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Expert Teachers at KSEEBSolutions.com has created KSEEB Karnataka SSLC Class 10 Science Important Questions and Answers Pdf Free Download in English Medium and Kannada Medium of Karnataka 10th Standard Science Questions and Answers, Notes Pdf are part of KSEEB SSLC Class 10 Solutions.

Here we have given KTBS Karnataka State Board Syllabus Class 10 Science Important Questions and Answers based on NCERT Syllabus. Students can also read Karnataka SSLC KSEEB Solutions for Class 10 Science.

Karnataka SSLC Class 10 Science Important Questions and Answers

10th Standard Science Important Questions Karnataka State Syllabus

• Chapter 1 Chemical Reactions and Equations
• Chapter 2 Acids, Bases and Salts
• Chapter 3 Metals and Non-metals
• Chapter 4 Carbon and Its Compounds
• Chapter 5 Periodic Classification of Elements
• Chapter 6 Life Processes
• Chapter 7 Control and Coordination
• Chapter 8 How do Organisms Reproduce?
• Chapter 9 Heredity and Evolution
• Chapter 10 Light Reflection and Refraction
• Chapter 11 Human Eye and Colourful World
• Chapter 12 Electricity
• Chapter 13 Magnetic Effects of Electric Current
• Chapter 14 Sources of Energy
• Chapter 15 Our Environment
• Chapter 16 Sustainable Management of Natural Resources

We hope the given KSEEB Karnataka SSLC Class 10 Science Important Questions and Answers Pdf Free Download in English Medium and Kannada Medium of Karnataka 10th Std Science Questions and Answers will help you.

If you have any queries regarding KTBS Karnataka State Board Syllabus Class 10 Science Important Questions and Answers based on NCERT Syllabus, drop a comment below and we will get back to you at the earliest.

Expert Teachers at KSEEBSolutions.com has created KSEEB Solutions for Class 4 all subjects Pdf Free Download in English Medium and Kannada Medium of 4th Standard Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material, are part of KSEEB Solutions. Here we have given KTBS Karnataka State Board Syllabus for Class 4 Textbook Solutions.

Karnataka State Board Syllabus for Class 4 Textbooks Solutions

• KSEEB Solutions for Class 4 Maths
• KSEEB Solutions for Class 4 EVS
• KSEEB Solutions for Class 4 English
• Savi Kannada Text Book Class 4 Solutions

We hope the given KSEEB Solutions for Class 4 all subjects Pdf Free Download in English Medium and Kannada Medium of 4th Std Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material will help you. If you have any queries regarding KTBS Karnataka State Board Syllabus for Class 4 Textbooks Solutions, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3

Arithmetic Progression Class 10 Exercise 1.3 Solutions KSEEB Question 1.
FInd the sum of the following APs:
i) 2, 7, 12, ……… to 10 terms
ii) -37, -33, -29,………. to 12 terms.
iii) 0.8, 1.7, 2.8 to 100 terms.
iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11\) terms
Solution:
i) Sum of 2, 7, 12, ………. 10?
a = 2, d = 7 – 2 = 5, n = 10
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(s_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]\)
= 5[4 + 9 × 5]
= 5[4 + 45]
= 5 × 49
∴ S₁₀ = 245

ii) -37, -33, -29, to 12……. terms.
Solution:
a = -37, d = -33 – (-37) = -33 + 37
n = 12, d = 4
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{12}=\frac{12}{2}[2 \times -37+(12-1) 4]\)
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6 × -30
∴ S₁₂ = -180.

iii) 0.6, 1.7, 2.8,……… , to 100 terms.
a = 0.6, d = 1.7 – 0.6 = 1.1
n = 100, S₁₀₀ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{100}=\frac{100}{2}[2 \times 0.6+(100-1)(1.1)]\)
= 50[1.2 + 99(1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1
∴ S₁₀₀ = 5505

iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11\) terms
Solution:

Arithmetic Progression Exercise 1.3 KSEEB Soltion Question 2.
Find the sums given below:
i) \(7+10 \frac{1}{2}+14+\ldots \ldots+84\)
ii) 34 + 32 + 30 + ……….+ 10
iii) -5 + (-8) + (-11) + ……….. +(-230)
Solution:

ii) 34 + 32 + 30 + ………. + 10
Solution:
a = 34, d = 32 – 34 =-2, a_n = 10,
S_n =?
a + (n – 1)d = a_n
34 + (n – 1)(-2) = 10
34 – 2n + 2 = 10
-2n + 36 = 10
-2n = 10 – 36
-2n = -26
2n = 26
\(n=\frac{26}{2}\)
∴ n = 13
\(S_{n}=\frac{n}{2}[a+l]\)
\(\mathrm{S}_{23}=\frac{13}{2}[34+10]\)
\(=\frac{13}{2} \times 44\)
= 13 × 22
∴ S₂₃ = 286.

iii) -5 + (-8) + (-11) +……… + (-230)
Solution:
a = -5, d = -8 – (-5) = -8 + 5
a, = -230.
S_n =? d = -3
a+(n – 1)d = a_n
-5 + (n – 1)(-3) = -230
-5 – 3n + 3 = -230
-3n – 2 = -230
-3n = -230 +2
-3n = -228
3n = 228
∴ n = 228/3
∴ n = 76.
\(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]\)
\(S_{ 76 }=\frac { 76 }{ 2 } \left[ -5-230 \right]  \)
\(=\frac{76}{2} \times -235\)
= 38 × -235
∴ S₇₆= -8930.

KSEEB Solutions For Class 10 Maths Arithmetic Progression Exercise 1.3 Question 3.
In an AP:
i) given a = 5. d = 3. a_n = 50, find ‘n’ and S_n.
ii) given a = 7, a₁₃ = 35, find d’ and S₁₃.
iii) given a₁₂= 37. d = 3. find ‘a’ and S₁₂.
iv) given a₃ = 15, S₁₀ = 125. find ‘d’ and a₁₀.
v) given d = 5. S₉ = 72, fInd ‘a’ and a₉.
vi) given a = 2. d = 8, S_n = 90, fInd ‘n and a₁₁.
vii) given a = 8, a_n = 62. S_n = 210, find ‘n’ and ‘d’.
viii) given a_n = 4, d = 2, S_n = -14. find ‘n’ and a.
ix) given a = 3, n = 8, S = 192, find ‘d’.
x) given L = 28, S = 144 and there are total 9 terms. Find ‘a’.
Solution:
i) a = 5, d = 3, a_n = 50. n=?, S_n =?
a + (n – 1)d = a_n
5 + (n – 1) 3 = 50
5 + 3n – 3 = 50
3n + 2 = 50
3n = +50 – 2
3n = 48
\(n=\frac{48}{3}\)
∴ n = 16
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(S_{16}=\frac{16}{2}[5+50]\)
\(=\frac{16}{2}[55]\)
= 8 × 55
∴ S₁₆= 440.

ii) a = 7, a₁₃ = 35. d =?. S₁₃ =?
Solution:
a+(n – 1) d = a_n
7 + (13 – 1) d = 35
7 + 12d = 35
12d = 35 – 7
12d = 28
\(d=\frac{28}{12}\)
\(d=\frac{7}{3}\)
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(S_{n}=\frac{7}{3} \times \frac{1}{2}[7+35]\)
\(\mathrm{s}_{\mathrm{n}}=\frac{7}{6} \mathrm{I} 42 \mathrm{l}\)
= 7 × 7
∴ S_n = 49

iii) a₁₂= 37, d = 3, a =? S₁₂ =?
a + (n – 1)d = a_n
a + (12 -1) 3 = 37
a + 11 × 3 = 37
a + 33 = 37
∴ a = 37 – 33
∴ a = 4
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(\mathrm{S}_{12}=\frac{12}{2}[4+37]\)
= 6[4 + 37]
∴ S₂ = 246

