# KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2

Arithmetic Progression Exercise 1.2 Question 1.
Fill in the blanks In the following table,
given that ‘a’ is the first term, ‘d’ is the common difference and an the nth term of the A.P.

Solution:
(i) a = 7, d = 3, n = 8, an =?
an = a + (n – 1) d
a8= 7 + (8 – 1) 3
= 7 + 7 × 3
= 7 + 21
∴ a8 = 28

(ii) a = -18, d =?, n = 10, an = 0
an = a + (n – 1) d
0 = -18 + (10 – 1) d
0 = -18 + 9d
18 = 9d
9d = 18
$$\mathrm{d}=\frac{18}{9}=2$$

(iii) a =?, d = -3, n = 18, an = -5
an = a + (n – 1) d
-5 = a + (18 – 1) (-3)
= a + 17(-3)
-6 = 1 – 51
∴ a = -5 + 51 = 46

a

(iv) a = -18.9, d = 2.5, n =? an = 3.6
an = a + (n – 1) d
3.6= -18.9 + (n – 1) (2.5)
3.6= -18.9 + 2.5n – 2.5
3.6 = 2.5n – 21.4
2.5n = 3.6 + 21.4
2.5n = 25
$$n=\frac{25}{2.5}=\frac{250}{25}$$
∴ n = 10.

(v) a = 3.5, d = 0, n = 105, an =?
an = a + (n – 1) d
= 3.5 + (105 – 1) (0)
= 3.5+ 104 × 0
= 3.5 +0
∴ an = 3.5

Arithmetic Progression Class 10 Exercise 1.2 Solutions 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
Solution:
a = 10. d = 7 – 10 = -3, n = 30, a30 =?
an = a + (n – 1) d
a30 = 10 + (30 – 1)(-3)
= 10 + 29(-3)
= 10 – 87
∴ a30 = 77
∴ Ans: (C) -77

(ii) 11th term of the A.P.
$$-3,-\frac{1}{2}, 2, \dots \ldots,$$ is
A) 28
B) 22
C) -38
D) $$-48 \frac{1}{2}$$
Solution:
$$a=-3, \quad d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=2 \frac{1}{2}$$
n = 11, a11 =?
an = a + (n – 1) d
$$a_{11}=-3+(11-1)\left(2 \frac{1}{2}\right)$$
$$=-3+10\left(\frac{5}{2}\right)$$
= -3 +25
∴ a11 = 22
∴ Ans: (B) 22

Exercise 1.2 Class 10 Maths Arithmetic Progression Question 3.
In the following APs find the missing terms in the boxes:

Solution:
(i)
a = 2, a + d =?, a + 2d = 26
a + 26 = 26
2 + 2d = 26
2d = 24
∴ d=12 . .
∴ a + d = 2 + 12 = 14
∴ Ans: 14

(ii) Here, a =?, a + d = 13, a + 2d=?,
a + 3d = 3
a + d + 2d = 3
13 + 2d = 3
2d = 3 – 13
2d = -10
∴ d = -5
a + d = 13
a + (-5) = 13
a – 5 = 13
∴ a – 13 + 5 = 18
∴ a = 18
a + 2d =?
= 18 + 2(-5)
= 18 – 10
a + 2d = 8
∴ Ans: 18, 8

(iii) a = 5, a + d =?, a + 2d =?

(iv) a = -4, a +d=? a + 2d =?
a+3d = ? a + 4d =? a + 5d = 6
a + 5d = 6
-4 + 5d = 6
5d = 6 + 4
5d = 10
$$\mathrm{d}=\frac{10}{5}$$
∴ d=2.
a + d = -4 + 2 = -2
a + 2d = -4 + 2 (2) = -4 + 4 = 0
a + 3d = -4 + 3 (2) = -4 + 6 = 2
a + 4d = -4 + 4 (2) = -4 + 8 = 4
∴ Ans: -2, 0, 2, 4

(v) a =? a + d = 38, a + 2d =?
a + 3d =?, a + 4d =?a + 5d = 22

4d = -60
$$\mathrm{d}=-\frac{-60}{4} \quad=-15$$
a + d = 38
a – 15=38
a = 38 + 15 = 53
a + 2d = 53 + 2(-15) = 53 – 30 = 23
a + 3d = 53 + 3(-15) = 53 – 45 = 8
a + 4d =53 + 4(-15) = 53 – 60 = 7
∴ 53, 23, 8, -7.

