NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3:
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 focus on solving advanced problems related to Arithmetic Progressions (AP). This exercise emphasizes real-world applications of AP concepts, including calculating the sum of n terms and solving word problems that require using the nth term formula.
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 are structured to simplify complex scenarios, enhancing students' ability to apply mathematical reasoning. By practicing these solutions, students gain a deeper understanding of AP, refine their problem-solving skills, and prepare thoroughly for board exams and competitive tests.
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 Overview
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 are essential for mastering the advanced applications of Arithmetic Progressions (AP). This exercise covers the sum of n terms and word problems, helping students connect mathematical concepts to real-life scenarios. By practicing these solutions, students learn to approach problems systematically, improve logical thinking, and strengthen their grasp of AP formulas.
CBSE Class 10 Previous Year Question Papers PDF with Solutions
These concepts are foundational for higher studies and competitive exams, making these solutions indispensable for thorough preparation and boosting confidence in handling complex mathematical problems.
Important Questions For Class 10 Maths Chapter 5
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 PDF
Below, we have provided the PDF for NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 on Arithmetic Progressions. This PDF contains detailed step-by-step solutions to all questions, focusing on advanced concepts like the sum of n terms and solving real-world AP problems. NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 aim to simplify learning and enhance understanding, helping students practice effectively. Download the PDF to strengthen your preparation and excel in your exams.
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 PDF
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 Arithmetic Progressions
Below is the NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 Arithmetic Progressions -
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
Solutions:
(i)
Given, 2, 7, 12 ,…, to 10 terms
For this A.P.,
First term, a = 2
And common difference, d = a
2
− a
1
= 7−2 = 5
n = 10
We know that the formula for sum of nth term in AP series is,
S
n
= n/2 [2a +(n-1)d]
S
10
= 10/2 [2(2)+(10 -1)×5]
= 5[4+(9)×(5)]
= 5 × 49 = 245
(ii)
Given, −37, −33, −29 ,…, to 12 terms
For this A.P.,
First term, a = −37
And common difference, d = a
2
− a
1
d= (−33)−(−37)
= − 33 + 37 = 4
n = 12
We know that the formula for sum of nth term in AP series is,
S
n
= n/2 [2a+(n-1)d]
S
12
= 12/2 [2(-37)+(12-1)×4]
= 6[-74+11×4]
= 6[-74+44]
= 6(-30) = -180
(iii)
Given, 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
First term, a = 0.6
Common difference, d = a
2
− a
1
= 1.7 − 0.6 = 1.1
n = 100
We know that the formula for sum of nth term in AP series is,
S
n
= n/2[2a +(n-1)d]
S
12
= 50/2 [1.2+(99)×1.1]
= 50[1.2+108.9]
= 50[110.1]
= 5505
(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms
For this A.P.,
First term, a = 1/5
Common difference, d = a
2
–a
1
= (1/12)-(1/5) = 1/60
And number of terms n = 11
We know that the formula for sum of nth term in AP series is,
S
n
=
n
/2 [2a + (
n
– 1)
d
]
= 11/2(2/15 + 10/60)
= 11/2 (9/30)
= 33/20
2. Find the sums given below
:
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Solutions:
(i)
First term, a = 7
n
th
term, a
n
= 84
Let 84 be the
n
th
term of this A.P., then as per the n
th
term formula,
a
n
= a(n-1)d
84 = 7+(n – 1)×7/2
77 = (n-1)×7/2
22 = n−1
n = 23
We know that the sum of n term is;
S
n
= n/2 (a + l) , l = 84
S
n
= 23/2 (7+84)
S
n
= (23×91/2) = 2093/2
(ii)
Given, 34 + 32 + 30 + ……….. + 10
For this A.P.,
First term, a = 34
Common difference, d = a
2
−a
1
= 32−34 = −2
n
th
term, a
n
= 10
Let 10 be the
n
th
term of this A.P., therefore,
a
n
= a +(n−1)d
10 = 34+(n−1)(−2)
−24 = (n −1)(−2)
12 = n −1
n = 13
We know that the sum of n terms is;
S
n
= n/2 (a +l) , l = 10
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
(iii)
Given, (−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
First term, a = −5
nth term, a
n
= −230
Common difference, d = a
2
−a
1
= (−8)−(−5)
⇒d = − 8+5 = −3
Let −230 be the
n
th
term of this A.P., and by the n
th
term formula we know,
a
n
=
a
+(
n
−1)
d
−230 = − 5+(
n
−1)(−3)
−225 = (
n
−1)(−3)
(
n
−1) = 75
n
= 76
And the sum of n term,
S
n
=
n
/2 (
a
+
l
)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930
3. 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
= 90, 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 9 terms. Find
a
.
