GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

   

Gujarat Board GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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) \(\frac { 1 }{ 15 } \), \(\frac { 1 }{ 12 } \), \(\frac { 1 }{ 10 } \)to 11 terms.
Solution:
(i) 2, 7, 12, … to 10 terms
Here a = 2, d = 7 – 2 = 5
We know that
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S10 = \(\frac { 10 }{ 2 } \) [2 x 2 + (10 – 1) x 5 ]
S10 = 5 [4 + 9 x 5 ]
S10 = 5 x 49
S10 = 245
Hence the sum of 10 terms of AP is 245.

(ii) – 37, – 33, – 29, … to 12 terms
Here a = – 37, d = – 33 + 37 = 4
We know that sum of AP is
Sn= \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S12 = \(\frac { 12 }{ 2 } \) [2 x ( – 37) + (12 – 1) x 4]
S12 = 6 [ – 74 + 44]
S12 = 6 x – 30
S12 = -180
Hence the sum of 12 terms of AP is – 180.

(iii) 0.6, 1.7, 2.8, … to 100 terms
Here a = 0.6 and d = 1.7 – 0.6 = 1.1
We know the sum of AP is
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1) d]
S100 = \(\frac { 100 }{ 2 } \) [2 x 0.6 + (100 – 1) x 1.1]
S100 = 50 [1.2 + 99 x 1.1]
S100 = 50 [1.2 + 108.9]
S100 = 50 [110.1]
S100 = 5505
The um of loo terms of AP is 5505.

(iv) \(\frac { 1 }{ 15 } \), \(\frac { 1 }{ 12 } \), \(\frac { 1 }{ 10 } \), ……..to 11 terms
Here a = \(\frac { 1 }{ 15 } \)
and d = \(\frac { 1 }{ 12 } \) – \(\frac { 1 }{ 15 } \) = \(\frac { 5-4 }{ 60 } \)
d = \(\frac { 1 }{ 60 } \)
n = 11
We know that the sum of AP is
Sn = [2a + (n – 1)d]
S11 = \(\frac { 11 }{ 1 } \) \(\left[ 2x\frac { 1 }{ 15 } +(11-1)x\frac { 1 }{ 60 } \right] \)
S11 = \(\frac { 11 }{ 2 } \) \(\left[ \frac { 2 }{ 15 } +\frac { 10 }{ 60 } \right] \)
S11 = \(\frac { 11 }{ 2 } \) \(\left[ \frac { 2+10 }{ 60 } \right] \)
S11 = \(\frac { 11 }{ 2 } \) x \(\frac { 18 }{ 60 } \)
S11 = \(\frac { 33 }{ 20 } \)
Sum of 11 terms of AP is \(\frac { 33 }{ 20 } \).

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 2.
Find the sums given below:
(i) 7 + 10\(\frac { 1 }{ 2 } \) + 14 + ……… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) – 5 + (-8) + (-11) + ………… + (-230)
Solution:
(i) 7 + 10\(\frac { 1 }{ 2 } \) + 14 + ……… + 84
Here
a = 7, d = 10\(\frac { 1 }{ 2 } \) – 7
d = \(\frac { 21 }{ 2 } \) – 7 = \(\frac { 7 }{ 2 } \)
an = a + (n – 1) d
84 = 7 + (n – 1) \(\frac { 7 }{ 2 } \)
77 = (n – 1) \(\frac { 7 }{ 2 } \)
n – 1 = \(\frac { 77 x 2 }{ 7 } \)
n – 1 = 22
n = 23
Sum of AP, Sn = \(\frac { n }{ 2 } \) \(\frac { 1 }{ 2 } \)
S23 = \(\frac { 23 }{ 2 } \) \(\left[ 2×7+(23-1)x\frac { 7 }{ 2 } \right] \)
S23 = \(\frac { 23 }{ 2 } \) \(\left[ 14+22x\frac { 7 }{ 2 } \right] \)
S23 = \(\frac { 23 }{ 2 } \) \(\left[ 14+77 \right] \)
S23 = \(\frac { 23 }{ 2 } \) x 91
S23 = \(\frac { 2093 }{ 2 } \)
S23 = 1046\(\frac { 1 }{ 2 } \)
Hence the required sum is 1046\(\frac { 1 }{ 2 } \)