iv) a₃ = 15, S₁₀ = 125. d =?, a₁₀=?
Solution:
a₃ = a + 2d = 15
∴ a= 15 – 2d
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(s_{10}=\frac{10}{2}[2(15-2 d) \div(10-1) d]=125\)
= 5[30 – 4d + 9d] = 125
= 5[30 + 5d] = 125
150 + 25d = 125
25d = 125 + 150
25d = -25
\(\mathrm{d}=\frac{-25}{25}\)
∴ d = -1.
Substittuting the value of ‘d’
a = 15 – 2d
= 15 – 2(-1)
= 15 + 2
∴ a = 17.
∴ a₁₀ = a + 9d
= 17 + 9(-1)
= 17 – 9
∴ a₁₀ = 8

v) d = 5, S₉ = 72. a =? a₉ =?
Solution:
\(\mathrm{s}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\)
\(s_{9}=\frac{9}{2}[2 \times a+(9-1) 5]=72\)
\(\frac{9}{2}[2 a+8 \times 5]=72\)
\(\frac{9}{2}[2 a+40]=72\)
\(2 a+40=72 \times \frac{2}{9}\)
2a + 40 = 16
2a = 16 – 40
2a = -24
\(a=\frac{-24}{2}\)
∴ a = -12
a₉ = a + (n – 1) d
= -12 + (9 – 1) (5)
= -12 + 8 × 5
= -12 + 40
∴ a₉ = 28

vi) a = 2, d = 8 S_n = 90, n =?, a₃ =?
Solution:
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(=\frac{n}{2}[2 \times 2+(n-1) 8]=90\)
\(\frac{n}{2}[4+8 n-8]=90\)
n (8n – 4) = 90 × 2/1
8n² – 4n = 180
8n² – 4n = 1800
2n² – n – 45 = 0

2n²– 10n + 9n – 45 = 0
2n(n – 5) + 9(n – 5) = 0
(n – 5)(2n + 9) = 0
If n – 5 = 0 then n = 5
∴ a_n = a + (n – 1) d
a₅= 2 + (5 – 1) 8
= 2 + 4 × 8
= 2 + 32
∴ a₅ = 34

vii) a = 8, a_n = 62, S_n = 210, n =? d =?
Solution:
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210\)
\(\frac{n}{2}[8+62]=210\)
\(\frac{n}{2} \times 70=21\)
\(n=210 \times \frac{2}{70}\)
∴ n = 6
a_n = 62
a₆= a + 5d = 62
8 + 5d = 62
5d = 62 – 8
5d = 54
\(d=\frac{54}{5}\)

viii) a = 4, d = 2, S_n = -14, n =?, a =?
Solution:
a_n = a + (n – 1) d = 4
a + (n – 1) 2 = 4
a + 2n – 2 = 4
a + 2n = 4 + 2
a = 2n = 6 …………. (i)
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210\)
\(\frac{n}{2}[-2 n+6+4]=-14\)
\(\frac{n}{2}[-2 n+10]=-14\)
-2n² + 10n = -28
-2n² + 10n + 28 = 0
2n² – 10n – 28 = 0 – 14
²ⁿ² – 5n – 14 = 0

n² – 7n + 2n – 14 = 0
n(n – 7) + 2 (n – 7) = 0
(n – 7) (n + 2) = 0
If n – 7 = 0 then. n = 7
a = -2n + 6
= -2 × 7 + 6
= -14 + 6
∴ a = -8
∴ n = 7, a = -8.

ix) a = 3, n = 8, S = 192, d =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{8}=\frac{8}{2}[2 \times 3+(8-1) d]=192\)
\(\frac { 8 }{ 2 } [6+7\times d]=192\)
4 (6 + 7d) = 192
\(6+7 d=\frac{192}{4}\)
6 + 7d = 48
7d = 48 – 6
7d = 42
\(d=\frac{42}{7}\)
∴ d = 6

x) l = a_n = 28, S = 144, n = 9, a =?
Solution:
\(s_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(s_{9}=\frac{9}{2}[a+28]=144\)
\(\frac{9}{2}[a+28]=144\)
\(a+28=144 \times \frac{2}{9}\)
a + 28 = 32
a = 32 – 28
∴ a = 4.

