NCERT full solutions here you can find full detailed solutions for all the questions with proper format. class 10 maths ncert solutions chapter 5 exercise 5.3 .
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Complete 10th Class Maths Solution
- NCERT Class 10th – Chapter 4 – Exercise 4.1 Quadratic Equations
- NCERT Class 10th – Chapter 4 – Exercise 4.2 Quadratic Equations
- NCERT Class 10th – Chapter 4 – Exercise 4.3 Quadratic Equations
EXERCISE -5.3
QUE:-1. 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:-
(i) 2, 7, 12, . . ., to 10 terms.
a= 2 , d= 7-2 =5 , n=10
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S = 10/2[2×2 +(10-1)5]
S = 5 [ 4 +45]
S =250
(ii) –37, –33, –29, . . ., to 12 terms.
a= -37 , d= -33+37 =54, n=12
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S = 12/2[2(-37) +(12-1)4]
S = 6 [ -74 +44]
S =6 ×(-30) = -180
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.
a= 0.6 , d= 1.7-0.6 =1.1, n=100
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S = 100/2 [ 2×0.6 +(100-1)1.1]
S = 50 [1.2 +108.9]
S = 50 ×110.1
S = 5505
(iv) 1/15,1/12, 1/10, . . ., to 11 terms.
a= 1/15 , d= 1/12-1/15=3/180=1/60, n=11
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S = 11/2[2×1/15 +(11-1)1/60]
S = 11/2[2/15+1/6]
=11/2 ×27/90
= 11/2 ×3/10
=33/20
QUE:-2. Find the sums given below :
(i) 7 +10(1/2)+ 14 + . . . + 84 (ii) 34 + 32 + 30 + . . . + 10
(iii) –5 + (–8) + (–11) + . . . + (–230)
SOL:-
(i) 7 +10(1/2)+ 14 + . . . + 84
a =7 d =21/2-7 =7/2
a(n) =a+(n-1)d
84 = 7 +(n-1) 7/2
84 = 7 +7/2n -7/2
7/2n =77+7/2
7/2n = 161/2
n =23
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S =23/2 [2×7 +(22)7/2]
S =23/2[14+77]
S =23/2[91]
S =2093/2 =1046.5
(ii) 34 + 32 + 30 + . . . + 10
a =34 ,d = 32-34 =-2
a(n) = 34 +(n-1)(-2)
10 = 34 -2n +2
2n = 36-10
n =26/2=13
for sum of AP use formula
S =n/2(a + l )
S =13/2 [34 +10]
S = 13/2[44]
S = 286
(iii) –5 + (–8) + (–11) + . . . + (–230)
a =-5 ,d =-8+5 =-3
a(n) = -5 +(n-1)(-3)
-230 = -5 -3n +3
3n = 230-2
n =228/3=76
for sum of AP use formula
S =n/2(a + l )
S =76/2 [-5 -230]
S = 38[-235]
S = -8930
QUE:-3. In an AP:
(i) given a = 5, d = 3, a(n)= 50, find n and Sn.
(ii) given a = 7, a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3, find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(v) given d = 5, S9 = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii) given an = 4, d = 2, Sn = –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 9 terms. Find a.
SOL:-
(i)given a = 5, d = 3, a(n)= 50, find n and Sn.
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S = 50/2[2×5 +(50-1)3]
S= 25[10+147]
S = 25×157
S = 3625
(ii) given a = 7, a13 = 35, find d and S13.
a13 = a +(13-1)d
35 =7+(12)d
12d = 35-7
12d =28
d =28/12 =7/3
for sum of AP use formula
S =n/2(a + l )
S = 13/2 (7 +35)
S = 13/2×42
S = 273
(iii) given a12 = 37, d = 3, find a and S12.
