In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 3 Progressions for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 3 Progressions pdf, free KSEEB solutions for Class 10 Maths Chapter 3 Progressions book pdf download. Now you will get step by step solution to each question.

**KSEEB SSLC Solutions for Class 10 Maths – Progressions (English Medium)**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3** **Exercise 3.1**

**Question 1:**

Which of the following form a sequence?

i. 4, 11, 18, 25,……

ii. 43, 32, 21, 10,……

iii. 27, 19, 40, 70,…….

iv. 7, 21, 63, 189,…….

**Solution :**

i. 4, 11, 18, 25, ….

ii. T_{2} – T_{1} = 11 – 4 = 7

iii. T_{3} – T_{2 }= 18 – 11 = 7

iv. T_{4} – T_{3} = 25 – 18 = 7

There is a rule of adding 7 to each term to get the next terms.

∴ It is a sequence.

ii. 43, 32, 21, 10, …..

Here,

T_{1} – 11 = T_{2} T_{2} – 11 = T_{3} T_{3} – 11 = T_{4}

There is a rule of subtracting 11 from each term to get the next terms.

∴ It is a sequence.

iii. 27, 19, 40, 70, ….

Here there is no rule in writing the terms.

∴ It is not a sequence.

iv. 7, 21, 63, 189, …..

Here,

T_{1} × 3 = T_{2} T_{2} × 3 = T_{3} T_{3} × 3 = T_{4} There is a rule of multiplying by 3 to each term to get the next terms.

∴ It is a sequence.

**Question 2:**

**Solution :**

**Question 3:**

If T_{n} = 5 – 4n, find the first three term

**Solution :**

T_{n} = 5 – 4n

T_{1} = 5 – 4(1)

T_{1} = 5 – 4

T_{1} = 1

T_{2} = 5 – 4(2)

T_{2} = 5 – 8

T_{2} = -3

T_{3} = 5 – 4(3)

T_{3} = 5 – 12

T_{3} = -7

**Question 4:**

If T_{n} = 2n^{2} + 5, find

i. T_{3}ii. T_{10}

**Solution :**

i. T_{n} = 2n^{2} + 5

T_{3} = 2(3)^{2} + 5

T_{3} = 2 × 9 + 5

T_{3} = 23

ii. T^{n} = 2n^{2} + 5

T_{10} = 2(10)^{2} + 5

T_{10} = 2 × 100 + 5

T_{10} = 205

**Question 5:**

If T_{n} = n^{2} – 1, find

i. T_{n-1}ii. T_{n}_{+ 1}

**Solution :**

i. T_{n} = n^{2} -1

T_{n-1 }= (n-1)^{2} -1

= n^{2} – 2n + 1 – 1

= n^{2} – 2n

T_{n-1 }= n(n-2)

ii. T_{n+1} = (n+1)^{2} – 1

= n^{2} + 2n + 1 – 1

T_{n+1} = n(n+2)

**Question 6:**

If T_{n} = n^{2}+4 and T_{n} = 200, find the value of ‘n’.

**Solution :**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Exercise 3.2**

**Question 1(i):**

Write the next four terms of the following A.P

0, -3, -6, -9

**Solution :**

a = 0, d = T_{2 }– T_{1 }= -3 – 0 = -3

T_{n+1} = T_{n}+ d

T_{5} = T_{4 }+ d

T_{5} = -9 + (-3)

T_{5} = -12

T_{6} = T_{5 }+ d

T_{6} = -12 + (-3)

T_{6} = -15

T_{7} =T_{6 }+ d

T_{7} =-15 + (-3)

T_{7} =-18

T_{8} =T_{7 }+ d

T_{8} =-18 + (-3)

T_{8} =-21

**Question 1(ii):**

**Solution :**

**Question 1(iii):**

Write the next four terms of the following A.P

a + b, a – b, a – 3b,

**Solution :**

T_{1} = a + b d = T_{2} – T_{1} = a – b – a – b

= -2b

T_{n+1} = T_{n} + d

T_{4} = T_{3} + d

T_{4} = a – 3b – 2b

T_{4} = a – 5b

T_{5} = T_{4} + d

T_{5} = a – 5b – 2b

T_{5} = a – 7b

T_{6} = T_{5} + d

T_{6} = a – 7b – 2b

T_{6} = a – 9b

T_{7} =T_{6} + d

T_{7} = a – 9 – 2b

T_{7} = a – 11b

**Question 2:**

Find the sequence if,

i. T_{n} = 2n – 1

ii. T_{n} = 5 – 4n

iii. T_{n} = 5n + 1

**Solution :**

i. T_{1} = 2(1) – 1

T_{1} = 1

T_{2} = 2(2) – 1

T_{2} = 4 – 1

T_{2} = 3

T_{3} = 2(3) – 1

T_{3} = 6 – 1

T_{3} = 5

Sequence is 1, 3, 5, 7…….