Solutions:
(i)
Given that,
a
= 5,
d
= 3,
a
n
= 50
As we know, from the formula of the nth term in an AP,
a
n
=
a
+(
n
−1)
d
,
Therefore, putting the given values, we get,
⇒ 50 = 5+(
n
-1)×3
⇒ 3(
n
-1) = 45
⇒
n
-1 = 15
⇒
n
= 16
Now, sum of n terms,
S
n
=
n
/2 (
a
+
a
n
)
S
n
= 16/2 (5 + 50) = 440
(ii)
Given that,
a
= 7,
a
13
= 35
As we know, from the formula of the nth term in an AP,
a
n
=
a
+(
n
−1)
d
,
Therefore, putting the given values, we get,
⇒ 35 = 7+(13-1)
d
⇒ 12
d
= 28
⇒
d
= 28/12 = 2.33
Now,
S
n
=
n
/2 (
a
+
a
n
)
S
13
= 13/2 (7+35) = 273
(iii)
Given that,
a
12
= 37,
d
= 3
As we know, from the formula of the n
th
term in an AP,
a
n
=
a
+(
n
−1)
d
,
Therefore, putting the given values, we get,
⇒
a
12
=
a
+(12−1)3
⇒ 37 =
a
+33
⇒
a
= 4
Now, sum of nth term,
S
n
=
n
/2 (
a
+
a
n
)
S
n
=
12
/2 (4+37)
= 246
(iv)
Given that,
a
3
= 15,
S
10
= 125
As we know, from the formula of the nth term in an AP,
a
n
=
a
+(
n
−1)
d
,
Therefore, putting the given values, we get,
a
3
=
a
+(3−1)
d
15 =
a
+2
d
…………………………..
(i)
Sum of the nth term,
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
S
10
= 10/2 [2
a
+(10-1)
d
]
125 = 5(2
a
+9
d
)
25 = 2
a
+9
d
………………………..
(ii)
On multiplying equation
(i)
by
(ii)
, we will get;
30 = 2
a
+4
d
……………………………….
(iii)
By subtracting equation
(iii)
from
(ii)
, we get,
−5 = 5
d
d
= −1
From equation
(i)
,
15 =
a
+2(−1)
15 =
a
−2
a
= 17 = First term
a
10
=
a
+(10−1)
d
a
10
= 17+(9)(−1)
a
10
= 17−9 = 8
(v)
Given that,
d
= 5,
S
9
= 75
As, sum of n terms in AP is,
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
Therefore, the sum of first nine terms are;
S
9
= 9/2 [2
a
+(9-1)
5
]
25 = 3(
a
+20)
25 = 3
a
+60
3
a
= 25−60
a
= -35/3
As we know, the n
th
term can be written as;
a
n
=
a
+(
n
−1)
d
a
9
=
a
+(9−1)(5)
= -35/3+8(5)
= -35/3+40
= (35+120/3) = 85/3
(vi)
Given that,
a
= 2,
d
= 8,
S
n
= 90
As, sum of n terms in an AP is,
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
90 =
n
/2 [2
a
+(
n
-1)
d
]
⇒ 180 =
n
(4+8
n
-8) =
n
(8
n
-4) = 8
n
2
-4
n
⇒ 8
n
2
-4
n –
180 = 0
⇒ 2
n
2
–
n
-45 = 0
⇒ 2
n
2
-10
n
+9
n
-45 = 0
⇒ 2
n
(
n
-5)+9(
n
-5) = 0
⇒ (
n
-5)(2
n
+9) = 0
So,
n
= 5 (as n only is a positive integer)
∴
a
5
= 8+5×4 = 34
(vii)
Given that,
a
= 8,
a
n
= 62,
S
n
= 210
As, the sum of n terms in an AP is,
S
n
=
n
/2 (
a
+
a
n
)
210 =
n
/2 (8 +62)
⇒ 35
n
= 210
⇒
n
= 210/35 = 6
Now, 62 = 8+5
d
⇒ 5
d
= 62-8 = 54
⇒
d
= 54/5 = 10.8
(viii)
Given that, n
th
term,
a
n
= 4, common difference,
d
= 2, sum of n terms,
S
n
= −14.
As we know, from the formula of the n
th
term in an AP,
a
n
=
a
+(
n
−1)
d
,
Therefore, putting the given values, we get,
4 =
a
+(
n
−1)2
4 =
a
+2
n
−2
a
+2
n
= 6
a
= 6 − 2
n
………………………………………….
