Gujarat Board Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

 Gujarat Board Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

Gujarat Board Textbook Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

Ex 5.1

Question 1.
In which of the following situations does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional kilometre.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it cost ₹150, for the first metre and rises by ₹5O for each subsequent metre.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
Solution:
(i) We can observed that
taxi fare for first km = ₹15
taxi fare for first 2 km = ₹15 + ₹8 = ₹23
taxi fare for first 3 km = 15 + 2 x 8 = 15 + 16 = ₹31
taxi fare for first 4 km = 15 + 3 x 8
= 15 + 24 = 39
Hence the terms are 15, 23,31, 39 ……………. these terms form an AP because every term is 8 more than the preceding terms.

(iii) It is given that
the cost of digging for first metre = 150
the cost of digging for first 2 metres = a + d
= 150 + 50 = 200
the cost of digging for first 3 metres
= a + 2d = 250
= 150 + 2 x 50
the cost of digging for first 4 metres = (a + 3d)
= 150 + 3 x 50 = 300
We observe that, the term obtained 150, 200, 250. 300 form an AP because every term is 50 more than the preceding term.

(iv) Let P is the amount be deposted at r% compound interest per annum for n years then after n years money becomes

Question 2.
Write first four terms of the AP when the first term a and the common difference d are given as follows.
(i) a = 10, d = 10
(ii) a = 4, d = – 3
(iii) a = – 2, d = 0
(iv) a = -2, d = 1/2
(v) a = – 1.25, d = – 0.25
Solution:
(i) It is given that a = 10, d = 10
First four terms of the AP are a1, a2, a3, a4,
a1 = a
a1 = 10
a2 = a + d
a2 = 10 + 10 = 20
a3 = a + 2d = 10 + 2 x 10
= 10 + 20 = 30
a4 = a + 3d = 10 + 3 x 10
= 10 + 30 = 40
Therefore first four terms of the AP are: 10, 20, 30, 40.

(ii) As a1 = 4 and d = – 3
a2 = a + d = 4 + (- 3) = 1
a3 = a + 2d = 4 + 2 x ( – 3)
= 4 – 6 = – 2
a4 = a + 3d = 4+ 3 x (-3)
= 4 – 9 = – 5
So the first four terms of the AP are: 4, 1, -2, -5.

(iii) As a1 = a = – 2 and d = 0
a2 = a + d = – 2 + 0 = -2
a3 = a + 2d = – 2 + 2 x 0 = – 2
a4 = a + 3d = – 2 + 3 x 0 = – 2
Therefore the first four terms of the AP are:
– 2, – 2, – 2, – 2.

(v) As a = – 1.25 and d = – 0.25
a1 = a = – 1.25
a2 = a + d = – 1.25 + (-0.25)
= – 1.25 – 0.25 = – 1.50
a3 = a + 2d = – 1.25 + 2 x (- 0.25)
= – 1.25 – 0.50
a3 = – 1.75
a4 = a + 3d = – 1.25 + 3 x (- 0.25)
= – 1.25 – 0.75 = – 2.00
Hence the first four terms of the AP are:
– 1.25, – 1.50, – 1.75, – 2.00.

Question 3.
For the following APs, write the first term and the common difference.
(i) 3, 1, -1, -3, …
(ii) -5, -1, 3, 7, …
(iii) 1/3, 5/3, 9/3, 13/3
(iv) 0.6, 1.7, 2.8, 3.9, …
Solution:
(i) As the given AP is 3, 1,-1,-3, …..
Here first terms a = 3
and Common difference d = a2 – a1
= 1 – 3 = – 2
d = – 2

(ii) As the given AP is – 5, – 1, 3, 7, ………….
Here first term a = – 5
and Common difference d = a2 – a1
= – 1 – (- 5)
d = – 1 + 5 = 4
Common difference is 4.