(ii) 34 + 32 + 30 + ………. + 10
Here a = 34, d = 32 – 34 = – 2
l = 10
Let the number of terms of the AP be n.
Then l = a + (n – 1)d
10 = 34 + (n – 1) x ( – 2)
\(\frac { -24 }{ -2 } \) = n – 1
n – 1 = 12
n = 13
We know that
Sn = \(\frac { n }{ 2 } \) [a + l]
S13 = \(\frac { 13 }{ 2 } \) [34 + 10]
S13 = \(\frac { 13 }{ 2 } \) x 44
S13 = 13 x 22
S13 = 286
Hence required sum is 286.

(iii) -5 + (-8) + (-11) + ……….. + (-230)
Here a = – 5, d = – 8 – (- 5)
d = – 8 + 5 = – 3
l = – 230
Let n be the number of terms is AP. Then,
l = a + (n – 1)d
-230 = – 5 + (n – 1) (- 3)
(n – 1) x (- 3) = – 230 + 5
n – 1 = \(\frac { 225 }{ 3 } \) = 75
n = 76
The sum of AP is
Sn = \(\frac { n }{ 2 } \) [a + l]
S76 = \(\frac { 76 }{ 2 } \) [ – 5 + (- 230)]
S76 = \(\frac { 76 }{ 2 } \) [ – 5 – 230]
S76 = \(\frac { 76 }{ 2 } \) [- 235]
S76 = 38 x -235
S76 = – 8930
Hence the required sum is -8930.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 3.
In an AP:
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7,a13 = 35. find d and S13.
(iii) Given 12 = 37, d = 3, find a and S12.
(iv) Given a = 15, S10 = 125, find d and S12.
(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 thora nra total 9
terms, find a
Solution:
(i) Here a = 5
d = 3
an = 50
We know that
an = a + (n – 1) d
50 = 5 + (n – 1) x 3
n – 1 = \(\frac { 45 }{ 3 } \)
n – 1 = 15 + 1
n = 16
Sum of AP is
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1) d]
Sn = \(\frac { -16 }{ 2 } \) [2 x 5 + (16 – 1) x 3 ]
Sn = 8 [ 10 + 45 ]
Sn = 8 x 55
Sn = 440

(ii) Here
a = 7
a13 = 35
a + 12d = 35
7 + 12d = 35
12d = 28
d = \(\frac { 28 }{ 12 } \) = \(\frac { 7 }{ 3 } \)
d = \(\frac { 7 }{ 3 } \)
We know that
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S13 = \(\frac { 13 }{ 2 } \) [ 2 x 7 + (13 – 1) x \(\frac { 7 }{ 3 } \)]
S13 = \(\frac { 13 }{ 2 } \) [ 14 + 12 x \(\frac { 7 }{ 3 } \) ]
S13 = \(\frac { 13 }{ 2 } \) [14 + 28]
S13 = \(\frac { 13 }{ 2 } \) x 42
S13 = 13 x 21
S13 = 273

(iii) Here a12 = 37
Now, d = 3
a12 = 37
a + 11d = 37
a + 11 x 3 = 37
a = 37 – 33
a = 4
s12 = \(\frac { 17 }{ 2 } \) [ 2a + (n – 1)d]
s12 = \(\frac { 12 }{ 2 } \) [ 2 x 4 + ( 12 – 1) x 3]
s12 = 6 [ 8 + 11 x 3]
s12 = 6 x 41
s12 = 246

(iv) Here a3 = 15
Now, S10 = 125
a3 = 15
a + 2d = 15
and Sn = \(\frac { n }{ 2 } \) [ 2a + (n – 1) d]
S10 = \(\frac { 10 }{ 2 } \) [ 2 x a + (10 – 1) x d]
125 = 5 [ 2a + 9d]
25 = 2a + 9d
2a + 9d = 25
Solving eqn. (1) and (2)
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 img-1
Putting value of d = – 1 in eqn. (1)
a + 24 = 15
a + 2 x (- 1) = 15
a – 2 = 15
a = 15 + 2
a = 17
them a10 = a + 9d
a10 = 17 + 9 x ( – 1)
a10 = 17 – 9