Exercise 1.3 Class 10 Maths Arithmetic Progression KSEEB Solutions Question 4.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
9 + 17 + 25 + ……. + a_n = 636
a = 9, d = 17 – 9 = 8, S_n= 636, n =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{n}=\frac{n}{2}[2 \times 9+(n-1) 8]=636\)
\(\frac{n}{2}[18+8 n-8]=636\)
\(\frac{n}{2}[8 n+10]=636\)
n(8n + 10) = 636 × 2
8n² + 10n = 1272
8n²+ 10n – 1272 = 0
4n² + 5n – 636 = o
4n²+ 5n – 48n – 636 = 0

n(4n + 5) – 12(4n + 5) = 0
(4n + 5) (n – 12) = 0
1f n – 12 = 0 then, n = 12
∴ n = 12.

KSEEB Solutions For Class 10 Maths Arithmetic Progression 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:
a = 5, a_n = 45, S_n= 400, n =?, d =?
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(=\frac{n}{2}[5+45]=400\)
\(\frac{n}{2} \times 50=400\)
n × 50 = 800
\(n=800 \times \frac{1}{50}\)
∴ n = 16
a_n = a + (n – 1) d
a₁₆= 5 + (16 – 1)d = 45
5 + 15d = 45
15d = 45 – 5
15d = 40
\(d=\frac{40}{15}=\frac{8}{3}\)
\(\mathrm{n}=16, \mathrm{d}=\frac{8}{3}\)

Arithmetic Progression Class 10 Exercise 1.3 KSEEB Solutions 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:
a = 17, a_n= 350, d = 9, n =?, S_n =?
a_n = a + (n – 1) d
17 + (n – 1)9 = 350
17 + 9n – 9 = 350
9n + 8 = 350
9n = 350 – 8
9n = 342
\(\quad n=\frac{342}{9}\)
∴ n = 38
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(S_{38}=\frac{38}{2}[17+350]\)
= 19 × 367
∴ S₃₈ = 6973
∴ n = 38, S₃₈ = 6973

10th Maths Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
d = 7, a₂₂ = 149, S₂₂ =?
a = a + (n – 1)d
a + (22 – 1)7 = 149
a + 21 × 7 = 149
a + 147 = 149
=149 – 147
∴ a = 2
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(S_{22}=\frac{22}{2}[2+149]\)
= 11[151]
∴ S₂₂ = 1661

10th Standard Maths Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
If a₂ = 14, a₃ = 18, then S₁₁ =?
a₂ = 14
a + d = 14 …………. (i)
a₃ = 18
a + 2d = 18 …………… (ii)
from equation (1) – equatIon (11),

d =4.
Substituting the value of ‘d’ In eqn. (i)
a + d = 14
a + 4 = 14
a = 14 – 4
∴ a = 10
a = 10, d = 4
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{51}=\frac{51}{2}[2 \times 10+(51-1) 4]\)
\(=\frac{51}{2}[20+50 \times 4]\)
\(=\frac{51}{2}[20+200]\)
\(=\frac{51}{2} \times 220\)
= 51 × 110
∴ S₅₁ = 5610

10th Arithmetic Progression Exercise 1.3 KSEEB Solutions 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:
S₇ = 49, S₁₇ = 289, S₁₁ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{7}=\frac{7}{2}[2 a+(7-1) d]\)
\(=\frac{7}{2}[2 a+6 d]=49\)
[latex2+6 d=49 \times \frac{2}{7}[/latex]
2a + 6d = 14 …………… (i)
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{17}=\frac{17}{2}[2 a+(n-1) d]\)
\(=\frac{17}{2}[2 \mathrm{a}+16 \mathrm{d}]\)
\(\frac{17}{2}[2 a+16 d]=289\)
\(2 a+16 d=289 \times \frac{2}{17}\)
∴ 2a + 16d = 34 …………. (ii)