(iv) As the given AP is 0.6, 1.7, 2.8, 3.9, ……………
Here a = 0.6
and Common difference d = a2 – a1
d = 1.7 – 0.6 = 1.1

Question 4.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
 

(iii) The given sequence is -1.2, -3.2, -5.2, -7.2,
a3 – a1 = – 3.2 – (- 1.2)
= – 3.2 + 1.2 = – 2
a3 – a2 = – 5.2 – (- 3.2)
= – 5.2 + 3.2 = – 2
a4 – a3 = – 7.2 – (- 5.2)
= – 7.2 + 5.2 – 2
Common difference = – 2 is same so the
given sequence is in AP.
a5= a + 4d = -1.2 + 4( – 2)
= – 1.2 – 8 = – 9.2
a6 = 0 +5d = – 1.2 + 5 x (- 2)
= – 1.2 – 10 = – 11.2
and a7 = a + 6d = – 1.2 + 6 x (- 2)
= – 1.2 – 12 = – 13.2
Hence, three terms of AP are – 9.2, – 11.2, -13.2

(iv) Given sequence is – 10, – 6, – 2, 2, …………..
a2 – a1 = – 6 – ( – 10) = – 6 + 10 = 4
a3 – a2 = – 2 – ( – 6) = – 2 + 6 = 4
a4 – a3 = 2 – ( – 2) = 2 + 2 = 4
Since given sequence has same common
difference. Hence the given sequence is in AP.
Next three terms of AP are.
a5 = a + 4d = -10 + 4 x 4
= – 10 + 16 = 6
a6 = a + 5d = – 10 + 5 x 4
= – 10 + 20 = 10
a7 = a + 6d = – 10 + 6 x 4
= – 10 + 24 = 14

(vi) Given sequence is 0.2, 0.22, 0.222, 0.2222, ……….
a2 – a1 = 0.22 – 0.2 = 0.02
a3 – a2 = 0.222 – 0.22 = 0.002
Since a2 – a1 ≠ a3 – a2 therefore given sequence does not form an AP.

(vii) Given sequence is 0, – 4, – 8, – 12
a2 – a1 = – 4 – 0 = – 4
a3 – a2 = – 8 – ( – 4) = – 4
a4 – a3= – 12 – ( – 8) = – 4
Since difference between two consecutive
terms is same therefore the sequence is in AP.
Next three terms are

(ix) Given sequence is 1, 3, 9, 27, ………
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9 – 3 = 6
Since a2 – a1 ≠ a3 – a2 therefore given sequence does not form an AP.

(x) Given sequence is a, 2a, 3a, 4a, …..
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
Since common difference is same. Hence the given sequence is in AP.
a1 = a
d = a
Next three terms of AP are
a5 = a + 4d = a + 4a = 5a
a6 = a + 5d = a + 5a = 6a
a7 = a + 6d = a + 6a = 7a

(xi) Given sequence is a, a2, a3, a4, …………..
a2 – a1 = a2 – = a (a – 1)
a3 – a2 = a3 – a2 = a2 (a – 1)
Since a2 – a1 ≠ a3 – a2therefore given sequence is not in AP.

 

(xiv) Given sequence is 12, 52, 72, …………
a2 – a1 = 32 – 12 = 9 – 1 = 8
a3 – a2 = 52 – 32 = 9 – 1 = 16
Since a2 – a1 ≠ a3 – a2 therefore the given sequence does not form an AP.

(xv) Given sequence is 12, 52, 73, …………
a = 12 = 1
a2 – a1 = 52 – 12 = 25 – 1 = 24
a3 – a2 = 72 – 52 = 49 – 25 = 24
a4 – a3 = 73 – 72 = 73 – 49 = 24
Since common difference is same then sequence is in AP.
Next terms are
a5 = a + 4d = 1 + 4 x 24 = 97
a6 = a + 5d 1 + 5 x 24 = 121
a7 = a + 6d = 1 + 6 x 24 = 145.

Ex 5.2

Question 1.
Fill in the blanks in the following table. Given that a is the first term d the common difference and a the nth term of the AP.

