GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

   

Gujarat Board GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 5 Arithmetic Progressions 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.
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-1
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 = \(\frac { 22.5 }{ 2.5 } \)
n – 1 = \(\frac { 225 }{ 25 } \)
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

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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, –\(\frac { 1 }{ 2 } \), 2, ….. is
(A) 28 (B) 22 (C) – 38 (D) 48 – \(\frac { 1 }{ 2 } \)
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).

(ii) In the given AP,
a = – 3
and
d = – \(\frac { 1 }{ 2 } \) – ( – 3) = \(\frac { -1 }{ 2 } \) + 3 = \(\frac { 5 }{ 2 } \)
a11 = a + (n – 1)d
= – 3 + (11 – 1) x \(\frac { 5 }{ 2 } \)
= – 3 + 10 x \(\frac { 5 }{ 2 } \)
= – 3 + 25
a11 = 22
Hence the correct choice is (B)

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Question 3.
In the following APs, find the missing terms in the boxes.
(j) 2, GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 26
(ii) GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 13, GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 3
(iii) 5, GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 9\(\frac { 1 }{ 2 } \)
(iv) – 4, GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 6
(v) GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 38, GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-2 – 22
Solution:
(i) Let the common difference of given AP is d
a = 2, a3 = 26
a3 = a + 2d = 26
a + 2d = 26
2 + 2d = 26
2d = 26 – 2
2d = 24
d = 12
Hence second term a2 = a + d
a2 = 2 + 12 = 14
Hence the’missing term in the box is GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-4.

(ii)
Let a be the first term and d be the common difference of given AP.
Then, a + d = 13
a4 = a + 3d = 3
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-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
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-5 and GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-6

(iii) Let common difference be d and a = 5
a4 = a + 3d = 9\(\frac { 1 }{ 2 } \)
a + 3d = \(\frac { 19 }{ 2 } \)
5 + 3d = \(\frac { 19 }{ 2 } \)
= 3d = \(\frac { 19 }{ 2 } \) – 5
3d = \(\frac { 19 }{ 2 } \)
d = \(\frac { 9 }{ 2×3 } \) = \(\frac { 3 }{ 2 } \)
Now, a2 = a + d = 5 + \(\frac { 3 }{ 2 } \)
= \(\frac { 10+3 }{ 2 } \) = \(\frac { 13 }{ 2 } \) = 6\(\frac { 1 }{ 2 } \)
a3 = a + 2d = 5 + 2 x \(\frac { 3 }{ 2 } \)
a3 = 8
Hence the missing terms are GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-7 and GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-6 in the boxes.

(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
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-8

(v) Let common difference be d.
a2 = a + d = 38
= a + d = 38 ……………(1)
a6 = a + 5d = – 22 ……………. (2)
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-12
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
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-9 and GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-10

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 75 }{ 5 } \)
n – 1 = 15
n = 16
Hence 16th term of the given AP is 78.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Question 5.
Find the number of terms in each of the following APs.
(i) 7 13, 19, …, 205
(ii) 18, 15\(\frac { 1 }{ 2 } \),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
\(\frac { 198 }{ 6 } \) = n – 1
= n – 1 = 33
n = 34
Hence the number of terms in the given AP is 34.

(ii) Given AP is 18, 15\(\frac { 1 }{ 2 } \), 13, …,- 47
a = 18, d = \(\frac { 31 }{ 2 } \) – 18 = \(\frac { -5 }{ 2 } \)
Let an = – 47
then an = a + (n – 1) d
– 47 = 18 + (n – 1) \(\frac { -5 }{ 2 } \)
– 47 = 18 – \(\frac { -5 }{ 2 } \) (n – 1)
7 – 18 = \(\frac { -5 }{ 2 } \)(n – 1)
– 65 = \(\frac { -5 }{ 2 } \)(n – 1)
\(\frac { 65×2 }{ 5 } \) = n – 1
26 = n – 1
= n – 1 = 26
= n = 26 + 1
n = 27
Hence the number of terms in the given AP is 27.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 164 }{ 3 } \)
But n must be positive integer therefore – 150 is not a term of given AP.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-11
d = \(\frac { 35 }{ 5 } \) = 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,

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 94 }{ 47 } \)
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

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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),
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-13
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.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 768 }{ 12 } \)
= n – 1 = 64
n = 65
Hence 65th term will be 132 more than 54th term.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 889 }{ 7 } \)
n – 1 = 127
n = 128
Hence, 128 three digit numbers are divisible by 7.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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, …………

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 250 }{ 5 } \)
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.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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
GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 img-14
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

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 2000 }{ 200 } \)
n – 1 = 10
n = 11
In 11th year his salary becomes ₹ 7000.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 = \(\frac { 15.75 }{ 1.75 } \)
n – 1 = \(\frac { 1575 }{ 175 } \)
n – 1 = 9
n = 10
Hence required value of n is 10.

GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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