(i)
As we know, the sum of n terms is;
S
n
=
n
/2 (
a
+
a
n
)
-14 =
n
/2 (
a
+
4
)
−28 =
n
(
a
+4)
−28 =
n
(6 −2
n
+4) {From equation
(i)
}
−28 =
n
(− 2
n
+10)
−28 = − 2
n
2
+10
n
2
n
2
−10
n
− 28 = 0
n
2
−5
n
−14 = 0
n
2
−7
n+
2
n
−14 = 0
n
(
n
−7)+2(
n
−7) = 0
(
n
−7)(
n
+2) = 0
Either
n
− 7 = 0 or
n
+ 2 = 0
n
= 7 or
n
= −2
However,
n
can neither be negative nor fractional.
Therefore,
n
= 7
From equation
(i)
, we get
a
= 6−2
n
a
= 6−2(7)
= 6−14
= −8
(ix) Given that, first term,
a
= 3,
Number of terms,
n
= 8
And sum of n terms,
S
= 192
As we know,
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
192 = 8/2 [2×3+(8 -1)
d
]
192 = 4[6 +7
d
]
48 = 6+7
d
42 = 7
d
d =
6
(x) Given that,
l
= 28,
S
= 144 and there are total of 9 terms.
Sum of n terms formula,
S
n
=
n
/2 (
a
+
l
)
144 = 9/2(
a
+28)
(16)×(2) =
a
+28
32 =
a
+28
a
= 4
4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Solutions:
Let there be
n
terms of the AP. 9, 17, 25 …
For this A.P.,
First term,
a
= 9
Common difference, d
=
a
2
−
a
1
= 17−9 = 8
As, the sum of n terms, is;
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
636 =
n
/2 [2×
a
+(8-1)×8]
636 =
n
/2 [18+(
n
-1)×8]
636 =
n
[9 +4
n
−4]
636 =
n
(4
n
+5)
4
n
2
+5
n
−636 = 0
4
n
2
+53
n
−48
n
−636 = 0
n
(4
n
+ 53)−12 (4
n
+ 53) = 0
(4
n
+53)(
n
−12) = 0
Either 4
n
+53 = 0 or
n
−12 = 0
n
= (-53/4) or
n
= 12
n
cannot be negative or fraction, therefore,
n
= 12 only.
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.
Solution:
Given that,
first term, a
= 5
last term, l
= 45
Sum of the AP, S
n
= 400
As we know, the sum of AP formula is;
S
n
=
n
/2 (
a
+
l
)
400 =
n
/2(5+45)
400 =
n
/2(50)
Number of terms, n
=16
As we know, the last term of AP series can be written as;
l = a+
(
n
−1)
d
45 = 5 +(16 −1)
d
40 = 15
d
Common difference, d
= 40/15 = 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?
Solution:
Given that,
First term, a
= 17
Last term, l
= 350
Common difference, d
= 9
Let there be
n
terms in the A.P., thus the formula for last term can be written as;
l = a+
(
n
−1)
d
350 = 17+(
n
−1)9
333 = (
n
−1)9
(
n
−1) = 37
n
= 38
S
n
=
n
/2 (
a
+
l
)
S
38
= 38/2 (17+350)
= 19×367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
7. Find the sum of first 22 terms of an AP in which
d
= 7 and 22
nd
term is 149.
Solution:
Given,
Common difference, d = 7
22
nd
term, a
22
= 149
Sum of first 22 term, S
22
= ?
By the formula of nth term,
a
n
=
a
+(
n
−1)
d
a
22
=
a
+(22−1)
d
149 =
a
+21×7
149 =
a
+147
a
= 2 = First term
Sum of n terms,
S
n
=
n
/2(
a
+
a
n
)
S
22
= 22/2 (2+149)
= 11×151
= 1661
8. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively.
Solution:
Given that,
Second term, a
2
= 14
Third term, a
3
= 18
Common difference, d
=
a
3
−
a
2
= 18−14 = 4
a
2
=
a
+
d
14 =
a
+4
a
= 10 = First term
Sum of n terms;
S
n
=
n
/2 [2
a
+ (
n
– 1)
d
]
S
51
= 51/2 [2×10 (51-1) 4]
= 51/2 [20+(50)×4]
= 51 × 220/2
= 51 × 110
= 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.
