**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 Class 10 Exercise 1.2 Solutions Question 1.**

Fill in the blanks In the following table,

given that ‘a’ is the first term, ‘d’ is the common difference and a_{n} the n^{th} term of the A.P.

Solution:

(i) a = 7, d = 3, n = 8, a_{n} =?

a_{n} = a + (n – 1) d

a_{8}= 7 + (8 – 1) 3

= 7 + 7 × 3

= 7 + 21

∴ a_{8} = 28

(ii) a = -18, d =?, n = 10, a_{n} = 0

a_{n} = 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, a_{n} = -5

a_{n} = a + (n – 1) d

-5 = a + (18 – 1) (-3)

= a + 17(-3)

-6 = 1 – 51

∴ a = -5 + 51 = 46

(iv) a = -18.9, d = 2.5, n =? a_{n} = 3.6

a_{n} = 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, a_{n} =?

a_{n} = a + (n – 1) d

= 3.5 + (105 – 1) (0)

= 3.5+ 104 × 0

= 3.5 +0

∴ a_{n} = 3.5

**Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 2.**

Choose the correct choice In the following and justify:

(i) 30^{th} 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, a_{30} =?

a_{n} = a + (n – 1) d

a_{30} = 10 + (30 – 1)(-3)

= 10 + 29(-3)

= 10 – 87

∴ a_{30} = 77

∴ Ans: (C) -77

(ii) 11^{th} 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, a_{11} =?

a_{n} = a + (n – 1) d

\(a_{11}=-3+(11-1)\left(2 \frac{1}{2}\right)\)

\(=-3+10\left(\frac{5}{2}\right)\)

= -3 +25

∴ a_{11} = 22

∴ Ans: (B) 22

**KSEEB Solutions For Class 10 Maths Arithmetic Progression Exercise 1.2 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.

**Exercise 1.2 Class 10 Maths Arithmetic Progression KSEEB Solutions Question 4.**

Which term of the AP : 3, 8, 13. 18, ………. is 78?

Solution:

a = 3, d = 8 – 3 = 5, a_{n} = 78, n =?

a_{n} = 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 KSEEB Solutions 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, a_{n} = 201. n =?

a + (n – 1) d = a_{n}

3 + (n – 1) 6 = 201

3 + 6n – 6 = 201

6n – 3 = 201

6n = 203

\(n=\frac{204}{6}\)

∴ n = 34

(ii) \(18,15 \frac{1}{2}, 13, \dots \dots 47\)

\(a=18, d=a_{2}-a_{1}=\frac{31-36}{2}=\frac{-5}{2}\)

a_{n} = -47, n =?

a_{n} = 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

**KSEEB Solutions For Class 10 Maths Arithmetic Progression 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. a_{n} = -150.

a + (n – 1) d = a_{n}

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.

**Arithmetic Progression Class 10 Exercise 1.2 KSEEB Solutions Question 7.**

Find the 31^{st} term of an AP whose 11^{th} term Is 38 and the 16^{th} term is 73.

Solution:

a = 38, a_{16} = 83 a_{31} =?

a_{n} = a + (n – 1) d

a_{16} = a + (16 – 1) d

a + 15d = 83

38 + 15d = 83

15d = 83 – 38

15d = 45

\(d=\frac{45}{15}\)

∴ d=3.

∴ a_{n} = a + (n – 1)d

a_{31} = 38 + (31 – 1) 3

= 38 + 30 × 3

= 38 + 90

∴ a = 128.

**Arithmetic Progression Class 10 KSEEB Solutions Question 8.**

An AP consists of 50 terms of which 3^{rd} term is 12 and the last term is 106. Find the 29^{th} term.

Solution:

n = 50, a_{3} = 12, a_{n} = 106, a_{29} =?

a_{11} = a + (n – 1) d

a_{50} = a + (50 – 1) d = 106

∴ a + 49d = 106 ……………… (1)

a_{3 }= 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.