Arithmetic Progression Class 10 Exercise 1.2 Question 4.
Which term of the AP : 3, 8, 13. 18, ………. is 78?
Solution:
a = 3, d = 8 – 3 = 5, an = 78, n =?
an = a + (n – 1) d
78 = 3+(n—1)(5)
78 = 3 + 5n – 5
78 = 5n – 2
5n = 78 + 2
$$n=\frac{80}{5}$$
∴ n = 16

10th Maths Arithmetic Progression Exercise 1.2 Question 5.
Find the number of terms In each of the following APs:
(i) 7, 13, 19, ………… 201
(ii) $$18,15 \frac{1}{2}, 13, \dots \dots 47$$
Solution:
(i) 7, 13, 19, ………. 201
a = 3, d = 13 – 7 = 6, an = 201. n =?
a + (n – 1) d = an
3 + (n – 1) 6 = 201
3 + 6n – 6 = 201
6n – 3 = 201
6n = 203
$$n=\frac{204}{6}$$
∴ n = 34

(ii) $$18,15 \frac{1}{2}, 13, \dots \dots 47$$
$$a=18, d=a_{2}-a_{1}=\frac{31-36}{2}=\frac{-5}{2}$$
an = -47, n =?
an = a + (n – 1) d
$$-47=18+(n-1)\left(\frac{-5}{2}\right)$$
$$-47-18=(n-1)\left(\frac{-5}{2}\right)$$
$$(n-1)\left(\frac{-5}{2}\right)=-65$$
$$n-1=-65 \times \frac{-2}{5}$$
n – 1 = -13 × -2
n – 1 = + 26
∴ n = 26 + 1
∴ n = 27

Arithmetic Progression Class 10 KSEEB Solutions Question 6.
Check whether -150 is a term of the A.P: 11, 8, 5, 2, ………..
Solution:
11, 8, 5, 2, ……….. -150
a = 11, d = 8 11 = -3. an = -150.
a + (n – 1) d = an
11 + (n – 1) (-3) = -150
11 – 3n + 3 = -150
-3n + 14 = -150
-3n = -150 – 14
-3n = -164
3n = 164
$$n=\frac{164}{3}$$
Here value of ‘n’ is not perfect. Hence -150 is not a term of the A.P.

Maths Arithmetic Progression Exercise 1.2 Question 7.
Find the 31st term of an AP whose 11th term Is 38 and the 16th term is 73.
Solution:
a = 38, a16 = 83 a31 =?
an = a + (n – 1) d
a16 = a + (16 – 1) d
a + 15d = 83
38 + 15d = 83
15d = 83 – 38
15d = 45
$$d=\frac{45}{15}$$
∴ d=3.
∴ an = a + (n – 1)d
a31 = 38 + (31 – 1) 3
= 38 + 30 × 3
= 38 + 90
∴ a = 128.

Arithmetic Progression Exercise 1.2 Solution Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
n = 50, a3 = 12, an = 106, a29 =?
a11 = a + (n – 1) d
a50 = a + (50 – 1) d = 106
∴ a + 49d = 106 ……………… (1)
a3 = a + 2d = 12 ………………. (2)
Subtracting equation (2) in equation (1).

Substituting the value of d.
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
∴ a = 12 – 4
a = 8.
∴ an = a + (n – 1) d
a29 = 8 + (29 – 1) 2
= 8 + 28 × 2
= 8 + 56
∴ a29 = 64

Arithmetic Progression Class 10 Notes KSEEB Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
a3 = 4, a9 = -8, an = 0, n =?
a3 = a + 2d = 4 ………….. (1)
a9 = a + 8d = -8 …………. (2)
From equation (1) – equation (2).