Solution:
Given that,
S
7
= 49
S
17
= 289
We know, sum of n terms;
S
n
=
n
/2 [2
a
+ (
n
– 1)
d
]
Therefore,
S
7
=
7
/2 [2
a
+(
n
-1)
d
]
S
7
= 7/2 [2
a
+ (7 -1)
d
]
49 = 7/2 [2
a
+ 6
d
]
7 = (
a
+3
d
)
a
+ 3
d
= 7 ………………………………….
(i)
In the same way,
S
17
= 17/2 [2
a
+(17-1)
d
]
289 = 17/2 (2
a
+16
d
)
17 = (
a
+8
d
)
a
+8
d
= 17 ……………………………….
(ii)
Subtracting equation
(i)
from equation
(ii)
,
5
d
= 10
d
= 2
From equation
(i)
, we can write it as;
a
+3(2) = 7
a+
6 = 7
a =
1
Hence,
S
n
=
n
/2[2
a
+(
n
-1)
d
]
=
n
/2[2(1)+(
n
– 1)×2]
=
n
/2(2+2
n
-2)
=
n
/2(2
n
)
=
n
2
10. Show that
a
1
,
a
2
… ,
a
n
, … form an AP where
a
n
is defined as below
(i)
a
n
= 3+4
n
(ii)
a
n
= 9−5
n
Also, find the sum of the first 15 terms in each case.
Solutions:
(i)
a
n
= 3+4
n
a
1
= 3+4(1) = 7
a
2
= 3+4(2) = 3+8 = 11
a
3
= 3+4(3) = 3+12 = 15
a
4
= 3+4(4) = 3+16 = 19
We can see here, the common difference between the terms are;
a
2
−
a
1
= 11−7 = 4
a
3
−
a
2
= 15−11 = 4
a
4
−
a
3
= 19−15 = 4
Hence,
a
k
+ 1
−
a
k
is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Now, we know, the sum of nth term is;
S
n
=
n
/2[2
a
+(
n
-1)
d
]
S
15
= 15/2[2(7)+(15-1)×4]
= 15/2[(14)+56]
= 15/2(70)
= 15×35
= 525
(ii)
a
n
= 9−5
n
a
1
= 9−5×1 = 9−5 = 4
a
2
= 9−5×2 = 9−10 = −1
a
3
= 9−5×3 = 9−15 = −6
a
4
= 9−5×4 = 9−20 = −11
We can see here, the common difference between the terms are;
a
2
−
a
1
= −1−4 = −5
a
3
−
a
2
= −6−(−1) = −5
a
4
−
a
3
= −11−(−6) = −5
Hence,
a
k
+ 1
−
a
k
is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
Now, we know, the sum of nth term is;
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
S
15
= 15/2[2(4) +(15 -1)(-5)]
= 15/2[8 +14(-5)]
= 15/2(8-70)
= 15/2(-62)
= 15(-31)
= -465
11. If the sum of the first
n
terms of an AP is 4
n
−
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.
Solution:
Given that,
S
n
= 4
n
−
n
2
First term,
a
=
S
1
= 4(1) − (1)
2
= 4−1 = 3
Sum of first two terms =
S
2
= 4(2)−(2)
2
= 8−4 = 4
Second term,
a
2
=
S
2
−
S
1
= 4−3 = 1
Common difference, d
=
a
2
−
a
= 1−3 = −2
N
th
term,
a
n
=
a
+(
n
−1)
d
= 3+(
n
−1)(−2)
= 3−2
n
+2
= 5−2
n
Therefore,
a
3
= 5−2(3) = 5-6 = −1
a
10
= 5−2(10) = 5−20 = −15
Hence, the sum of first two terms is 4. The second term is 1.
The 3
rd
, the 10
th
, and the
n
th
terms are −1, −15, and 5 − 2
n
respectively.
12. Find the sum of the first 40 positive integers divisible by 6.
Solution:
The positive integers that are divisible by 6 are 6, 12, 18, 24 ….
We can see here, that this series forms an A.P. whose first term is 6 and the common difference is 6.
a
= 6
d
= 6
S
40
=
?
By the formula of sum of n terms, we know,
S
n
=
n
/2 [2
a
+(
n
– 1)
d
]
Therefore, putting n = 40, we get,
S
40
= 40/2 [2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20(12+234)
= 20×246
= 4920
13. Find the sum of first 15 multiples of 8.
Solution:
The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP, having first term as 8 and common difference as 8.
Therefore,
a
= 8
d
= 8
S
15
= ?
By the formula of sum of nth term, we know,
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
S
15
= 15/2 [2(8) + (15-1)8]
= 15/2[16 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 64
= 960
14. Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.
Therefore, we can see that these odd numbers are in the form of A.P.