Solution:
(i) a = 7, d = 3, n = 8, a = n ?
than an = a + (n – 1) d
an = 7 + (8 – 1) 3
an = 7 + 7 x 3
an = 7 + 21
an = 28

(ii) a = – 18, d = ?, n = 10, an = 0
an = a + (n – 1) d
0 = – 18 + (10 – 1)d
= 18 = 9d
d = 2

(iii) a = ?, d = – 3, n = 18, an = – 5
an = a + (n – 1) d
– 5 = a + (18 – 1)(- 3)
– 5 = a + (- 51)
51 – 5 = a
a = 46

(iv) a = – 18.9, d = 2.5, n = ?, an = 3.6
an = a + (n – 1) d
3.6 = – 18.9 + (n – 1) x 2.5
3.6 + 18.9 = 2.5 (n – 1)
22.5 = 2.5 (n – 1)
n – 1 = 22.5/2.5
n – 1 = 225/25
n – 1 = 9
n = 9 + 1
n = 10

(v) a = 3.5, d = 0, n = 105, an = ?
an = a + (n – 1) d
an = 3.5 + (105 – 1) 0
an = 3.5

Question 2.
Choose the correct choice in the following and justify.
(i) 30th term of the AP: 10, 7, 4,….. is
(A) 97 (B) 77 (C) – 77 (D) – 87
(ii) 11th term of the AP: – 3, –1/2, 2, ….. is
(A) 28 (B) 22 (C) – 38 (D) 48 – 1/2
Solution:
(i) The given AP is 10, 7, 4, …
a = 10, d = 7 – 10 = – 3
n = 30
an = a + (n — 1) d
a30 = 10 + (30 – 1) x (- 3)
= 10 + 29 x (- 3)
= 10 – 87
a30 = – 77
hence the correct choice is (C).

Question 3.
In the following APs, find the missing terms in the boxes.

(ii) Let a be the first term and d be the common difference of given AP.
Then, a + d = 13
a4 = a + 3d = 3

d = – 5
Putting value of d in equation (1)
a + d = 13
a + ( – 5) = 13
a = 13 + 5
a = 18
a3 = a + 2d
= 18 + 2x ( – 5)
= 18 – 10
Hence the missing terms in the boxes are
 and 

(iv) a = – 4 and we let d be the common difference.
Then a6 = a + (n – 1) d
a6 = – 4 + (6 – 1) x d
6 = – 4 + 5d
10 = 5d
d = 2
a2 = a + d = – 4 + 2 = – 2
a3 = a + 2d = – 4 + 2 x 2
= 4 + 4 = 0
a4 = a + 3d = – 4 + 3 x 2
= – 4 + 6 = 2
a5 = a + 4d = – 4 + 4 x 2
= – 4 + 8 = 4
Hence the missing terms in the boxes are

(v) Let common difference be d.
a2 = a + d = 38
= a + d = 38 ……………(1)
a6 = a + 5d = – 22 ……………. (2)

d = – 15
Putting value of d in equation (1)
a + d = 38
= a – 15 = 38
a = 38 + 15
a = 53
a3 = a + 2d
a3 = 53 + 2 ( – 15)
a3 = 53 – 30
a3 = 23
a4 = a + 3d
a4 = 53 + 3 x ( – 15)
a4 = 53 – 45 = 8
a5 = a + 4d
a5 = 53 + 4 x ( – 15)
a5 = 53 – 60
a5 = – 7
Hence the missing terms in the boxes are
 and 

Question 4.
Which term of the AP 3, 8, 13, 18, ……. is 78 ?
Solution:
The given AP is 3, 8, 13, 18, ……………
a = 3
d = 8 – 3 = 5
Let the term of given AP is 78.
Hence an = a + (n – 1) d
78 = 3 + (n – 1) x 5
75 = (n – 1) x 5
n – 1 = 75/5
n – 1 = 15
n = 16
Hence 16th term of the given AP is 78.

Question 5.
Find the number of terms in each of the following APs.
(i) 7 13, 19, …, 205
(ii) 18, 15 1/2,13,…,- 47
Solution:
(i) Given AP is 7, 13, 19, ………….., 205
a = 7, d = 13 – 7 = 6
Let n be the number of terms
then an = 205
an = a + (n – 1) d
205 = 7 + (n – 1) 6
= 205 – 7 = (n – 1) x 6
198 = (n – 1) x 6
198/6 = n – 1
= n – 1 = 33
n = 34
Hence the number of terms in the given AP is 34.

