+(n-1)2]$

$S_n=n/2[2+2n-2]$

$S_n=n/2[2n]$

$S_n=n^2$

Answer :

$d=2, a=1,$ and

$S_n=n^2.$

Q10. Show that $a_1, a_2, ...., a_n,...$ form an AP where $a_n$ is defined as below :

i) $a_n=3+4n$ ii) $a_n=9-5n$

Also find the sum of the first $15$ terms in each case.

Sol. :

$a_n=3+4n$

$a_1=3+4(1)=3+4=7$

$a_2=3+4(2)=3+8=11$

$a_3=3+4(3)=3+12=15$

$....................................$

$a_n=3+4n$

AP is

$7, 11, 15, .........(3+4n)$

where

$a=7, d=4$

$a_1, a_2, ...., a_n,...$ form an AP if

$a_n=3+4n$ Proved.

$S_n=n/2[2a+(n-1)d]$

$S_15=15/2[2(7)+(15-1)4]$

$S_15=15/2[14+14times4]$

$S_15=15/2[14+56]$

$S_15=15/2[70]$

$S_15=15times35$

$S_15=525$

Answer :

$S_15=525.$

ii)

$a_n=9-5n$

$a_1=9-5(1)=9-5=4$

$a_2=9-5(2)=9-10=-1$

$a_3=9-5(3)=9-15=-6$

$................................$

$a_n=9-5n$

AP is

$4, -1, -6, .......(9-5n).$

where

$a=4, d=-5$

$a_1, a_2, ...., a_n,...$ form an AP if

$a_n=9-5n$ Proved.

$S_n=n/2[2a+(n-1)d]$

$S_15=15/2[2(4)+(15-1)(-5)]$

$S_15=15/2[8-(14)5]$

$S_n=15/2[8-70]$

$S_n=15/2[-62]$

$S_n=-15times31$

$S_n=-465$

Answer :

$S_n=-465.$

Q11. 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 $n$ th terms.

Sol. :

Given

$S_n=4n-n^2$

$S_1=4(1)-(1)^2=4-1=3$

$a_1=3$

$S_2=4(2)-(2)^2=8-4=4$

$a_2=S_2-S_1$

$a_2=4-3=1$

$d=a_2-a_1=1-3=-2$

$a_3=a+2d$

$a_3=3+2(-2)=3-6=-3$

$a_10=a+9d$

$a_10=3+9(-2)=3-18=-15$

$a_n=a+(n-1)d$

$a_n=3+(n-1)(-2)=3-2n+2=5-2n$

Answer :

$a=S_1=3, S_2=4, a_2=1, a_3=-1, a_10=-15,$ and

$a_n=5-2n.$

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

Sol. :

AP is

$6, 12, 18, ......, 40$

where

$a=6, d=12-6=6,$ and

$n=40$

$S_n=n/2[2a+(n-1)d]$

$S_40=40/2[2(6)+(40-1)(6)]$

$S_40=20[12+39times6]$

$S_40=20[12+234]=20times246=4920$

Answer :

$S_40=4920.$

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

Sol. :

AP is 8, 16, 24, .....15`

where

$a=8, d=16-8=8,$ and

$n=15$

$S_n=n/2[2a+(n-1)d]$

$S_15=15/2[2(8)+(15-1)8]$

$S_15=15/2[16+14times8]$

$S_15=15/2[16+112]$

$S_15=15/2[128]$

$S_15=15times64$

$S_15=960$

Answer :

$S_15=960.$

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

Sol. :

AP is

$1, 3, 5, 7......, 47, 49.$

where

$a=1, d=3-1=2,$ and

$a_n=49$

$a_n=a+(n-1)d$

$49=1+(n-1)(2)$

$49=1+2n-2$

$49=2n-1$

$2n=49+1$

$2n=50$

$n=50/2$

$n=25$

$S_n=n/2[2a+(n-1)d]$

$S_25=25/2[2(1)+(25-1)2]$

$S_25=25/2[2+24times2]$

$S_25=25/2[2+48]$

$S_25=25/2[50]$

$S_25=25times25$

$S_25=625$

Answer :

$S_25=625.$

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

Sol. :

AP is

$200, 250, 300, ......a_30$

where

$a=200, d=250-200=50,$ and

$n=30$

$S_n=n/2[2a+(n-1)d]$

$S_30=30/2[2(200)+(30-1)50]$

$S_30=15[400+29times50]$

$S_30=15[400+1450]$

$S_30=15[1850]$

$S_30=27750$

Answer :

Penalty is ₹

$27750.$

Q16. A sum of ₹ $700$ is to be used to give seven cash prize is ₹ $20$ less than its preceding prize, find the value of each of the prizes.