6d = 12
$${d}=\frac{-12}{6}$$
∴ d = -2
a + 2d = 4
a – 2(2) = 4
a – 4 =4
∴ a = 4 + 4
∴ a = 8
an = a + (n – 1) d
= 8 + (n – 1) (-2)
= 8 – 2n + 2
= 10 – 2n = 0 ∵ an = 0
$$n=\frac{10}{2}$$
∴ n = 5
∴ 5th term of this AP is Zero.

KSEEB Solutions For Class 10 Maths Arithmetic Progression Question 10.
The 1 7th term of an AP exceeds its 17th term by 7. Find the common difference.
Solution:
a17 = a10 + 7, d =?
a + 16d = a + 9d + 7
a +1 6d – a – 9d = 7
7d = 7
$$\mathrm{d}=\frac{7}{7}$$
∴ = 1

10th Standard Maths Arithmetic Progression Exercise 1.2 Question 11.
Which term of the AP: 3, 15, 27. 39, … will be 132 more than its 54th term?
Solution:
3, 15, 27, 39, ………… an =?, n =?
an = a54 + 132
a = 3, d = 15 – 3 = 12
an = a54 + 132
an = a + 53d + 132
3 + 53(12) + 132
= 3 + 636 + 132
∴ an = 771
an = a + (n – 1) d = 771
= 3 + (n – 1)12 = 771
3 + 12n — 12 = 771
12n – 9 = 771
12n = 771 + 9
12n = 780
$$n=\frac{780}{12}$$
∴ n = 65.
∴ 65th term is 132 more than its 54th term.

10th Arithmetic Progression Exercise 1.2 Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100. what is the difference between their 1 000th terms?
Solution:
Having common difference ‘d’, the APs
1st set a, a+d, a+2d
2nd set b, b + d, b + 2d
100th term of 1st set – 100th term of 2nd set = 100
∴ a + 99d – (b + 99d) = 100
a + 99d – b – 99d = 100
a – b= 100
Similarly.
1000th term of 1st set = 1 + 999d
1000 th term of 2nd set = b + 999d
Their difference
= a + 999d – (b + 999d)
= a + 999d – b – 999d
= a – b.

KSEEB Solutions For Class 10 Maths Arithmetic Progression Exercise 1.2 Question 13.
How many three-digit numbers are divisible by 7?
Solution:
First three-digit number divisible by 7
105 and the List number Is 994.
∴ AP is 105, 112. 119 994.
a = 105, d = 112 – 105 = 7. an = 994.
n =?
a + (n – 1)d = an
105 + (n – 1) 7 = 994
105 + 7n – 7 = 994
7n + 98 = 994
7n = 994 – 98
7n = 896
$$n=\frac{896}{7}$$
∴ n = 128.
∴ Numbers with 3 digits divisible by 7 are 128.

Exercise 1.2 Arithmetic Progression Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4. after 10 are 12, 16, 20….
Multiples of 4 upto 250 is 248
∴ A.P. is 12, 16, 20, …….. 248
a = 12, d = 16 – 12 = 4
n =?
a = a + (n – 1) d = 248
12 + 4n – 4 = 248
4n + 8 = 248
4n = 248 – 8
4n = 240
$$n=\frac{240}{4}$$
∴ n = 60
∴ Multiples of 4 lie between 10 and 250 is 60.

Arithmetic Progression 1.2 Exercise Question 15.
For what value of n’. are the nth terms of two APs: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?
Solution;
63, 65, 67,……….
a = 63. d = 65 – 63 =2. an =?
nth term of this is
an = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
an= 2n + 61 …………….. (i)
3, 10, 17, ………….
a = 3, d = 10 – 3 = 7, an =?
an = a + (n – 1)d
= 3 + (n – 1) 7
= 3 + 7n – 7
an = 7n — 4 ………….(ii)
Here. nth terms of second AP are equal.
∴ equation (i) = equation (ii)
2n + 61 = 7n – 4
2n – 7n = -4 – 61
5n = 65
5n =65
$$n=\frac{65}{5}$$
∴ n = 13
∴13th terms of the two given APs are equal.