(v) Here d = 5
S9 = 75
We know that
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S9 = \(\frac { 9 }{ 2 } \) [2 x a + (9 – 1) x 5]
75 = \(\frac { 9 }{ 2 } \) [ 2a + 8 x 5]
150 = 9 x [2a + 40]
2a = \(\frac { 50 }{ 3 } \) – 40
2a = \(\frac { 50 – 120 }{ 3 } \)
2a = \(\frac { -70 }{ 3 } \)
a = \(\frac { -70 }{ 3×2 } \)
a = \(\frac { -35 }{ 3 } \)
Again we know that
an = a + (n – 1) d
a9 = a + (9 – 1)d
a9 = a + 8d
a9 = \(\frac { -35 }{ 3 } \) + 8 x 5
a9 = \(\frac { -35 }{ 3 } \) + 40
a9 = \(\frac { -35+120 }{ 3 } \)
a9 = \(\frac { 85 }{ 3 } \)

(vi) Here a = 2, d = 8,
and Sn = 90
We know that
Sn = \(\frac { n }{ 2 } \) [ 2a + (n – 1) x 8]
90 = \(\frac { n }{ 2 } \) [2 x 2 + (n – 1) x 8]
180 = n [4 + 8n – 8]
180 = n [8n – 4]
180 = 4n [2n – 1]
\(\frac { 180 }{ 4 } \) = 2n2 – n
45 = 2n2 – n
2n2 – n – 45 = 0
2n(n – 5) + 9(n – 5) = 0
(n – 5)(2n + 9) = 0
n = 5 or n = \(\frac { -9 }{ 2 } \)
It is number of term. So it will be a natural number. Hence, n = 5

(vii) Here a = 8, an = 62, Sn = 210
We know that
an = a + (n – 1)d
62 = 8 + (n – 1) x d
54 = ( n – 1) d
(n – 1) d = 54 …………… (1)
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
210 = \(\frac { n }{ 2 } \) [2 x 8 + (n – 1) d] …………(2)
From eqn. (1) and (2) we get
210 = \(\frac { n }{ 2 } \) [16 + 54]
210 = \(\frac { n }{ 2 } \) x [70]
n = \(\frac { 210 }{ 35 } \)
n = 6
Putting n = 6 in eqn. (1), we get(n – 1)d = 54
(6 – 1)d = 54
5d = 54
d = \(\frac { 54 }{ 5 } \)
(viii) Herr an = 4
d = 2
Sn = – 14
We know that
an = a + (n – 1)d
4 = a + (n – 1) 2
4 = a + 2n – 2
a + 2n = 6
Now Sn = – 14
We know that
Sn = \(\frac { n }{ 2 } \)[2a + (n – 1)d]
– 14 = \(\frac { n }{ 2 } \) [2a + (n – 1) 2]
– 14 = \(\frac { n }{ 2 } \) [a + n – 1]
– 14 = n[6 – n – 1]
[From eqn. (1) a + 2n = 6 ⇒ a + n = 6 – n]
– 14 = n[5 – n]
– 14 = 5n – n2
n2 – 5n – 14 = 0
n2 – 7n + 2n – 14 = 0
n2 – 7n + 2(n – 7) = 0
n(n – 7) + 2(n – 2) = 0
(n – 7) (n + 2) = 0
n = 7 or n = – 2
n = – 2 is impossible because n is a number of terms.
Therefore, n = 7
Putting value of in eqn. (1)
a + 2n = 6
a +2 x 7 = 6
a = 6 – 14
a = – 8

(ix) It is given that
a = 3
n = 8
S = 192
We know that
8 = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
192 = \(\frac { 8 }{ 2 } \)[2 x 3 + (8 – 1) x d]
\(\frac { 192 }{ 4 } \) = [16 + 7d]
48 = 6 + 7d
7d = 42
d = 6