Sol. :

Let the first prize ₹

$a,$ and

$d=-$ ₹

$20$ ,

$n=7,$ and

$S_7=$ ₹

$700$

$S_n=n/2[2a+(n-1)d]$

$700=7/2[2a+(7-1)(-20)]$

$700=7/2[2a-120]$

$700/7=a-60$

$100=a-60$

$a=100+60$

$a=160$

$a_2=a+d$

$a_2=160-20=140$

$a_3=a+2d$

$a_3=160+2(-20)=160-40=120$

$a_4=a+3d$

$a_4=160+3(-20)=160-60=100$

$a_5=a+4d$

$a_5=160+4(-20)=160-80=80$

$a_6=a+5d$

$a_6=160+5(-20)=160-100=60$

$a_7=a+6d$

$a_7=160+6(-20)=160-120=40$

Answer :

The prizes are ₹

$160, 140, 120, 100, 80, 60,$ and

$40.$

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 $1$ tree a section of Class $II$ will plant $2$ trees and so on till Class $XII$ . There are three sections of each class. How many trees will be planted by the students?

Sol. :

Class I will be planting

$1times3=3$

Class II will be planting

$2times3=6$

Class III will be planting

$3times3=9$

$...................................................$

Class XII will be planting

$12times3=36$

AP is `3, 6, 9, ....., 36.

Where

$a=3, d=6-3=3,$ and

$n=12$

$S_n=n/2[2a+(n-1)d]$

$S_12=12/2[2(3)+(12-1)3]$

$S_12=6[6+11times 3]$

$S_12=6[6+33]$

$S_12=6times39$

$S_12=234$

Answer :

$234$ trees will be planted by the students.

Q18. A spiral is made up of successive semicircles, with certres alternately at A and B, starting with centre at A, of radii $0.5$ cm, $1.0$ cm, $1.5$ cm, $2.0$ cm,... as shown in Figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $pi=22/7$ )

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

Sol. :

Let length of successive semicircles is

$l_1, l_2, l_3, l_4, ..$ with centres at A, B, A, B,..., respectively.

$l_1=0.5pi$ cm,

$l_2=1.0pi$ cm,

$l_3=1.5pi$ cm,

$l_4=2.0pi$ cm,

$......................$

$13$ terms form an AP

where

$a=0.5pi$ cm,

$d=1.0pi-0.5pi=0.5pi$ cm, and

$n=13$

$S_n=n/2[2a+(n-1)d]$

$S_13=13/2[2(0.5pi)+(13-1)0.5pi]$

$S_13=13/2[1.0pi+12times0.5pi]$

$S_13=13/2[1.0pi+6.0pi]$

$S_13=13/2[7.0pi]=13/2[7.0times22/7]$

$S_13=13/2[22]$

$S_13=13times11$

$S_13=143$

Answer :

The total length of such a spiral made up of

$13$ consecutive semicircles is

$143$ cm.

Q19. $200$ 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 (see figure). In how many rows are the $200$ logs placed and how many logs are in the top row?

Sol. :

According to question

AP is

$20, 19, 18, ...., n.$

where

$a=20, d=19-20=-1,$ and

$S_n=200$

$S_n=n/2[2a+(n-1)d]$

$200=n/2[2(20)+(n-1)(-1)]$

$200=n/2[40-(n-1)]$

$200=n/2[40-n+1]$

$200=n/2[41-n]$

$400=41n-n^2$

$n^2-41n+400=0$

$n^2-25n-16n+400=0$

$n(n-25)-16(n-25)=0$

$(n-25)(n-16)=0$

$n-25=0$

$n=25$

$a_25=20+24(-1)=20-24=-4$ [ it is not possible ]

$n-16=0$

$n=16$

$a_16=20+15(-1)=20-15=5$

Answer :

$16$ rows are the

$200$ logs placed and

$5$ logs are in the top row.

Q20. In a potato race, a bucket is placed at the starting point, which is $5$ 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 figure).

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 meters) run by a competitor is $2times5+2times(5+3)$ ]

Sol. :

The distance travelled to put and drop the potatoes are

$a_1=2times5=10$ m,

$a_2=2times(5+3)=16$ m,

$a_3=2times(8+3)=22$ m,

AP is

$10, 16, 22,....$

where

$a=10, d=16-10=6,$ and

$n=10$

$S_n= n/2[2a+(n-1)d]$

$S_10=10/2[2(10)+(10-1)6]$

$S_10=5[20+9times6]$

$S_10=5[20+54]$

$S_10=5[74]$

$S_10=370$

Answer :

The total distance the competitor has to run

$370$ m.

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10th Maths 5.3

Chapter 5

Arithmetic Progressions

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NCERT Class 10th solution of Exercise 5.1

NCERT Class 10th solution of Exercise 5.2

NCERT Class 10th solution of Exercise 6.1

NCERT Class 10th Maths Projects

Exercise 5.3

NCERT Class 10th solution of Exercise 6.1

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