10d = 20
∴ d=2
Substituting the value of d in equation (i), we have
2a + 6d = 14
2a + 6 × 2 = 14
2a + 12 = 14
2a = 14 – 12
2a = 2
\(a=\frac{2}{2}=1\)
a = 1, d = 2, S_n =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{n}=\frac{n}{2}[2 \times 1+(n-1) 2]\)
\(=\frac{\mathrm{n}}{2}[2+2 \mathrm{n}-2]\)
\(=\quad \frac{\mathrm{n}}{2}[2 \mathrm{n}]\)
∴ S_n = n²

Maths Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 10.
Show that a₁, a₂, a₃, … a₁₁. … form an AP where a11 is defIned as below:
i) a = 3 + 4n
ii) a = 9 – 5n
Solution:
i) a_n= 3 + 4n
a₁ = 3 + 4(1)
= 3 + 4
∴ a_n = 7
∴ a = 7
a₁ = 3 + 4n
a₂ = 3 + 4 × 2
= 3 + 8
∴ a₂ = 11
a_n = 3 + 4n
a₃ = 3 + 4 × 3
= 3 + 12
a₃ = 15
∴ a₁, a₂, a₃, ………….
7, 11, 15, ……….
d = a₂ – a₁ = 11 – 7 = 4
d = a₃ – a₂= 15 – 11 = 4
Here, the value of ‘d’ is constant.
∴ a_n = 3 + 4n forms an Arithmetic Progression.

ii) a_n = 9 – 5n
a₁= 9 – 5 × 1
= 9 – 5
∴ a₁ = 4
a₁ = 9 – 5n
a₁ = 9 – 5 × 2
= 9 – 10
∴ a₂ = -1
a_n= 9 – 5n
a₃ = 9 – 5 × 3
= 9 – 15
∴ a₃ = -6
a₁, a₂, a₃, …………… a_n
4, -1, -6, …………
d = a₂ – a₁ = -1 – 4 = -5
d=a₃ – a₂ = -6 – (-1) = -5
Here, the value of ‘d’ is constant.
∴ a_n = 9 – 5n form an Arithmetic Progression.

Arithmetic Progression Class 10 Notes KSEEB Question 11.
lf the sum of the first n terms of an AP is 4n – n², what is the first term (that is SI)? What is the sum of first two termš? What is the second term? Similarly. find the 3^rdrd the 10^th and the n^th terms.
Solution:
If S_n = 4n – n², then
i) S₁ = a =?
ii) S₂ =?
iii) a₂ =?
iv) a₃ =?
v) a₁₀ =?
vi) a_n =?
(i) S_n = 4n – n²
S₁ = 4(1) – 1²
= 4 – 1
S₁ = 3
∴ S₁ = a = 3.

(ii) S₂ = 4n – n²
S₂ = 4(2) – 2²
= 8 – 4
S₂ = 4
∴ S₂ = 4

(iii) We have S₂ = a₁ + a₂ = 4
= 3 + a₂ = 4
∴ a₂ = 4 – 3
∴a₂ = 1

(iv) S_n = 4n – n²
S₃= 4(3) – 3²
= 12 – 9
S₃ = 3
a₁ + a₂ + a₃ = 3
3 + 1 + a₃ = 3
4 + a₃= 3
∴ a₃ = 3 – 4
∴ a ₃ = -1

(v) d = a₃ – a₃ = -1 -1 = -2
a₁₀= a + 9d
= 3 + 9(-2)
= 3 – 18
a₁₀= -15
∴ a₁₀ = -15

(vi) a_n = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 + 2n + 2
∴ a_n = 5 – 2n
∴ a_n = 5 – 2n

Arithmetic Progression Class 10 KSEEB Solutions Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
6 + 12 + 18 + 24 + 40 term
Here a = 6, d = ₂ – a₁ = 12 – 6 = 6
n = 40, S₄₀ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{40}=\frac{40}{2}[2 \times 6+(40-1) 6]\)
= 20[12 + 39 × 6]
= 20[12 + 234]
= 20 × 246
∴ S₄₀ = 4920.