a12 = a+(n-1)d
37 = a +(12-1)3
a = 37-33=4
for sum of AP use formula
S = n/2[2a + (n – 1) d]
S = 12/2[ 2×4 +(12 -1)3]
S = 6[8+33]
S= 246
(iv) given a3 = 15, S10 = 125, find d and a10.
a3= a+(n-1)d
15 = a +2d
a =15-2d …………..(1)
S10 = n/2[2a + (n – 1) d]
125= 10/2[ 2a +(10-1)d}
125 = 5(2a +9d)
125= 10a +45d ………(2)
putting the value of a in equation (2)
125= 10(15-2d) +45d
125=150 -20d +45d
25d =125-150
d =-25/25=-1
putting the value of d in equ(1)
a =15-2(-1)
a =17
a10 = a+(10-1)(-1)
= 17 -9=8
a10 =8
(v) given d = 5, S9 = 75, find a and a9.
S9 =75
S9 = n/2[2a + (n – 1) d]
75 = 9/2(2a +(9-1)5)
75 =9a +180
9a =75-180
a = -105/9 =-35/3
a9 =-35/3 +(9-1)5
a9 =-35/3 +40
a9 =85/3
(vi) given a = 2, d = 8, Sn = 90, find n and an.
Sn =90 , a=2 ,d=8
S9 = n/2[2a + (n – 1) d]
90 = n/2 [ 2×2 +(n-1)8]
90 =2n +4n² -4n
4n²-2n-90=0
2n²-n -45=0
2n²-10n+9n-45=0
2n(n-5) +9(n-5)=0
(n-5)(2n +9)=0
n=5
a5 = 2 +(5-1)8
a5 =2 +32
a5 =34
(vii) given a = 8, an = 62, Sn = 210, find n and d.
Sn =n/2[a +an]
210 =n/2[8 +62]
210 =35n
n =210/35
n =6
an =a6
a6 = a+(6-1)d
62 = 8 +5d
5d = 62-8
d =54/5
(viii) given an = 4, d = 2, Sn = –14, find n and a
a(n) = a +(n-1)d
4 = a +2n -2
6 = a +2n
a = 6-2n………….(1)
Sn = n/2[2a + (n – 1) d]
putting the value of a from equ(1)
-14=n/2[2(6-2n)+(n-1)2)]
-14 =n/2[12-4n +2n -2]
-14 =n/2[10-2n]
-14=5n -n²
n²-5n -14=0
n²-7n +2n -14=0
n(n-7) +2(n-7)=0
(n-7)(n+2)=0
n = 7 (n ≠ -2 because n never negetive)
putting the value of n in equation (1)
a = 6-2×7
a = -8
(ix)given a = 3, n = 8, S = 192, find d.
S = n/2[2a + (n – 1) d]
192 =8/2[2×3 +(8-1)d]
192 = 4[6+7d]
6+7d =192/4
7d =48-6
d =42/7=6
d =6
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
S =n/2(a + l )
144 = 9/2(a +28)
a+28=288/9
a =32-28 =4
QUE:-4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?
SOL:-
AP : 9, 17, 25, . . .
a= 9 , d=17-9=8, S=636
S = n/2[2a + (n – 1) d]
636 =n/2[2×9 +(n-1)8]
636 =9n +4n² -4n
4n² +5n -636=0
4n² +53n -48n -636=0
n(4n+53)-12(4n-53)=0
(4n+53)(n-12) =0
n=12
so 12 terms of AP taken to give sum 636 .
QUE:-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.
SOL:-
a =5 , a(n)=45 , Sn=400 n=? , d=?
