KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Progressions

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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. T₂ – T₁ = 11 – 4 = 7
iii. T₃ – T₂ = 18 – 11 = 7
iv. T₄ – T₃ = 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,
T₁ – 11 = T₂
      T₂ – 11 = T₃
      T₃ – 11 = T₄
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,
T₁ × 3 = T₂
      T₂ × 3 = T₃
      T₃ × 3 = T₄
      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
T₁ = 5 – 4(1)
T₁ = 5 – 4
T₁ = 1

T₂ = 5 – 4(2)
T₂ = 5 – 8
T₂ = -3

T₃ = 5 – 4(3)
T₃ = 5 – 12
T₃ = -7

Question 4:

If T_n = 2n² + 5, find
i. T₃
ii. T₁₀

Solution :

i. T_n = 2n² + 5
T₃ = 2(3)² + 5
T₃ = 2 × 9 + 5
T₃ = 23

ii. Tⁿ = 2n² + 5
T₁₀ = 2(10)² + 5
T₁₀ = 2 × 100 + 5
T₁₀ = 205

Question 5:

If T_n = n² – 1, find
i. T_n-1
ii. T_{n+ 1}

Solution :

i. T_n = n² -1
T_n-1 = (n-1)² -1
= n² – 2n + 1 – 1
= n² – 2n
T_n-1 = n(n-2)

ii. T_n+1 = (n+1)² – 1
= n² + 2n + 1 – 1
T_n+1 = n(n+2)

Question 6:

If T_n = n²+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 = T₂ – T₁ = -3 – 0 = -3
T_n+1 = T_n+ d
T₅ = T₄ + d
T₅ = -9 + (-3)
T₅ = -12

T₆ = T₅ + d
T₆ = -12 + (-3)
T₆ = -15

T₇ =T₆ + d
T₇ =-15 + (-3)
T₇ =-18

T₈ =T₇ + d
T₈ =-18 + (-3)
T₈ =-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 :

T₁ = a + b            d = T₂ – T₁
                             = a – b – a – b
= -2b

T_n+1 = T_n + d
T₄ = T₃ + d
T₄ = a – 3b – 2b
T₄ = a – 5b

T₅ = T₄ + d
T₅ = a – 5b – 2b
T₅ = a – 7b

T₆ = T₅ + d
T₆ = a – 7b – 2b
T₆ = a – 9b

T₇ =T₆ + d
T₇ = a – 9 – 2b
T₇ = 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. T₁ = 2(1) – 1
T₁ = 1

T₂ = 2(2) – 1
T₂ = 4 – 1
T₂ = 3

T₃ = 2(3) – 1
T₃ = 6 – 1
T₃ = 5
Sequence is 1, 3, 5, 7…….

ii. T_n = 5 – 4n
T₁ = 5 – 4 (1)
T₁ = 5 – 4
T₁ = 1

T₂ = 5 – 4(2)
T₂ = 5 – 8
T₂ =-3

T₃ = 5 – 4(3)
T₃ = 5 – 12
T₃ = -7
The sequence is 1, -3, -7 ….

iii. T_n = 5n + 1
T₁ = 5(1) + 1
T₁ =5 + 1
T₁ =6

T₂ = 5(2) + 1
T₂ = 10 + 1
T₂ = 11

T₃ = 5(3) + 1
T₃ = 15 + 1
T₃ = 16
The sequence is 6, 11, 16….

Question 3(i):

In an A.P,
if a = 5, d = 3, find T₁₀

Solution :

T_n = a + (n – 1)d
T₁₀ = 5 + (10 – 1)3
= 5 + 9 × 3
= 5 + 27
T₁₀ = 32

Question 3(ii):

In an A.P,
if a = -7, d = 5 find T₁₂

Solution :

T_n = a + (n – 1)d
T₁₂ = -7 + (12 – 1)5
= -7 + 11 × 5
= -7 + 55
T₁₂ = 48

Question 3(iii):

In an A.P,
a = -1, d = -3, find T₅₀

Solution :

T_n = a + (n – 1)d
T₅₀ = -1 + (50 – 1) (-3)
= – 1 + 49 × (-3)
= – 1 – 147
T₅₀ = -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, T₂₂ = -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, T₁₅ = 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 S₃

Solution :

;

Question 2:

If T_n = n² – 1 find
i. S₅
ii. S₂

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. S₂₀
ii. S₅₀ – S₄₀
iii. S₃₀ + S₁₅

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 T₂₀ = -112. Find S₂₀.

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 S₁ = 1
S₂ = 3 (3 – 1 = 2)
S₃ = 6 (6 – 3 = 3)
S₄ = 10 (10 – 6 = 4)
S₅ = 15 (15 – 10 = 5)
S₆ = 21
.
.
.
.
S₁₃ = 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 T₂ – T₁ = T₃ – T₂ = T₄ – T₃
        (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 T₂ – T₁ = 6 – 2 = 4
T₃ – T₂ = 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 T₂ – T₁ = 7 – 3 = 4
T₃ – T₂ = 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 T₂ – T₁ = 2 – 1 = 1
T₃ – T₂ = 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 T₇
Solution :

Question 3:

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

Solution :

T₅ = 64 = ar⁴
 r = 2
T₁₂= ?
T_n = ar^n-1
T₁₂ = ar^12-1
T₁₂ = ar¹¹
T₁₂ = ar⁴ × r⁷
T₁₂ = T₅ × r⁷ [since T₅ = ar⁴]
T₁₂ = 64 × 2⁷ [given T₅ = 64]
T₁₂ = 64 × 128
T₁₂ = 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, 4², 4³ …… 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
∴ b² = ca, c² = db, bc = ad
Now LHS = (b – c)² + (c – a)² + (d – b)²
   = b² + c² – 2bc + c² + a² – 2ca + d² + b² -2bd
= 2b² + 2c² – 2bc – 2ca – 2bd + a² + d²
Substituting the valves of b², c² and bc
= 2ca + 2bd – 2bc – 2ca – 2bd + a² + d²
 = a² + d² – 2bc
= a² + d² – 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
∴ b² = ca, c² = db, bc = ad

LHS = (a – b + c) (b + c + d)
= ab + ac + ad – b² – bc – bd + bc + c² + cd
= ab + bc + cd + ac + ad – b² – bc – bd + c²
Substituting the values of b², c² 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

∴ b² = ca, c² = db, bc = ad
(b + c)² = (a + b) (c + d)
(b + c)² = ac + ad + bc + bd
b² + c² + 2bc = b² + bc + bc + c²
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 S₈ = 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  S₂: S₄ for the series 5 + 10 + 20 + ………

Solution :

Question 5(ii):

Find  S₄: S₈ for the series 4 + 12 + 36 + ……..

Solution :

Question 6(i):

Find the G.P if S₆ : S₃ = 126 : 1 and T₄ = 125

Solution :

Question 6(ii):

Find the G.P if S₁₀ : S₅ = 33 : 32 and T₅ = 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

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All Subject KSEEB Solutions For Class 10

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