Arithmetic Progression 1.3 KSEEB Solutions Question 13.
Find the sum of first 15 multiples of 8,
Solution:
Sum of the first 15 multiples of 8 ?
8 + 16 + 24+ ……… 15 terms.
Here, a = 8, d = a₂ – a₁= 16 – 8 = 8
n = 15, S₁₅ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{15}=\frac{15}{2}[2 \times 8+(15-1) 8]\)
\(=\frac{15}{2}[16+14 \times 8]\)
\(=\frac{15}{2}[16+112]\)
\(=\frac{15}{2} \times 128\)
= 15 × 64
∴ S₁₅ = 960

Arithmetic Progression Exercise 1.3 Solution Question 14.
Fnd the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are
1 + 3 + 5 + 7 + ……… + 49 =?
Here. a = 1, d = a₃ – a₁= 3 – 1 = 2
a_n = 49, n =?, S =?
a_n = a + (n – 1) d = 49
= 1 + (n – 1)2 = 49
1 + 2n – 2 = 49
2n – 1 = 49
2n = 49 + 1
2n = 50
\(n=\frac{50}{2}\)
∴ n = 25
\(S_{n}=\frac{n}{2}\left[a+a_{n}\right]\)
\(\mathrm{s}_{25}=\frac{25}{2}[1+49]\)
\(=\frac{25}{2} \times 50\)
= 25 × 25
∴ S₂₅ = 625

Class 10 Maths Chapter 1 Arithmetic Progression KSEEB Solutions Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day. Rs. 250 for the second day. Rs. 300 for the third day. etc.. the penalty for each succeeding day being Rs. 50 more than for the preceeding day. How much money the contractor has to pay as penalty. If he has delayed the work by 30 days?
Solution:
Penalty for the 1st Day 2nd day 3rd day … 30^th day
Rs. 200 Rs. 250 Rs. 300 Rs.?
200 + 250 + 300 + 30 days
a = 200, d = 250 – 200 = 50, n = 20,
S₃₀ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(\dot{\mathrm{S}}_{30}=\frac{30}{2}[2 \times 200+(30-1) 50]\)
= 15[400 + 29 × 50]
= 15 [400 + 1450]
= 15 × 1850
∴ S₃₀= Rs. 27750.

KSEEB Solutions For Class 10 Maths 1.3 Question 16.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for thler overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the first prize be ‘a’.
and Second prize is a – 20 and
the third prize Is a 40.
a, (a – 20), (a – 40) ……….. n = 7
a = a, d = a₂ – a₁ = a – 20 – a d = -20
n = 7, S₇ =?
\(\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]=\mathrm{Sn}\)
\(\frac{7}{2}[2 \times a+(7-1)(-20)]=700\)
\(\frac{7}{2}[2 a+6(-20)]=700\)
\(\frac{7}{2}[2 a-120]=700\)
\(2 a-120=700 \times \frac{2}{7}\)
2a – 120 = 200
∴ 2a = 200 + 120
2a = 320
\(\quad a=\frac{320}{2}=R s .160\)
∴ Each prizes are
Rs. 160, 140, 120, 100, 80, 60, 40

Arithmetic Progression KSEEB Solutions 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 Il 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:

a = 3. d = 6 – 3 = 3, n = 12, S₁₂ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(\mathrm{S}_{12}=\frac{12}{2}[2 \times 3+(12-1) 3]\)
= 6[6 + 11 × 3]
= 6[6 + 33]
= 6 × 39
∴ S₁₂ = 234
∴ Total number of trees from 3 sections of each class upto 12 class is 234.

Arithmetic Progression Class 10 Exercise 1.3 KSEEB Solutions Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm. 1.0 cm. 1.5 cm, 2.0 cm as shows In fig. What Is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi=\frac{22}{7}\))

[Hint: Length of successive semicircles is l₁, l₂, l₃, l₄ with centres at A, B, A. B respectively.]
Solution :
\(: l_{1}=\pi \times \frac{1}{2}, 12=\pi \times 1.13=\pi \times \frac{3}{2}\)
\(l_{1}=\frac{\pi}{2}, \quad l_{2}=\mathrm{p}, \quad l_{3}=\frac{3}{2} \pi\)
∴ Arithmetic Progression,
l₁, l₂, l₃, l₄, ………..