Sn = n/2 [ a +a(n)]
400 = n/2[ 5+45]
400 = 25n
n = 400/25=16
a16 =45
a16 = a+(n-1)d
45 = 5 + 15d
d =40/15
d = 8/3
so n=16 and d=8/3
QUE:-6. 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:-
an =350
an =a +(n-1)d
350 = 17 +(n-1)9
8+9n =350
9n =350-8
n = 342/ 9 = 38
S = n/2[2a + (n – 1) d]
S =38/2[2×17+(38-1)9]
S = 19[34+333]
S = 19 × 367 =6973
QUE:-7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
SOL:-
S22 =? , a22 =149 , d=7 , a=?
a22 =a +(n-1)d
149 = a +(22-1)7
149 = a +147
a =149-147=2
Sn = n/2 [ a +a(n)]
S22 = 22/2 [ 2+149]
S22 = 11× 151 =1661
QUE:-8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18
respectively.
a 2 =14 ,a3=18 , d =18-14 =4
an = a +(n-1)d
14 = a +(2-1)4
a =14-4 =10
S51 = n/2[2a + (n – 1) d]
S51= 51/2[2×10 +(51-1)4]
S51 = 51/2 [ 20+200]
S51 = 51 ×110
S51 =5610
QUE:-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.
SOL:-
Sn= n/2[2a + (n – 1) d]
S7 =7/2[2a +(7-1)d]
49= 7/2 [2a +6d ]
7a +21d =49
a +3d =7
a = 7-3d …………(1)
S17 = 17/2[2a + (17 – 1) d]
289 =17a +16d
17a +16d =289
putting the value of a from equ(1)
17 (7-3d )+136d =289
119 -51d +136d =289
85d = 289-119
d = 170/85=2
putting the value of d in equ(1)
a = 7-3×2
a = 7-6 = 1
Sn =n/2[2 +(n-1)2]
QUE:-10. Show that a1, a2, . . ., an, . . . 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:-
(i) a(n) = 3 + 4n
a1 = 3+4×1 =7
a2 = 3 +4×2 =11
d = 11-7 =4
Sn= n/2[2a + (n – 1) d]
S15 =15/2 [2×7 +(15-1)4]
= 105+420 =525
(ii) a(n )= 9 – 5n
a1 = 9-5×1 = 4
a2 =9-5×2 = -1
d = -1-4 =-5
Sn= n/2[2a + (n – 1) d]
S15 =15/2 [2×4+(15-1)(-5)]
= 60 -525
=-465
QUE:-11. If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S1)? What
is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and
the nth terms.
SOL:-
Sn =4n – n²
n =1
S1 = 4n – n² =4×1 -1² =3 =a1
S2 =4×2 – 2² =8-4=4
S2 =a1+a2
4= 3 +a2
a2 = 1
d = a2-a1 = 1-3 =-3
a10 = a +(10-1)d
a10 =4 -18 =-14
a3 = a+(3-1)d
a3 = 4 -4 =0
an = a +(n-1)d
an = 3+(n-1)(-2)
an = 3 -2n +2
an =5-2n
QUE:-12. Find the sum of the first 40 positive integers divisible by 6.
SOL:-
a =6 , d =6 , n =40
Sn= n/2[2a + (n – 1) d]
S40 = 40/2[2×6 +(40-1)6]
S40 = 20[12 +234] =4920
so the sum of first 40 positive integers divisible by 6 is =4920
QUE:-13. Find the sum of the first 15 multiples of 8.
SOL:-
a =8 , d=8 ,n =15
Sn= n/2[2a + (n – 1) d]
S15 = 15/2[2×8 +(15-1)8]
S15 = 15/2 [16 +112]
= 15/2 ×128 =960
QUE:-14. Find the sum of the odd numbers between 0 and 50.
SOL:-
According to question AP 0 to 50 ; 1,3,5,6,………49.
a = 1 , d =3-1 =2 , n= 25
Sn= n/2[2a + (n – 1) d]
Sn = 25/2 [2×1 + (25-1)2]
Sn = 25/2 [ 2 +48] = 25×25 =625
QUE:-15. 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 work by 30
days?