Hence,
First term, a = 1
Common difference, d = 2
Last term,
l
= 49
By the formula of last term, we know,
l
=
a
+(
n
−1)
d
49 = 1+(
n
−1)2
48 = 2(
n
− 1)
n
− 1 = 24
n
= 25 = Number of terms
By the formula of sum of nth term, we know,
S
n
=
n
/2(
a
+
l
)
S
25
= 25/2 (1+49)
= 25(50)/2
=(25)(25)
= 625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days.
Solution:
We can see that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.
Therefore, a
= 200 and
d
= 50
The penalty that has to be paid if the contractor has delayed the work by 30 days =
S
30
By the formula of sum of nth term, we know,
S
n
=
n
/2[2
a
+(
n
-1)
d
]
Therefore,
S
30
= 30/2[2(200)+(30 – 1)50]
= 15[400+1450]
= 15(1850)
= 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.
Solution:
Let the cost of 1
st
prize be Rs.
P
.
Cost of 2
nd
prize = Rs.
P
− 20
And cost of 3
rd
prize = Rs.
P
− 40
We can see that the cost of these prizes are in the form of A.P., having common difference as −20 and first term as
P
.
Thus, a
=
P and d
= −20
Given that,
S
7
= 700
By the formula of sum of nth term, we know,
S
n
=
n
/2 [2
a
+ (
n
– 1)
d
]
7/2 [2
a
+ (7 – 1)
d
] = 700
a
+ 3(−20) = 100
a
−60 = 100
a
= 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 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 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?
Solution:
It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term,
a
= 1
Common difference,
d
= 2−1 = 1
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
S
12
= 12/2 [2(1)+(12-1)(1)]
= 6(2+11)
= 6(13)
= 78
Therefore, the number of trees planted by 1 section of the classes = 78
The number of trees planted by 3 sections of the classes = 3×78 = 234
Therefore, 234 trees will be planted by the students.
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, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Solution:
We know,
Perimeter of a semi-circle = π
r
Therefore,
P
1
= π(0.5) = π/2 cm
P
2
= π(1) = π cm
P
3
= π(1.5) = 3π/2 cm
Where, P
1,
P
2
,
P
3
are the lengths of the semi-circles.
Hence we got a series here, as,
π/2, π, 3π/2, 2π, ….
P
1
= π/2 cm
P
2
= π cm
Common difference, d
=
P
2
– P
1
= π – π/2 = π/2
First term = P
1
=
a
= π/2 cm
By the sum of n term formula, we know,
S
n
=
n
/2 [2
a
+ (
n
– 1)
d
]
Therefor, the sum of the length of 13 consecutive circles is;
S
13
=
13
/2 [2(π/2) + (13 – 1)π/2]
=
13
/2 [π + 6π]
=13
/2 (7π)
=
13
/2 × 7 ×
22
/7
= 143 cm
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. In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
We can see that the numbers of logs in rows are in the form of an A.P. 20, 19, 18…
For the given A.P.,
First term, a
= 20 and common difference,
d
=
a
2
−
a
1
= 19−20 = −1
Let a total of 200 logs be placed in
n
rows.
Thus,
S
n
= 200
By the sum of nth term formula,
S
n
=
n
/2 [2
a
+(
n
-1)
d
]
S
12
= 12/2 [2(20)+(
n
-1)(-1)]
400 =
n
(40−
n
+1)
400 =
n
(41-
n
)
400 = 41
n
−
n
2
n
2
−41
n
+ 400 = 0
n
2
−16
n
−25
n
+400 = 0
n
(
n
−16)−25(
n
−16) = 0
(
n
−16)(
n
−25) = 0
Either (
n
−16) = 0 or
n
−25 = 0
n
= 16 or
n
= 25
By the nth term formula,
a
n
=
a
+(
n
−1)
d
a
16
= 20+(16−1)(−1)
a
16
= 20−15
a
16
= 5
Similarly, the 25
th
term could be written as;
a
25
= 20+(25−1)(−1)
a
25
= 20−24
= −
4
It can be seen, the number of logs in the 16
th
row is 5 as the numbers cannot be negative.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16
th
row is 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 2×5+2×(5+3)]
Solution:
The distance of the potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.
Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.
Therefore, the distance to be run w.r.t distances of potatoes, could be written as;
10, 16, 22, 28, 34,……….
Hence, the first term, a
= 10 and
d
= 16−10 = 6
S
10
=?
By the formula of sum of n terms, we know,
S
10
= 10/2 [2(10)+(10 -1)(6)]
= 5[20+54]
= 5(74)
= 370
Therefore, the competitor will run a total distance of 370 m.
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