ii. T_{n} = 5 – 4n

T_{1} = 5 – 4 (1)

T_{1} = 5 – 4

T_{1} = 1

T_{2} = 5 – 4(2)

T_{2} = 5 – 8

T_{2} =-3

T_{3} = 5 – 4(3)

T_{3} = 5 – 12

T_{3} = -7

The sequence is 1, -3, -7 ….

iii. T_{n} = 5n + 1

T_{1} = 5(1) + 1

T_{1} =5 + 1

T_{1} =6

T_{2} = 5(2) + 1

T_{2} = 10 + 1

T_{2} = 11

T_{3} = 5(3) + 1

T_{3} = 15 + 1

T_{3} = 16

The sequence is 6, 11, 16….

**Question 3(i):**

In an A.P,

if a = 5, d = 3, find T_{10}

**Solution :**

T_{n} = a + (n – 1)d

T_{10} = 5 + (10 – 1)3

= 5 + 9 × 3

= 5 + 27

T_{10} = 32

**Question 3(ii):**

In an A.P,

if a = -7, d = 5 find T_{12}

**Solution :**

T_{n} = a + (n – 1)d

T_{12} = -7 + (12 – 1)5

= -7 + 11 × 5

= -7 + 55

T_{12} = 48

**Question 3(iii):**

In an A.P,

a = -1, d = -3, find T_{50}

**Solution :**

T_{n} = a + (n – 1)d

T_{50} = -1 + (50 – 1) (-3)

= – 1 + 49 × (-3)

= – 1 – 147

T_{50} = -148

**Question 3(iv):**

In an A.P,

if a = 12, d = 4, T_{n} = 76, find n

**Solution :**

**Question 3(v):**

In an A.P,

if d = -2, T_{22} = -39, find a

**Solution :**

T_{n} = a + (n – 1)d

-39 = a +(22 – 1)(-2)

-39 = a + 21 × (-2)

-39 = a – 42

a = -39 + 42

a = 3

**Question 3(vi):**

In an A.P,

if a = 13, T_{15} = 55, find ‘d’

**Solution :**

**Question 4:**

Find the number of terms in the A.P, 100, 96, 92,…, 12.

**Solution :**

**Question 5:**

The angles of a triangle are in A.P. If the smallest angle is 50°, find the other two angles.

**Solution :**

**Question 6:**

If a, b, c, d, e are in A.P, prove that a + e = b + d = 2c

**Solution :**

∴ b – a = c – b = d – c = e – d

Now b – a = e – d

∴ b + d = a + e ….. (1)

consider c – b = d – c

b + d = c + c

b + d = 2c ….. (2)

comparing (1) and (2)

b + d = a + e = 2c

**Question 7:**

Which term of the A.P is 12, 10, 8,…… is -48?

**Solution :**

**Question 8:**

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

**Solution :**

**Question 9:**

The sum of 4^{th} and 8^{th} terms of an A.P is 24 and the sum of 6^{th} and 10^{th} terms of the same A.P is 44. Find the first three terms.

**Solution :**

**Question 10:**

The ratio of 7^{th} to 3^{rd} term of an A.P is 12 : 5. Find the ratio of 13^{th} to 4^{th} term.

**Solution :**

**Question 11:**

A company employed 400 persons in the year 2001 and each year increased by 35 persons. In which year the number of employees in the company will be 785?

**Solution :**

**Question 12:**

If the p^{th} term of an A.P is q and the q^{th} term is p, prove that the n^{th} term equal to (p + q – n).

**Solution :**

**Question 13:**

Find four numbers in A.P such that the sum of 2^{nd} and 3^{rd} terms is 22 and the product of 1^{st} and 4^{th} terms is 85.

**Solution :**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3** **Exercise 3.3**

**Question 1:**

If T_{n} = 2n + 3, find S_{3}

**Solution :**

;

**Question 2:**

If T_{n} = n^{2} – 1 find

i. S_{5}ii. S_{2}

**Solution :**

**Question 3(i):**

Find the sum of 3 + 7 + 11 + ……. to 25 terms.