∴ a_{n} = a + (n – 1) d

a_{29} = 8 + (29 – 1) 2

= 8 + 28 × 2

= 8 + 56

∴ a_{29} = 64

**10th Standard Maths Arithmetic Progression Exercise 1.2 KSEEB Solutions 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:

a_{3} = 4, a_{9} = -8, a_{n} = 0, n =?

a_{3} = a + 2d = 4 ………….. (1)

a_{9} = 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

a_{n} = a + (n – 1) d

= 8 + (n – 1) (-2)

= 8 – 2n + 2

= 10 – 2n = 0 ∵ a_{n} = 0

\(n=\frac{10}{2}\)

∴ n = 5

∴ 5^{th} term of this AP is Zero.

**Maths Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 10.**

The 1 7^{th} term of an AP exceeds its 17^{th} term by 7. Find the common difference.

Solution:

a_{17} = a_{10} + 7, d =?

a + 16d = a + 9d + 7

a +1 6d – a – 9d = 7

7d = 7

\(\mathrm{d}=\frac{7}{7}\)

∴ = 1

**10th Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 11.**

Which term of the AP: 3, 15, 27. 39, … will be 132 more than its 54^{th} term?

Solution:

3, 15, 27, 39, ………… a_{n} =?, n =?

a_{n} = a_{54} + 132

a = 3, d = 15 – 3 = 12

a_{n} = a_{54} + 132

a_{n} = a + 53d + 132

3 + 53(12) + 132

= 3 + 636 + 132

∴ a_{n} = 771

a_{n} = 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.

∴ 65^{th} term is 132 more than its 54th term.

**Arithmetic Progression Class 10 Notes KSEEB 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

100^{th} term of 1st set – 100^{th} 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.

**Arithmetic Progression 1.2 Exercise KSEEB Solutions 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. a_{n} = 994.

n =?

a + (n – 1)d = a_{n}

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.

**Arithmetic Progression Exercise 1.2 Solution KSEEB Solutions 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.

**Exercise 1.2 Arithmetic Progression KSEEB Solutions Question 15.**

For what value of n’. are the n^{th} terms of two APs: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?

Solution;

63, 65, 67,……….

a = 63. d = 65 – 63 =2. a_{n} =?

n^{th} term of this is

a_{n} = a + (n – 1)d

= 63 + (n – 1) 2

= 63 + 2n – 2

a_{n}= 2n + 61 …………….. (i)

3, 10, 17, ………….

a = 3, d = 10 – 3 = 7, a_{n} =?

a_{n} = a + (n – 1)d

= 3 + (n – 1) 7

= 3 + 7n – 7

a_{n} = 7n — 4 ………….(ii)

Here. n^{th} 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 Ex 1.2 KSEEB Solutions Question 16.**

DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.

Solution:

a = 16, a_{7} = a_{5} + 12, A.P =?

a_{7} = a_{5} + 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, ………….

**Arithmetic Progression 1.2 KSEEB Solutions 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, a_{n} = 253

20th term from the last term of the AP starting from 253 =?

253, 258, 263, ………… a_{20} =?

a = 253, d = 258 – 253 = 5, n = 20

a_{n} = a + (n – 1) d

a_{20}= 253 + (20 – 1) 5

= 253 + 19 × 5

= 253 + 95

∴ a_{20} = 348

∴ 20^{th} term from the last term of the AP is 348.

**Ex 1.2 Class 10 Arithmetic Progression KSEEB Solutions 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:

a_{4} + a_{8} = 24 …………. (1)

a_{6} + a_{10} = 44 ………… (2)

But A.P is a, a + d, a + 2d

from equation (1).

a_{4} + a_{8} = 24

a + 3d + a + 7d = 24

2a + 10d = 24 ………….. (3)

from equation (2).

a_{6} + a_{10} = 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, …………

**10th Standard Arithmetic Progression Exercise 1.2 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,

a_{n} = 7000, n=?

a + (n – 1)d = a_{n}

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.

**Ap Exercise 1.2 KSEEB Solutions 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 ‘n^{th} week is Rs. 20.75

∴ A.P. 5, 6, 75 , …………. , 20.75

a = 5, d = 6.75 – 5 = 1.75, a_{n} = 20.75.

n =?

a + (n – 1) d = a_{n}

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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