Arithmetic Progression Class 10 Notes Exercise 1.2 Question 16.
DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution:
a = 16, a7 = a5 + 12, A.P =?
a7 = a5 + 12
a + 6d = a + 4d + 12
a + 6d – a – 4d = 12
2d = 12 ∴ d = 6.
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
∴ a = 16 – 12 ∴ a = 4
a = 4, d = 6.
∴ A.P.:
a, a + d, a + 2d, …………………
4, 4 + 6, 4 + 12, …………….
4, 10, 16, ………….

Karnataka State Syllabus 10th Class Maths Textbook Pdf Question 17
Find the 20th term from the Last term of theAP: 3, 8, 13, …………., 253.
Solution:
3, 8, 13, ………….., 253
a = 3. d = 8 – 3 = 5, an = 253
20th term from the last term of the AP starting from 253 =?
253, 258, 263, ………… a20 =?
a = 253, d = 258 – 253 = 5, n = 20
an = a + (n – 1) d
a20= 253 + (20 – 1) 5
= 253 + 19 × 5
= 253 + 95
∴ a20 = 348
∴ 20th term from the last term of the AP is 348.

10th Standard Karnataka State Syllabus Maths Guide Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. FInd the first three terms of the AP.
Solution:
a4 + a8 = 24 …………. (1)
a6 + a10 = 44 ………… (2)
But A.P is a, a + d, a + 2d
from equation (1).
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24 ………….. (3)
from equation (2).
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44 ………… (4)
Subtracting eqn. (4) from equation (3)

4d = 20
$$d=20 / 4$$
∴ d = 5
Substituting the value of d in equation (3)
2a + 10d = 24
2a + 10(5) = 24
2a + 50 = 24
2a = 24 – 50
2a = -26
a = 26/2 ,
∴ a = -13 .
∴ AP: a, a + d, a + 2d, ………..
-13, -13 + 5, -13 + 2(5), ………
-14, -8, -3, …………

KSEEB Solutions For Class 10 Question 19.
Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an Increment of Rs. 200 each year. in which year did his income reach Rs. 7000?
Solution:
Payment of Subba Rao in the year 1995 = Rs. 5000
increment = Rs. 200
∴ The payment he received in the year 1996 is Rs. 5.200

a = 5000, d = 5200 – 5000 = 200,
an = 7000, n=?
a + (n – 1)d = an
5000 + (n – 1) 200 = 7000
5000 + 200n – 200 = 7000
200n + 4800 = 7000
200n = 7000 – 4800
200n = 2200
$$n=\frac{2200}{200}$$
∴ n = 11
Value of ‘n’ is 11
∴ From 1995 to 10 years means 2005. his salary becomes Rs. 7,000.

Arithmetic Progression Ex 1.2 Question 20.
Ramkall saved Rs. 5 In the first week of a year and then Increased her weekly savings by Rs. 1.75. If in the nüI week. her weekly savings become Rs. 20.75. find ‘n’.
Solution:
Savings in the First week = Rs. 5.
Savings in the Second week 5 + 1.75 = Rs. 6.75
Savings in ‘nth week is Rs. 20.75
∴ A.P. 5, 6, 75 , …………. , 20.75
a = 5, d = 6.75 – 5 = 1.75, an = 20.75.
n =?
a + (n – 1) d = an
5 + (n – 1) (1.75) = 20.75
5 + 1.75n – 1.75 = 20.75
1 .75n + 3.25 = 20.75
1.75n = 20.75 – 3.25
1.75n = 17.5
$$n=\frac{17.5}{1.75}$$
$$n=\frac{1750}{175}$$
∴ n = 10
∴ In the 10th week, her savings becomes Rs. 20.75.

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