**Solution :**

**Question 3(ii):**

Fine the sum of -3, 1,5 …….. to 17 terms.

**Solution :**

**Question 3(iii):**

Find the sum of 3a, a, -a …… to a terms.

**Solution :**

**Question 3(iv):**

**Solution :**

**Question 3(v):**

**Solution :**

**Question 3(vi):**

Find the sum of p, 0, -p …… to p terms.

**Solution :**

**Question 4:**

**Solution :**

**Question 5(a):**

For a sequence of natural numbers,

Find

i. S_{20}ii. S_{50} – S_{40}iii. S_{30} + S_{15}

**Solution :**

**Question 5(b):**

For a sequence of natural numbers,

Find n if

i. S_{n} = 55

ii. S_{n} = 15

**Solution :**

**Question 6:**

Find the sum of all the first ‘n’ odd natural numbers.

**Solution :**

**Question 7:**

Find the sum of all natural numbers between 1 and 201 which are divisible by 5.

**Solution :**

**Question 8:**

Find the sum of all natural numbers between 1 and 201 which are divisible by 5.

If the smallest angle, ∠AOB = 20°, find ∠BOC and ∠COD.

**Solution :**

**Question 9:**

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**Solution :**

**Question 10:**

How many terms of the A.P 1, 4, 7,…… are needed to make the sum 51?

**Solution :**

**Question 11(i):**

Find three numbers in AP whose sum and products are respectively. 21 and 231

**Solution :**

**Question 11(ii):**

Find three numbers in AP whose sum and products are respectively. 36 and 1620.

**Solution :**

**Question 12:**

The sum of 6 terms which form an A.P is 345. The difference between the first and last terms is 55. Find the terms.

**Solution :**

**Question 13:**

In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that T_{20} = -112. Find S_{20}.

**Solution :**

**Question 14:**

The third term of an A.P is 8 and the ninth term of the A.P exceeds three times the third term by 2. Find the sum of its first 19 terms.

**Solution :**

**Question 15:**

**Solution :**

**Question 16:**

A child wishes to build up a triangular pile of toy bricks so as to have 1 brick in the top rows, 2 in the second, 3 in the third and so on. If he has 100 bricks, how many rows can be completed and how many bricks will remain un utilized?

**Solution :**

Total number of bricks = 100 = S_{n} when the bricks are arranged in row 1 + 2 + 3 + ….. = 100

Here S_{1} = 1

S_{2} = 3 (3 – 1 = 2)

S_{3 }= 6 (6 – 3 = 3)

S_{4 }= 10 (10 – 6 = 4)

S_{5 }= 15 (15 – 10 = 5)

S_{6 }= 21

.

.

.

.

S_{13} = 91

∴ 13 rows can be completed and 9 bricks are left.

**Question 17:**

In a game, a basket and 16 potatoes are placed in line at equal intervals of 6 ft. How long will a player take to bring the potatoes one by one into the basket, if he starts from the basket and runs at an average speed of 12 feet a second?

**Solution :**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3** **Exercise 3.4**

**Question 1(i):**

**Solution :**

Reciprocals of the given Harmonic progression are in Arithmetic progression.

The reciprocal of the given sequence

1, 4, 7, 10 ……

Here T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} (4 – 1) = (7 – 4) = (10 – 7) = 3

∴ Reciprocals of the given sequence form an Arithmetic Progression.

∴ It is a H.P

**Question 1(ii):**

**Solution :**

**Question 1(iii):**

**Solution :**

The reciprocal of the given sequence is

2, 6, 18

Here T_{2} – T_{1} = 6 – 2 = 4

T_{3} – T_{2} = 18 – 6 = 12

There is no constant difference between the consecutive terms.

∴ Reciprocals do not form an Arithmetic Progression.

∴ It is not a H.P

**Question 1(iv):**

**Solution :**

The reciprocal of the given sequence is,

3, 7, 11…….

Here T_{2} – T_{1} = 7 – 3 = 4

T_{3} – T_{2} = 11 – 7 = 4

There is a constant difference between the consecutive terms.

∴ Reciprocals of the given sequence form an Arithmetic Progression.

∴ It is a H.P

**Question 1(v):**

**Solution :**

**Question 1(vi):**

**Solution :**

The reciprocal of the given sequence is,

1, 2, 4…….

Here T_{2} – T_{1} = 2 – 1 = 1

T_{3} – T_{2} = 4 – 2 = 2

There is no constant difference in the sequence.