(x) It is given that
l = 28
S = 144
n = 9
We know that
S = \(\frac { n }{ 2 } \) [a + l]
144 = \(\frac { 9 }{ 2 } \) [a + 28]
\(\frac { 144×2 }{ 9 } \) = a + 28
32 = a + 28
a = 4

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 4.
How many terms of the AP: 9, 17., 25, ……………. must be taken to give s sum of 636?
Solution:
Given AP is 9, 17, 25, …………..
Here a = 9, d = 17 – 9 = 8
Let n terms must be taken, then
Sn = 636
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1) d]
\(\frac { n }{ 2 } \)[2a x 9 + (n – 1) x 8] = 636
n[9 + 4n – 4] = 636
n[4n + 5] = 636
4n2 + 5n = 636
4n2 + 5n – 636 = 0
4n2 + 53n – 48n – 636 = 0
n(4n + 53) – 12(4n + 53) = 0
(4n + 53)(n – 12) = 0
4n + 53 = 0 or n – 12 = 0
n = \(\frac { -53 }{ 4 } \) (Rejected) or n = 12
Therefore. 12 terms of the AP must be taken.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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:
It is given that
a = 5
l = 45
S = 400
We know that
S = \(\frac { n }{ 2 } \) [a + l]
400 = \(\frac { n }{ 2 } \) [5 + 45]
800 = n [50]
n = \(\frac { 800 }{ 50 } \) = 16
Hence the number of terms is 16. We also know
that
l = a + (n – 1 )d
45 = 5 + (16 – 1) x d
40 = 15d
d = \(\frac { 40 }{ 15 } \)
d = \(\frac { 8 }{ 3 } \)
Hence the common difference is \(\frac { 8 }{ 3 } \)

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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?
Solution:
Here a = 17
l = 350
d = 9
We know that
l = a + (n – 1)d
350 = 17 + (n – l) x 9
350 – 17 = (n – 1)9
(n – 1) = \(\frac { 333 }{ 9 } \)
n – l = 37
n = 38
So, there are 38 terms. We know that
Sn= \(\frac { n }{ 2 } \) (a.I)
S38 = \(\frac { 38 }{ 2 } \) (17 + 350)
S38= 19 x 367
S38 = 6973
Hence sum is 6973.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here d = 7
a22 = 149
We know that
an = a + (n – 1)d
a22 = a + (22 – 1)d
149 = a + 21 x 7
a = 149 – 147
a = 2
Sum of first 22 terms
S22 = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S22 = \(\frac { 22 }{ 2 } \) [2 x 2 + (22 – 1) x 7]
S22 = 11[4 + 21 x 7]
S22 = 11 [4 + 147]
S22 = 11 x 151
S22 = 1661
Hence sum of 22 terms is 1661.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 8.
Find the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let a and d are the first term and the common difference respectively.
Second term = 14
(an = a + (n – 1) d)
a + d = 14 …………… (1)
Third term = 18
a + 2d = 18 ……………(2)
Solving eqn. (1) and (2)
a = 10 and d = 4
Now sum of first 51 terms of the AP
S51 = [2a + (51 – 1) d]
(Sn = \(\frac { n }{ 2 } \)[2a + (n – 1) d])
S51= \(\frac { 51 }{ 2 } \) [2a + 50 d]
S51 = 51 [a + 25d]
= 51 [10 + 25 x 4 ]
= 51 [10 + 100]
= 51 x 110
= 5610

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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:
Let a be the first term and d be the common difference of the AP.
Sum of first seven terms = 49
S7 = 49
(Sn = \(\frac { n }{ 2 } \)[2a + (n – 1) d])
= \(\frac { 7 }{ 2 } \) [2a + (7 – 1)d] = 49
\(\frac { 17 }{ 2 } \) [2a + 6d] = 49
a + 3d = 7 …………. (1)
Sum of first 17 terms – 289
S17 = 289
\(\frac { 17 }{ 2 } \) [ 2a + (17 – 1)d] = 289
17 [a + 8d] = 289
a + 8d = \(\frac { 289 }{ 17 } \)
a + 8d = 17 ………….. (2)
Solving eqn (1) and (2), we get
a = 1 and d = 2
Sum of first n terms
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
= \(\frac { n }{ 2 } \) [2 x 1 + (n – 1) x 2]
=n [1 + n – 1]
= n2