Question 6.
Check whether – 150 is a term of the AP: 11, 8, 5, 2, ………….
Solution:
The given list of numbers is 11, 8, 5, 2, ……..
a2 – a1 = 8 – 11 = – 3
Here a = 11, d = – 3
Let – 150 be the an term of the given AP.
then
an = a + (n – 1)d
– 150 = 11 + (n – 1)( – 3)
– 150 = 11 – 3n + 3
3n = 150 + 14
n = 164/3
But n must be positive integer therefore – 150 is not a term of given AP.

Question 7.
Find the 31st term of AP whose 11th term is 38 and 16th term is 73.
Solution:
Let a be the first term and cl is the common difference of the AP.
11th term, a11 = 38 [a11 = a + (11 – 1)d]
a + 10 d = 38 ……………….. (1)
And 16th term = 73

d = 35/5 = 7
d = 7
Putting value of d an equation (1)
a + 10 d = 38
a + 10 x 7 = 38
a + 70 = 38
a = 38 – 70
a = – 32
there fore an = a + (n – 1)d
a31 = – 32 + (31 – 1) x 7
a31 = – 32 + 210
a31 = 178
Hence 31 term of the AP is 178,

Question 8.
An AP cvneists of 50 terms ofwhich third tarms is 12 and last term is 106. Find the 29th terms.
Solution:
Let a be the first term and d be the common difference of AP.
Third terni a3 = 12
a + 2d = 12
last term = 106 (l = 50th)
l = a + (n – 1)d
106 = a + (50 – 1) d
106 = a + 49d
a + 49d = 106 ………. (2)
Solving equations (1) and (2), we get
47d = 94
d = 94/47
Putting value of din equation (1)
a + 2d = 12
a + 2 x 2 = 12
a = 12 – 4
a = 8
Therefore 29th term
a29 = a + 28d
= 8 + 28 x 2
= 8 + 56 = 64

Question 9.
If the 3rd and the Wit terms of an AP are 4 and -8 respectively, which term of this AP is zero?
solution:
Let a and d be the first term and common difference of an AP.
Third term a3 = 4
a + 2d = 4
(an = a + (n – 1)d)
Ninth term a9 = – 8
a + 8d = – 8 …………… (2)
Solving equations (1) and (2),

d = – 2
Putting value old in equation (2),
a + 8d = – 8
a + 8 x (- 2) = – 8
a – 16 = – 8
a = 8
Let the the nth term of the AP be zero
then a + (n – 1)d = 0
(an = a + (n – 1)d)
8 + (n – 1)( – 2) = 0
8 – 2n + 2 = 0
2n = 10
n = 5
Hence 5’ term of the AP is zero.

Question 10.
The 17th term of an AP exceeds 10th term by 7. Find the common difference.
Solution:
Let a be the first term and d be the common difference of the AP.
17th term = a + 16d
(a + (17 – 1) d)
and 10th term = a + 9d
According to problem
a17 = a10 + 7
a + 16d = a + 9d +7
7d = 7
d = 1
Hence common difference is 1.

Question 11.
Which term of the AP: 3, 15, 27, 39, ….. will be 132 more than its 84th term?
Solution:
Given AP is 3, 15, 27, 39, ……….
Here a = 3 and d = 15 – 3 = 12
4th term = a + 53d (a + (54 – 1)d)
According to problem
nth term = 54th term + 132
a + (n – 1)d = a + 53d + 132
3 + (n – 1) x 12 = 3 + 53 x 12 + 132
(n – 1) x 12 = 636 + 132
n – 1 = 768/12
= n – 1 = 64
n = 65
Hence 65th term will be 132 more than 54th term.

Question 12.
Two APs have the same common difference The difference between their 100th term is 100. What is the difference between 1000th terms?
Solution:
Let a and a be the Gral term of two APs respectively and ci be the saine cximrnon difference of the two APs.
Then 100th term of first AP = a1 + 99d
(a100 = a1 + (100 – 1) d)
100th term of second AP = a2 + 99d
According to problem
a1 + 99d (a2 + 99d) = 100
a1 – a2 = 100
Now we have
1000th term of lint AP = a1 + 999d
and 1000th term of second AP
= a + 999d
Hence the difference of 1000th terms of both AP’s
a1 + 999d – (a2 + 999d)
= a1 – a2
= 100
Hence the difference of 1000th term of both the APs is also 100.