∴ Reciprocals of the given sequence do not form an Arithmetic Progression.

∴ It is not a Harmonic Progression.

**Question 2(i):**

**Solution :**

**Question 2(ii):**

**Solution :**

**Question 3:**

**Solution :**

**Question 4:**

**Solution :**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Exercise 3.5**

**Question 1:**

**Solution :**

**Question 2(i):**

**Solution :**

**Question 2(ii):**

Do as directed In the G.P 729, 243, 81, ……. Find T_{7}**Solution :**

**Question 3:**

Find the 12^{th} term of a G.P whose 5^{th} term is 64 and common ratio is 2.

**Solution :**

T_{5} = 64 = ar^{4} r = 2

T_{12}= ?

T_{n} = ar^{n-1}T_{12} = ar^{12-1}T_{12} = ar^{11}T_{12} = ar^{4} × r^{7}T_{12} = T_{5 }× r^{7} [since T_{5} = ar^{4}]

T_{12} = 64 × 2^{7} [given T_{5} = 64]

T_{12} = 64 × 128

T_{12} = 8192

**Question 4(i):**

Find the following 5^{th} and 8^{th} terms of the G.P. 3, 6, 12, …………

**Solution :**

**Question 4(ii):**

Find the following 10^{th} and 16^{th} terms of the G.P. 256, 128, 64, …………

**Solution :**

**Question 4(iii):**

Find the following 8^{th} and 12^{th} terms of the G.P. 81, -27, 9, …………

**Solution :**

**Question 4(iv):**

**Solution :**

**Question 4(v):**

Find the following 4^{th} and 8^{th} terms of the G.P. 0.008, 0.04, 0.2 …………

**Solution :**

**Question 5(i):**

Find the last term of the following series : 2, 4, 8 …… to 9 terms

**Solution :**

**Question 5(ii):**

Find the last term of the following series : 4, 4^{2}, 4^{3} …… to 2n terms.

**Solution :**

**Question 5(iii):**

**Solution :**

**Question 5(iv):**

**Solution :**

**Question 6:**

**Solution :**

**Question 7:**

The half life period of a certain radioactive material is 1 hour. If the initial sample weighed 500 gm, find the mass of the sample remaining at the end of 5^{th} hour.

**Solution :**

**Question 8:**

Which term of the sequence 3, 6, 12, ….. is 1536?

**Solution :**

**Question 9:**

If the 4^{th} and 8^{th} terms of a GP are 24 and 384 respectively, find the first term and common ratio.

**Solution :**

**Question 10(i):**

Find the G.P. in which the 10th term is 320 and the 6th term is 20.

**Solution :**

**Question 10(ii):**

**Solution :**

**Question 10(iii):**

**Solution :**

**Question 11(i):**

If a, b, c, d are is geometric sequence, then prove that (b-c)^{2} + (c-a)^{2} + (d-b)^{2} = (a-d)^{2}

**Solution :**

(b – c)^{2} + (c – a)^{2} + (d – b)^{2} = (a – d)^{2}a, b, c, d are in G.P

∴ b/a = c/b = d/c

∴ b^{2} = ca, c^{2} = db, bc = ad

Now LHS = (b – c)^{2} + (c – a)^{2} + (d – b)^{2} = b^{2} + c^{2} – 2bc + c^{2} + a^{2} – 2ca + d^{2} + b^{2} -2bd

= 2b^{2} + 2c^{2} – 2bc – 2ca – 2bd + a^{2} + d^{2}Substituting the valves of b^{2}, c^{2} and bc

= 2ca + 2bd – 2bc – 2ca – 2bd + a^{2} + d^{2} = a^{2} + d^{2} – 2bc

= a^{2} + d^{2} – 2ad

RHS = (a – d)^{2}∴ LHS = RHS.

**Question 11(ii):**

If a, b, c, d are is geometric sequence, then prove that (a-b+c) (b+c+d) = ab+bc+cd

**Solution :**

a, b, c, d are in G.P

∴ b/a = c/b = d/c

∴ b^{2} = ca, c^{2} = db, bc = ad

LHS = (a – b + c) (b + c + d)

= ab + ac + ad – b^{2} – bc – bd + bc + c^{2} + cd

= ab + bc + cd + ac + ad – b^{2} – bc – bd + c^{2}Substituting the values of b^{2}, c^{2} and bc

= ab + bc + cd + ac + ad – ca – ad – bd +bd

RHS = ab + bc + cd

∴ LHS = RHS.