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 10.
Show that a1, a2, a3, ………., an form an AP where an,
is defined as below.
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
(i) We have
an = 3 + 4n
Putting n = 1, 2, 3, 4 we get
a1 = 3 + 4 x 1 = 7
a2 = 3 + 4 x 2 = 11
a3 = 3 + 4 x 3 = 15
a4 = 3 + 4 x 4 = 19
…………………………………
…………………………………
…………………………………
a2 – a1 = 11 – 7 = 4
a3 – a2 = 15 – 11 = 4
a4 – a3 = 19 – 15 = 4
Heice the sequence is in AP with
a = 7, d = 4
Sum of first 15 terms
S15= \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S15 = \(\frac { 15 }{ 2 } \) [2 x 7 + (15 – 1) 4]
= \(\frac { 15 }{ 2 } \) [14 + 14 x 4]
= \(\frac { 15 }{ 2 } \) x [14 + 56]
= \(\frac { 15 }{ 2 } \) x 70 = 15 x 35 = 525

(ii) We have
an = 9 – 5n
Putting n = 1, 2, 3, 4 we get
a1 = 9 – 5 x 1 = 4
a2 = 9 – 5 x 2 = – 1
a3 = 9 – 5 x 3 = – 6
a4 = 9 – 5 x 4 = – 11
…………………………………
…………………………………
………………………………….
a2 – a1 = – 1 – 4 = 4
a3 – a2 = – 6 – ( – 1)
= – 6 + 1 = – 5
a4 – a3 = – 11 – (- 6)
= – 11 + 6 = – 5
Since the given sequence have same common differences hence it is in AP.
a = 4, d = – 5
Sum of first n term = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S15 = \(\frac { 15 }{ 2 } \) (2 x 4 + (15 – 1) x (- 5))
= \(\frac { 15 }{ 2 } \) [8 + 14 x (-5)]
= \(\frac { 15 }{ 2 } \) [8 – 70] = \(\frac { 15 }{ 2 } \) x – 62
= 15 x (-31) = – 465

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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 terms? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
We have
Sum of the first n term = 4n – n2
i.e. Sn = 4n – n2
Putting n = 1, 2, 3, ……….
S1 = 4 x 1 – 12 = 4 – 1 = 3
Therefore, a1 = S1 = 3
Hence the first term is 3
For n = 2
S2 = 4 x 2 x 22 = 8 – 4 = 4
S2 is the sum of first two terms
Second term = S2 – S1
a2 = 4 – 3 = 1
For, n = 3
S3 = 4 x 3 – 32
= 12 – 9 = 3
Therefore, third term = S3 – S2
a3 = 3 – 4 = – 1
d = a2 – a1
d = 1 – 3 = – 2
10th term = a10 = a + 9d
a10 = 3 + 9 x (-2)
= 3 – 2n + 2
= 5 – 2n

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution.
The first 40 integers divisible by 6 are 6, 12, 18, 24.
Here a2 – a1 = 126 = 6
a3 – a2 = 18 – 12 = 6
a4 – a3 = 24 – 18 = 6
Since difference of two consecutive terms is
same, so the list of integers divisible by fi is in AP.
Here a = 6; d = 6 and n = 40
Sum of first 40 positive integers = S40
S40 = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S40 = \(\frac { 40 }{ 2 } \) [2 x 6 + (40 – 1) x 6]
= 20[12 + 39 x 6]
= 20[12 + 234]
= 20 x 248
= 4920