Question 13.
How many three.digit numbers are divisible by 7 ?
Solution:
Three digit numbers which are divisible by 7
are: 105, 112, 119, 126, …………, 994
a2 – a1 = 112 – 105 = 7
a3 – a2 = 119 – 112 = 7
a4 – a3 = 126 – 119 = 7
Since difference, between two consecutive terms is 7. So the three-digit numbers which an divisible by 7 are in AP.
a = 105
last term = 994
nth term = l
l = a + (n – 1) d
994 = 105 + (n – 1) x 7
994 – 105 = (n – 1) x 7
(n – 1) x 7 = 889
n – 1 = 889/7
n – 1 = 127
n = 128
Hence, 128 three digit numbers are divisible by 7.

Question 14.
How many multiples of 4 Lie between 10 and 250?
Solution:
Multiples at 4 which lies between 10 and 250
are 12, 16, 20, 24, ……….., 248
a2 – a1 = 16 – 12 = 4
a3 – a2 = 20 – 16 = 4
Since difference between two consecutive terms is 4 i.e. same so these numbers form an AP.
Here a = 12, and l = 248
l = a + (n – 1) d
12 + (n – 1) x 4 = 248
(n – 1) x 4 = 248 – 12
n – 1 = 59
n = 60
Hence 60 multiples of 4 between 10 and 250.

Question 15.
For what value of is, are the e terms of two
APs 63, 65, 67, ….. and 3, 10, 17, ……….. equal?
Solution:
First AP is 63. 65,67, ………
Here a = 63, and d = 65 – 63 = 2
nth term a + (n – 1)d
(an = a + (n – 1) d)
= 63 (n – 1) x 2
Second AP is 3,10, 17, ……….
a = 3 and d = 10 – 3 = 7
nth term = a + (n – 1) d
= 3 + (n – 1) = 7
According to problem nth terms are same.
63 + (n – 1) x 2 = 3 + (n – 1) x 7
63 + 2n – 2 = 3 + 7n – 7
61 + 2o = 7n – 4
– 5n = – 65
n = 13
Hence 13th term of both APa are equal.

Question 16.
Determine the AP whose third term 16 and the 7th term exceeds the 5th, term by 12.
Solution:
Let a and d be the first and the common difference of the AP respectively.
Third term a3 = 16
n3 = a + 2d
a + 2d = 16
and 7th term = a + 6d
5th term = a + 21d
According the problem
a7 = a5 + 12
a + 6d = a + 4d + 12
2d = 12
d = 6
Putting value old in eqn.(1)
a + 2d = 16
a + 2 x 6 = 16
a = 4
Hence required AP is.
a, a + d, a + 2d, a + 3d, ……….
4, 4 + 6, 4 + 2 x 6, 4 + 3 x 6, ……….
4, 10, 16, 22, …………

Question 17.
Find the 20 term from the la.t term of the AP: 3, 8, 13, …………. 253.
Solution:
Given AP is
3, 8, 13, ……….., 253
Here a = 3 and d = 8 – 3 = 5
20th term from the last = l – (n – 1)d
= 253 – (120 – 1) x 5
= 253 – 19 x 5
= 253 – 95 = 158
Hence the 20th term from the last term of the given AP is 158.
Alternate method
Given AP is 3, 8, 13, …………. 253
Here a =3 and d = 8 – 3 = 5
l = a + (n – 1)d
253 = 3 + (n – 1) x 5
250 = (n – 1) x 5
n – 1 = 250/5
n – 1 = 50
n = 51
term from the last = (51 – 20 + 1) term from the beginning.
= 32 term from the beginning
an term = a + (n – 1)d
a32 = 3 + (32 – 1) x 5
= 3 + 31 x 5
= 3 + 155 = 158
Hence 20th term from the last term of the given AP is 158.