**Question 11(iii):**

If a, b, c, d are in geometric sequence, then prove that (a + b), (b + c), (c + d) are also in G.P.

**Solution :**

a, b, c, d are in G.P

∴ b/a = c/b = d/c

∴ b^{2} = ca, c^{2} = db, bc = ad

(b + c)^{2} = (a + b) (c + d)

(b + c)^{2 }= ac + ad + bc + bd

b^{2} + c^{2} + 2bc = b^{2} + bc + bc + c^{2}2bc = 2bc

∴ (a + b), (b + c) and (c + d) are in G.P.

∴ LHS = RHS

∴ (a + b), (b + c), (c + d) are also in G.P.

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3** **Exercise 3.6**

**Question 1(i):**

Find the sum of the following geometric series. 1 + 2 + 4 + …. upto 10 terms.**Solution :**

**Question 1(ii):**

Find the sum of the following geometric series. 2 – 4 + 8 – …. upto 6 terms.

**Solution :**

&

**Question 1(iii):**

**Solution :**

**Question 1(iv):**

**Solution :**

**Question 2:**

Find the first term of a G.P in which S_{8} = 510 and r = 2**Solution :**

**Question 3(i):**

Find the sum of the G.P 1 + 2 + 4 + ….. + 512

**Solution :**

**Question 3(ii):**

**Solution :**

**Question 4:**

How many terms of the series 2 + 4 + 8 + ……make the sum 1022?

**Solution :**

**Question 5(i):**

Find S_{2}: S_{4} for the series 5 + 10 + 20 + ………

**Solution :**

**Question 5(ii):**

Find S_{4}: S_{8} for the series 4 + 12 + 36 + ……..

**Solution :**

**Question 6(i):**

Find the G.P if S_{6} : S_{3} = 126 : 1 and T_{4} = 125

**Solution :**

**Question 6(ii):**

Find the G.P if S_{10} : S_{5} = 33 : 32 and T_{5} = 64

**Solution :**

**Question 7:**

The first term of an infinite geometric series is 6 and its sum is 8. Find the G.P.

**Solution :**

**Question 8(i):**

Find 3 terms in G.P whose sum and product respectively are 7 and 8

**Solution :**

**Question 8(ii):**

Find 3 terms in G.P whose sum and product respectively are 21 and 216

**Solution :**

**Question 8(iii):**

Find 3 terms in G.P whose sum and product respectively are 19 and 216

**Solution :**

**Question 9:**

A person saved every year half as much he saved the previous year. If he totally saved Rs. 19,375 in 5 years, how much did he save the first year?

**Solution :**

**Question 10(a):**

A bouncing ball rebounds each time to a height half the height of the previous bounce. If it is dropped from a height of 10m, find the total distance it has travelled when it hits the ground for the 10th time.

**Solution :**

**Question 10(b):**

A bouncing ball rebounds each time to a height half the height of the previous bounce. If it is dropped from a height of 10m, find the total distance it travels before coming to rest.

**Solution :**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Exercise 3.7**

**Question 1(i):**

Find the A.M, G.M and H.M between 12 and 30**Solution :**

**Question 1(ii):**

**Solution :**

**Question 1(iii):**

Find the A.M, G.M and H.M between -8 and -42

**Solution :**

**Question 1(iv):**

Find the A.M, G.M and H.M between 9 and 18

**Solution :**

**Question 2:**

Find x if 5, 8, x are in H.P

**Solution :**

**Question 3:**

Find x if the following are in A.P

i. 5, (x – 1), 0

ii. (a + b)^{2}, x, (a – b)^{2}

**Solution :**

**Question 4:**

The product of two numbers is 119 and their AM is 12. Find the numbers.

**Solution :**

**Question 5:**

**Solution :**

**Question 6:**

The arithmetic mean of two numbers is 17 and their geometric mean is 15. Find the numbers.

**Solution :**

**Question 7:**

**Solution :**

**Question 8:**

Find two numbers whose arithmetic mean exceeds their geometric mean by 2, and whose harmonic mean is one-fifth of the larger number.

**Solution :**

**Question 9:**

**Solution :**

**Question 10:**

If ‘a’ be the arithmetic mean between b and c, and ‘b’ the geometric mean between a and c, then prove that ‘c’ will be the harmonic mean between a and b.

**Solution :**

**All Chapter KSEEB Solutions For Class 10 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

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