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are
16, 24, 32, 40, …………
Here
a2 – a1 = 16 – 8 = 8
= 24 – 16 = 8
a3 – a2 = 32 – 24 = 8
Difference between the two consecutive terms
is same. Hence the List of multiples of 8 is in AP.
It is given
a = 8; d = 8 and n = 15
Sum of firs 15 multiples of 8 = S15
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S15 = \(\frac { 15 }{ 2 } \) [2 x 8 + (15 – 1) x 8]
S15 = \(\frac { 15 }{ 2 } \) [16 + 14 x 8]
S15 = \(\frac { 15 }{ 2 } \) [16 + 112]
= \(\frac { 15 }{ 2 } \) [128]
= 15 x 64
= 960

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 14.
Find the sum of odd numbers between O and 50.
Solution:
The odd numbers between 0 and 50 are 1, 3, 5, 7, ……………. 49
Here
a2 – a1 = 3 – 1 = 2
a3 – a2 = 5 – 3 = 2
a4 – a3 = 7 – 5 = 2
Difference between the two consecutive terms is same, Hence the list of odd numbers between 0 and 50 is in AP.
We have a = 1;d = 2 and l = 49
1 = a + (n – 1)d
49 = 1 + (n – 1) x 2
48 = (n – 1) x 2
n – 1 = \(\frac { 48 }{ 2 } \)
n – 1 = 24
n = 25
Sum of odd numbers between 0 to 50
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S25 = \(\frac { 25 }{ 2 } \) [2 x 1 + (25 – 1) x 2]
= \(\frac { 25 }{ 2 } \) x 50
= \(\frac { 25 }{ 2 } \) x 25
= 625

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 15.
A contract an construction job specifies a penalty for delay of completion beyond a rertain date as follows: ₹ 200 for the first day, ₹ 250 for the second, ₹ 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 delays the work by 30 days?
Solution:
Penalty increases per day by ₹ 50.
We have 200, 250, 300, …………..
Here the list of penalty form an AP
a = 200.
d = 250 – 200 = 50
n = 30
Total money paid by contractor,
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S30 = \(\frac { 30 }{ 2 } \) [2 x 200 + (30 – 1) x 50]
= ₹ 15 [400 + 29 x 50]
= ₹ 15 x [400 + 1450]
= ₹ 15 x 1850
= ₹ 27750

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 16.
A sum of ₹ 700 is to be used to gives even 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.
Solution:
Since value of each prize decreases by ₹ 20,
hence the value of seven succeeding cash prize form an AP.
It tint prize be ₹ x.
Then Second prize = x – 20
Third prize = x – 40
Fourth prize = x – 60
i.e. list of prize, x, x – 20, x – 40, z – 60, ……………
Here
a = x
d = x – 20 – x = – 20
n = 7
S7 = 700
We know that
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S7 = \(\frac { 7 }{ 2 } \) [2 x x + (7 – 1) x ( – 20)]
700 = \(\frac { 7 }{ 2 } \) [2x – 120]
700 = 7 [x – 60]
x – 60 = \(\frac { 100 }{ 7 } \)
x – 60 = 100
x = ₹ 160
Value of first prize = ₹ 160
Value of second prize
=x – 20 = 160 – 20
= ₹ 140
Value of third prize
= ₹ (140 – 20) = ₹ 120
Value of fourth prize
= ₹ (120 – 20) = ₹ 100
Value of fifth prize
= ₹ (100 – 20)= ₹ 80
Value of sixth prize
= ₹ (80 – 20) = ₹ 60
Value of seventh prize
= ₹ (60 – 20) = ₹ 40

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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:
In class I there are three sections.
So number of trees planted by class 1 = 3.
Number of trees planted by class II
= 2 + 2 + 2 = 6
Number of trees planted by class III
= 3 + 3 + 3 = 9
Number of trees planted by class IV
= 4 + 4 + 4 = 12
In the same way,
Number of trees planted by class XII
= 12 + 12 + 12 = 36
We have 3, 6, 9, 12, …., 36
We observed that number of trees planted in
each class having same difference between two
consecutive classes. Therefore it forms an AP.
Here a = 3,d = 6 – 3 = 3 and n = 12
Total number of trees planted
Sn = \(\frac { n }{ 2 } \) [2a + (n – 1) d]
S12 = \(\frac { 12 }{ 2 } \) = [2 x 3 + (12 – 1) x 3]
= 6[6 + 33] = 6 x 39 = 234