Question 18.
The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three term of the AP.
Solution:
Let a be the first term and d be the common difference
4th term = a + 3d [a4 = a + (4 – 1)d]
8th term = a + 7d [a8 = a + (8 – 1)d]
According to the problem.
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10 d = 24
a + 5d = 12
6th term = a + 5d [a6 = a + (6 – 1)d]
10th term = a + 9d [a10 = a + (10 – 1)d)
According to problem,
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44

d = 5
Putting value, old in equ. (1)
a + 5d = 12
a + 5 x 5 = 12
a = 12 – 25
a = – 13
Therefore, first three terms of AP art
a, a + d and a + 2d
i.e.,-13, – 13 + 5 and -13 + 2 x 5
or, – 13, – 8 and -3

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
Here a = ₹ 5000
d = ₹ 200
l = ₹ 7000
Suppose in n years his income reached ₹ 7000,
Then
l – a + (n – 1)d
7000 = 5000 + (n – 1) 200
(n – 1) x 200 = 7000 – 5000
(n – 1) x 200 = 2000
n – 1 = 2000/200
n – 1 = 10
n = 11
In 11th year his salary becomes ₹ 7000.

Question 20.
Ramkali saved ₹ 5 In the Brat week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week her weekly savings becomes ₹ 20,75, find n.
Solution:
Herr
a = ₹ 5
d = ₹ 1.75
an = 20.75
term of an AP.
an = a + (n – 1)d
20.75 = 5 + (n – 1)1.75
(n – 1)(1.75) = 15.75
n – 1 = 15.75/1.75
n – 1 = 1575/175
n – 1 = 9
n = 10
Hence required value of n is 10.

Ex 5.3

Question 1.
Find the sum of the following APs

  

Question 2.
Find the sums given below:

  

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 = 45/3
n – 1 = 15 + 1
n = 16
Sum of AP is
Sn = n/2 [2a + (n – 1) d]
Sn = −16/2 [2 x 5 + (16 – 1) x 3 ]
Sn = 8 [ 10 + 45 ]
Sn = 8 x 55
Sn = 440

(iii) Here a12 = 37
Now, d = 3
a12 = 37
a + 11d = 37
a + 11 x 3 = 37
a = 37 – 33
a = 4
s12 = 17/2 [ 2a + (n – 1)d]
s12 = 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 = n/2 [ 2a + (n – 1) d]
S10 = 10/2 [ 2 x a + (10 – 1) x d]
125 = 5 [ 2a + 9d]
25 = 2a + 9d
2a + 9d = 25
Solving eqn. (1) and (2)

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

   

[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

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 = n/2 [2a + (n – 1) d]
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 = −53/4 (Rejected) or n = 12
Therefore. 12 terms of the AP must be taken.

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 = n/2 [a + l]
400 = n/2 [5 + 45]
800 = n [50]
n = 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 = 40/15
d = 8/3
Hence the common difference is 8/3

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) = 333/9
n – l = 37
n = 38
So, there are 38 terms. We know that
Sn= n/2 (a.I)
S38 = 38/2 (17 + 350)
S38= 19 x 367
S38 = 6973
Hence sum is 6973.

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 = n/2 [2a + (n – 1)d]
S22 = 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.

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 = n/2[2a + (n – 1) d])
S51= 51/2 [2a + 50 d]
S51 = 51 [a + 25d]
= 51 [10 + 25 x 4 ]
= 51 [10 + 100]
= 51 x 110
= 5610

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 = n/2[2a + (n – 1) d])
= 7/2 [2a + (7 – 1)d] = 49
17/2 [2a + 6d] = 49
a + 3d = 7 …………. (1)
Sum of first 17 terms – 289
S17 = 289
17/2 [ 2a + (17 – 1)d] = 289
17 [a + 8d] = 289
a + 8d = 289/17
a + 8d = 17 ………….. (2)
Solving eqn (1) and (2), we get
a = 1 and d = 2
Sum of first n terms
Sn = n/2 [2a + (n – 1)d]
= n/2 [2 x 1 + (n – 1) x 2]
=n [1 + n – 1]
= n2

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= n/2 [2a + (n – 1)d]
S15 = 15/2 [2 x 7 + (15 – 1) 4]
= 15/2 [14 + 14 x 4]
= 15/2 x [14 + 56]
= 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 = n/2 [2a + (n – 1)d]
S15 = 15/2 (2 x 4 + (15 – 1) x (- 5))
= 15/2 [8 + 14 x (-5)]
= 15/2 [8 – 70] = 15/2 x – 62
= 15 x (-31) = – 465