SOL:-
a =200 , d=250-200=50 , n=30
Sn= n/2[2a + (n – 1) d]
Sn = 30/2 [ 2×200+(30-1)50]
Sn = 15 [ 400 +1450] =15×1850 =27750
so the penalty for 30 days to deleyed work =27750
QUE:-16. A sum of ` 700 is to be used to give seven cash prizes to students of a school for their
overall academic performance. If each prize is ` 20 less than its preceding prize, find the
value of each of the prizes.
SOL:-
Sn=700 , d =-20 ,n =7 a =?
Sn= n/2[2a + (n – 1) d]
700= 7/2[2a + (7 – 1) -20]
700 = 7a -420
7a = 700+420 =1120
a =1120/7 =160
so the value of each prize is 160 ,140,120,100,80,60,40 .
QUE:-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 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:-
Total no. of class is 12
tree planted by class 1 is 1×3=3
tree planted by class 2 is 2 ×3=6
tree planted by class 3 is 3×3=9
let AP is 3 , 6, 9 …………….36
so a = 3 d =6-3 =3 n =12
Sn= n/2[2a + (n – 1) d]
Sn= 12/2[2×3 + (12– 1) 3]
= 6 [ 6 +33]
=234
so the tree planted by students are 234 .
QUE:-18. A spiral is made up of successive semicircles, with centres 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
Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive
semicircles? (Take π =22/7)
[Hint : Length of successive semicircles is l1,l2,l3,l4,……….with centres at A, B, A, B, . . .,respectively.].
SOL:-
l1,l2,l3,l4………. are the circumfrence of semicircles
so cicumfrence of semicircle is =1/2(2πr)=πr
l1 =π 0.5
l2 =π1.0
l3 =π 1.05
a = π0.5 , d =π1.0-π0.5 =0.5 , n = 13
Sn= n/2[2a + (n – 1) d]
Sn =13/2[ 2π0.5 +(13-1)0.5π]
Sn = 13/2 [1.0π +6.0π]
Sn =13/2[7π]
Sn =13/2[7×22/7]
Sn = 13/2×154/7=143
so the length of spiral is 143 cm
QUE:-19. 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 (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?
SOL:-
logs in 1st row is =20
logs in second row is =19
logs in third row is =18
a =20 d =19-20 =-1, n =? Sn =200
Sn= n/2[2a + (n – 1) d]
200 = n/2[ 2×20+(n-1)(-1)]
200 = 20n -n²/2 +n/2
400 =40n -n² +n
n² -41n +400=0
n² -16n -25n +400=0
n(n-16) -25(n-16) =0
(n-16)(n-25) =0
n =16 ,n=25
a16 =a +(n-1)d
a16 = 20 +(16-1)(-1) = 20-16 =4
a25 = 20 +(25-1)(-1)=20 -24 =-4
QUE:-20. 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 Fig. 5.6).
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 2 × 5 + 2 × (5 + 3)]
SOL:-
total distance travel to pick first patato is =2×5 =10
total distance travel to pick second patato is =2×(5 +3)=16
total distance travel to pick third patato is =2×(5+6 )=22
so the AP is =10 ,16,22,28,………….
a =10 ,d =16-10=6, n =10
Sn= n/2[2a + (n – 1) d]
Sn = 10/2[2×10 +(10-1)6]
Sn =5 [ 20+54]
Sn = 370
so the total distance covered to pick all potato is 370m.
Summary
In this chapter, we have studied the following points :
1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by
adding a fixed number d to the preceding term, except the first term. The fixed number d
is called the common difference.
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
2. A given list of numbers a1, a2, a3, . . . is an AP, if the differences a2 – a1, a3 – a2,a4 – a3, . . ., give the same value, i.e., if a(k + 10) – ak is the same for different values of k.
3. In an AP with first term a and common difference d, the nth term (or the general term) is
given by an= a + (n – 1) d.
4. The sum of the first n terms of an AP is given by :
Sn= n/2[2a + (n – 1) d]
5. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP
is given by :
S =n/2(a + l )
You can see the solution for complete chapter here –
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