Arithmetic Progression Class 10 Karnataka State Board KSEEB Solutions 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 to it and so on (see the fig. given below). In how many rows are the 200 logs placed and how many logs are In the top row?

Solution:
20, 19, 18, …..
a = 20, d = 19 – 20 = -1
S_n = 200, n =?, a_n =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(200=\frac{\mathrm{n}}{2}[2 \times 20+(\mathrm{n}-1)(-1)]\)
\(200=\frac{\mathrm{n}}{2}[40-\mathrm{n}+1]\)
\(200=\frac{n}{2}[41-n]\)
∴ 400 = n(41 – n)
400 = 4n – n²
∴ n² – 41n + 400 = 0
n² – 25n – 16n + 400 = 0
n(n – 25) – 16(n – 25) = 0
(n – 25) (n – 16) =
1f n – 16 = 0 then, n = 16
∴ a_n = a + (n – 1) d
a₁₆= 20 + (16 – 1) (-1)
= 20 + 15(-1)
= 20 – 15
∴ a₁₆ = 5
∴ 200 logs are placed In 16 rows and there are 5 logs in the top row.

Exercise 1.3 Arithmetic Progression KSEEB Solutions 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 line. There are ten potatoes in the line (see fig, given below)

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops ¡tin 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 al) the potaotes 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 × 5 +2 × (5 + 3)]
Solution:
Total distance competitor taken to pick up the first potato = 5 + 5m
= 2 × 5m.
= 10m.
Total distance taken by compeUtor to pick up the second potato
= 5 + 3 + 3 + 5m.
= 2 × 5 + 2 × 3
= 2(5 + 3)
= 2 × 8
= 16m.
∴ 10m. 16m, 22m 10th potato
a = 10, d = 16 – 10 = 6m.
n = 10, S₁₀ =?
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{10}=\frac{10}{2}[2 \times 10+(10-1) 6]\)
= 5[20 + 9 × 6]
= 5 [20 + 54]
= 5 × 74
∴ S₁₀ = 370m.
∴ Total distance the competitor has to run to pick up 10 potatoers is 370 m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3, drop a comment below and we will get back to you at the earliest.

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
a₈= 7 + (8 – 1) 3
= 7 + 7 × 3
= 7 + 21
∴ a₈ = 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, a₃₀ =?
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1)(-3)
= 10 + 29(-3)
= 10 – 87
∴ a₃₀ = 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, a₁₁ =?
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
∴ a₁₁ = 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, a₁₆ = 83 a₃₁ =?
a_n = 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.
∴ a_n = a + (n – 1)d
a₃₁ = 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, a₃ = 12, a_n = 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.
∴ a_n = a + (n – 1) d
a₂₉ = 8 + (29 – 1) 2
= 8 + 28 × 2
= 8 + 56
∴ a₂₉ = 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:
a₃ = 4, a₉ = -8, a_n = 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
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:
a₁₇ = a₁₀ + 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 = a₅₄ + 132
a = 3, d = 15 – 3 = 12
a_n = a₅₄ + 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, 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, ………….

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, ………… a₂₀ =?
a = 253, d = 258 – 253 = 5, n = 20
a_n = a + (n – 1) d
a₂₀= 253 + (20 – 1) 5
= 253 + 19 × 5
= 253 + 95
∴ a₂₀ = 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:
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, …………

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.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2, drop a comment below and we will get back to you at the earliest.

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Karnataka SSLC 2nd Language Kannada Model Question Paper Design

Karnataka SSLC 3rd Language Kannada Model Question Paper Design

We hope the Karnataka State Board Syllabus KSEEB SSLC 10th Kannada Model Question Papers 2019-2020 with Key Answers Pdf Download of KSEEB Class 10th Std 1st and 2nd Language Kannada Previous Year Model Question Papers, Sample Papers will help you.

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