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 18.
A sapiral is made up of  successive semicircles, with centres alternatively at A and B. starting with centres at A of radii 0.5 cm. 1.0cm 1.5 cm. 2.0 cm, … as shown in figure, What is the total length of such a spiral made up thirteen consecutive semicircles?
[Hint: Length of successive semicircles is l1, l2, l3, l4 ……….. with centres A. B, A, B , ……….]
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 img-2
Solution:
Length of successive semicircles of radii, r1 – 0.5 cm, r2 = 1.0 cm, r3 = 1.5 cm, r4 = 2.0cm are l1, = 0.5π, l2 = 1 x π, l3 = 1.5π, l4 = 2π, …
Let a1 = 0.5π, a3 = 1.5π, ………
a2 – a1 = π – 0,5π = 0.5π cm
a3 – a2 = 1.5π – π = 0.5π cm
a4 – a3 = 2π – 1.5cm = 0.5cm
Hence difference between two consecutive
terms is saine So the length of semicircles is in AP.
Here a = 0.5π; d = 0.5π and π – 13
There fore total length
Sn= \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S13 = \(\frac { 13 }{ 2 } \) [2 x 0.5π + (13 – 1) x 0.5π] cm
= \(\frac { 13 }{ 2 } \) [π + 12 x 0.5π] cm
= \(\frac { 13 }{ 2 } \) [π + 6π] cm
= \(\frac { 13 }{ 2 } \) x 7π cm
= \(\frac { 13×7 }{ 2 } \) x \(\frac { 22 }{ 7 } \) cm
13 x 11 cm
= 143 cm

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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 Lo it and so on (see figure) in how many rows are the 200 logs placed and how many logs are in the top row?
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 img-3
Solution:
The number of logs in the bottom row, in the
next row, and in the next to it and so on are:
20, 19, 18, ………….
a2 – a1 = 19 – 20 = – 1
a3 – a2 = 18 – 19 = – 1
Hence two consecutive terms have same
difference So the above hat form an AP.
Here a = 20; d = – l and Sn = 200
We know that
Sn= \(\frac { n }{ 2 } \) [2a + (n – 1)d]
200 = \(\frac { n }{ 2 } \) [2 x 20 + (n – 1) x ( – 1)]
400 = n[40 – n + 1]
400 = n[41 – n]
400 = 41n – n2
n2 – 41n +400 = 0
n2 – 25n – 16n + 400 = 0
n (n – 25) – 16 (n – 25) = 0
(n – 25) (n – 16) = 0
n = 25, or n = 16
Hence number of rows either 25 or 16.
Number of logs in nth row = a + (n – 1) d
Number of Logs in 25th row
= 20 + 24 x (-1)
= 20 – 24 = – 4
Which is not possible, therefore number of
rows i.e. n = 16
Number of Logaintop row = a16
an = a + (n – 1)
a16 = a + (16 – 1)d
= 20 + 15 x (- 1)
= 20 – 15
= 5

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 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 unit. There are Len potatoes in the line (see Figure).
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 img-4
A competitor starts from the bucket, picks up the nearest potato, nina back with it, drop. 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 x 5 + 2 x (5 + 3)]
Solution:
To pick up the first potato, the distance run 2 x 5 m = 10 m
To pick up the second potato, the distance run = 2 x (5 + 3) = 16m
To pick up the third potato, the distance run = 2 (5 + 3 + 3) = 22 m and so on.
a2 – a1 = 16 – 10 = 6m
a3 – a2= 22 – 16 – 6m
Hence difference between two consecutive terms are same. Therefore the list of distance run is in AP.
Here a = 10 m; d = 6m and n = 10
Therefore total distance run byu competitor = S10
Sn= \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S10 = \(\frac { 10 }{ 2 } \) [2 x 10 + (10 – 1) 6]
S10 = 5 [20 + 9 x 6]
S10 = 5 x [20 + 54]
S10 = 5 x 74
S10 = 370 m

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

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