# 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

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) Given
a = 7, an = a + (n – 1 )d
d = 3, an = 7 + (8 – 1)3
n = 8, an = 7 + 7 x 3
an = ?, an = 7 + 21
an = 28
∴ nth term of an A.P = 28.

(ii) Given
a = – 18 an = a + (n – 1)d d = ? 0 = – 18 + (10 – 1)d
n = 10
an = 0
18 = 9d
d = $$\frac { 18 }{ 9 }$$
d = 2
Common difference = 2

(iii) Given
a = ?, a$$n$$ = a + (n – 1)d
d = – 3, – 15 = a + (18 – 1) – 3
n = 18, – 15 = a + 17 x – 3
a$$n$$ = – 5, – 15 = a – 51 = a
36 = a
a = 36
First term = 36.

(iv) Given
a = – 18.9, an a = a +(n – 1)d
d = 2.5, 3.6 = – 18.9 + (n – 1)2.5
n = ? 3.6 + 18.9 = (n – 1)2.5
n – 1 = $$\frac { 225 }{ 25 }$$
n – 1 = 9
n = 9 + 1
n = 10
an = 3.6
3.6 + 18.9 = (n – 1)2.5

(v) Given
a = 3.5, an = a + (n – 1)d
d = 0, an = 3.5 + (105 – 1)0
n = 105, an = 3.5 + 104 x 0
an = ?, 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, $$-\frac{1}{2}$$, 2, …. ,is
(A) 28
(B) 22
(C) -38
(D) -48$$\frac{1}{2}$$
Solution:
(i) (C): Here, a = 10, n = 30
∵ T10 = a + (n – 1)d and d = 7 – 10 = -3
∴ T30 = 10 + (30 – 1) × (-3)
⇒ T30 = 10 + 29 × (-3)
⇒ T30 = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

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) a2 = 13, a4 = 3, a? and a3 = ?
a2 = a + d
13 = a + d
a = 13 – d ➝ (1)
a4 = a + 3d
3 = a + 3d ➝ (2)
Putting eqn (1) in eqn (2)
3 = 13 – d + 3d
3 – 13 = 2d
2d = – 10
d = $$\frac{10}{2}$$
d= – 5
Put d = – 5 in eqn (1)
a = 13 – (- 5) = 13 + 5 = 18
a = 18
Third term a3 = a + 2d
a3 = 18 + 2 x (- 5)
= 18 – 10 = 8
a3 = 8
Hence missing terms are 18 and 8.

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

(iv) a = – 4, a6 = 6
a6 = a + 5d
6 = – 4 + 5d
6 + 4 = 5d
d = $$\frac{10}{5}$$ = 2
d = 2
a2 = a + d = – 4 + 2 = – 2
a3 = a + 2d = – 4 + 2(2) = – 4 + 40
a4 = a + 3d = – 4 + 3(2) = – 4 + 6 = 2
a5 = a + 4d = – 4 + 4(2) = – 4 + 8 = 4
Hence missing terms are – 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.

Question 4.
Which term of the AP: 3, 8, 13, 18, is 78?
Solution:
Let the nth term be 78
Here, a = 3 ⇒ T1 = 3 and T2 = 8
∴ d = T2 – T1 = 8 – 3 = 5
And, Tn = a + (n- 1 )d
⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5
⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16th term of the given AP.

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) Given a = 7
d = 13 – 7 = 6 = 205
an = a + (n – 1)d
205 = 7 + (n – 1)6
205 – 7 = 6n – 6
198 + 6 = 6n
6n = 198 + 6
6n = 204
n = $$\frac{204}{6}$$ = 34
n = 34
Hence, number of terms of a given AP is 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

Question 6.
Check whether -150 is a term of the AP:
11, 8, 5, 2…
Solution:
For the given AP,
we have a = 11, d = 8 -11 = -3
Let -150 be the nth term of the given AP
∴ Tn = a + (n – 1 )d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 – 11 = (n – 1) × (-3)
⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.
Thus, -150 is not a term of the given AP

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.

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:
Here, n = 50, T3 = 12, Tn = 106
⇒ T50 = 106
Let the first term = a and the common difference = d
Tn = a + (n – 1 )d
T3 = a + 2d = 12 …(1)
T50 = a + 49d = 106 …(2)
Subtracting (1) from (2), we get

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.

Question 10.
The 1 7th term of an AP exceeds its 17th term by 7. Find the common difference.
Solution:
a17 = a10 + 7
a + 16d = a + 9d + 7
a + 16d – a – 9d = 7
7d = 7
d = $$\frac{7}{7}$$ = 1
Common difference = d = 1

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.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let for the 1st AP, the first term = a and common difference = d

And for the 2nd AP, the first term = a and common difference = d

According to the condition,

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.

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.

Question 15.
For what value of n, are the nth terms of two APs: 63,65,67 … and 3,10,17, …. equal?
Solution:

Thus, the 13th terms of the two given AP’s are equal.

Question 16.
DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution:
a3 = a + 2d
16 = a + 2d ➝ (1)
7th term = 5th term + 12
a + 6d = a + 4d + 12
a – a + 6d – 4d = 12
2d = 12
d = $$\frac{12}{2}$$ = 6
put d = 6 in eqn (1)
16 = a + 2 (6)
16 – 12 = a
a = 4
Hence required AP a, a + d, a + 2d
4, 4 + 6, 4 + 2(6),…
4, 10, 16, 22, …..

Question 17
Find the 20th term from the Last term of the AP: 3, 8, 13, …………., 253.
Solution:
3, 8, 13, …., 253.
Given a = 3
d = a2 – a = 8 – 3
d = 5
an = 253
an= a + (n – 1)d
253 = 3 + (n – 1)5
253 – 3 (n – 1)5
n – 1 = $$\frac{250}{5}$$ = 50
n = 50 + 1 = 51
20th term from last term is 51 – 19 = 32
a32 = a + 31 d
= 3 + 31 x 5 = 3 + 155
a32 = 158
20th term from the last term is 158

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:
Let the first term = a and the common difference = d

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.

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:
Let the saving in the first week = ₹ 5
The savings in the next week = ₹ 5 + ₹ 175 = 6.75
saving in the next week = ₹ 6.75 + ₹ 1.75 = Rs.8.5
∴ AP, 5, 6.75, 8.5, …………. 20.75
a = 5, d = 1.75 an = 20.75 & n = ?
an = a + (n – 1)d
20.75 = 5 + (n – 1)1.75
20.75 – 5 = (n – 1)1.75
15.75 = (n – 1) 1.75
n – 1 = $$\frac{15.75}{1.75}$$
n – 1 = 9 = 9 + 1 = 10
n = 9 + 1
n = 10

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