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

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 = n/2 [2a + (n – 1)d]
S40 = 40/2 [2 x 6 + (40 – 1) x 6]
= 20[12 + 39 x 6]
= 20[12 + 234]
= 20 x 248
= 4920

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 = n/2 [2a + (n – 1)d]
S15 = 15/2 [2 x 8 + (15 – 1) x 8]
S15 = 15/2 [16 + 14 x 8]
S15 = 15/2 [16 + 112]
= 15/2 [128]
= 15 x 64
= 960

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 = 48/2
n – 1 = 24
n = 25
Sum of odd numbers between 0 to 50
Sn = n/2 [2a + (n – 1)d]
S25 = 25/2 [2 x 1 + (25 – 1) x 2]
= 25/2 x 50
= 25/2 x 25
= 625

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 = n/2 [2a + (n – 1)d]
S30 = 30/2 [2 x 200 + (30 – 1) x 50]
= ₹ 15 [400 + 29 x 50]
= ₹ 15 x [400 + 1450]
= ₹ 15 x 1850
= ₹ 27750

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 = n/2 [2a + (n – 1)d]
S7 = 7/2 [2 x x + (7 – 1) x ( – 20)]
700 = 7/2 [2x – 120]
700 = 7 [x – 60]
x – 60 = 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

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 = n/2 [2a + (n – 1) d]
S12 = 12/2 = [2 x 3 + (12 – 1) x 3]
= 6[6 + 33] = 6 x 39 = 234

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 , ……….]

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= n/2 [2a + (n – 1)d]
S13 = 13/2 [2 x 0.5π + (13 – 1) x 0.5π] cm
= 13/2 [π + 12 x 0.5π] cm
= 13/2 [π + 6π] cm
= 13/2 x 7π cm
= 13×7/2 x 22/7 cm
13 x 11 cm
= 143 cm

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?

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= n/2 [2a + (n – 1)d]
200 = 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

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).

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= n/2 [2a + (n – 1)d]
S10 = 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

Ex 5.4

Question 1.
Which term of AP 121, 117, 113, …………..is its first negative term?
[Hint: Find n for an < 0 ]
Solution:
The given AP is 121, 117, 113, ……………
Here a = 121,
d = 117 – 121 = – 4
Now we let nth term of the given AP be the negative term then
an < 0
a + (n – 1) d < 0
121 + (n – 1)( – 4) < 0
121 < (n – 1) x 4
n – 1 > 121/4
n > 125/4 + 1
= n > 31 1/4
Least Integral value of n is 32.
Hence 32nd term is first negative term of given AP.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first. sixteen terms of the AP, respectively.
Solution:
Let a and d be the first term and common difference of the given AP, respectively.
According to the problem,
Third term + seventh term = 6
a + 2d + a + 6d = 6
2a + 8d = 6
a + 4d = 3 …………… (1)
(dividing by 2)
and third term x seventh term = 8
(a + 2d)(a + 6d) = 8
From eqn. (1) we get
a = 3 – 44
Putting value of a in eqn. (2)
= (3 – 4d + 2d)(3 – 4d + 6d) = 8
(3 – 2d)(3 + 2d) = 8
9 – 4d2 = 8
 

Question 3.
A ladder has rungs 25 cm apart. (see Fig.). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 1/2 m apart, what is the length of the wood required for the rungs?

Solution:
Number of rungs = + 1
= 10 + 1
= 11
It is given that the length of rungs decreases
uniformly from 45 cm at bottom to 25 cm at the
top. So the length of rungs form an AP of first
term a = 45 cm and last term is 25 cm
Here a = 45 cm
d = 25cm
n = 11
Total length of wood required for rungs
= [a + 1] = – [45 + 25]
= 11/2 x 70 = 11 x 35
= 385 cm = 3.85 metre.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
Find this value of x.
[Hint: Sx-1 = S49 – Sx]
Solution:
The consecutive numbers on the houses of a row are 1, 2, 3 49. This list of numbers form an AP.
Here a = 1,
d = 2 – 1 = 1

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see figure). Calculate the total volume of concrete required to build the terrace.