KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Progressions

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. 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:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.1-2
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.1-2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.1-2s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.1-2s

Question 3:

If Tn = 5 – 4n, find the first three term

Solution :

Tn = 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 Tn = 2n2 + 5, find
i. T3
ii. T10

Solution :

i. Tn = 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 Tn = n2 – 1, find
i. Tn-1
ii. Tn+ 1

Solution :

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

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

Question 6:

If Tn = n2+4 and Tn = 200, find the value of ‘n’.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.1-6s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.1-6s

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
Tn+1 = Tn+ 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):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-1(ii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-1(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-1(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-1(ii)s

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

Tn+1 = Tn + 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. Tn = 2n – 1
ii. Tn = 5 – 4n
iii. Tn = 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. Tn = 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. Tn = 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 :

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

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-3(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-3(i)s

Question 3(ii):

In an A.P,
if a = -7, d = 5 find T12

Solution :

Tn = 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 :

Tn = 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, Tn = 76, find n

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-3(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-3(iv)s

Question 3(v):

In an A.P,
if d = -2, T22 = -39, find a

Solution :

Tn = 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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-3(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-3(vi)s

Question 4:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-4s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-4s

Question 5:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-5s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-5s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-7s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-7s

Question 8:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-8s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-8s

Question 9:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-9s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-9s

Question 10:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-10s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-10s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-11s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-11s

Question 12:

If the pth term of an A.P is q and the qth term is p, prove that the nth term equal to (p + q – n).

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-12s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-12s

Question 13:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-13s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.2-13s

KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Exercise 3.3

Question 1:

If Tn = 2n + 3, find S3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-1s;

Question 2:

If Tn = n2 – 1 find
i. S5
ii. S2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-2s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-2s

Question 3(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(i)s

Question 3(ii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(ii)s

Question 3(iii):

Find the sum of 3a, a, -a …… to a terms.
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iii)s

Question 3(iv):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iv)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(iv)s

Question 3(v):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(v)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(v)s

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(v)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(v)s

Question 3(vi):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-3(vi)s

Question 4:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-4
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-4

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-4s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-4s

Question 5(a):

For a sequence of natural numbers,
Find
i. S20
ii. S50 – S40
iii. S30 + S15

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-5(a)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-5(a)s

Question 5(b):

For a sequence of natural numbers,
Find n if
i. Sn = 55
ii. Sn = 15

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-5(b)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-5(b)s

Question 6:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-6s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-6s

Question 7:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-7s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-7s

Question 8:

Find the sum of all natural numbers between 1 and 201 which are divisible by 5.
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-8
If the smallest angle, ∠AOB = 20°, find ∠BOC and ∠COD.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-8s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-8s

Question 9:

<imgsrc=”https://c1.staticflickr.com/5/4280/35456652641_af244e7fe9_o.png” width=”251″ height=”56″ alt=”KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-9″>

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-9s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-9s

Question 10:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-10s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-10s

Question 11(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-11(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-11(i)s

Question 11(ii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-11(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-11(ii)s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-12s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-12s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-13s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-13s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-14s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-14s

Question 15:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-15
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-15

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-15s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-15s

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 = Sn 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 – Progressions -Ex-3.3-17s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.3-17s

KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Exercise 3.4

Question 1(i):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(i)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(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):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(ii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(ii)s

Question 1(iii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(iii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(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):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(iv)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(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):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(v)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(v)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(v)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(v)s

Question 1(vi):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(vi)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-1(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):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(i)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(i)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(i)s

Question 2(ii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(ii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-2(ii)s

Question 3:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-3
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-3s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-3s

Question 4:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-4
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-4

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-4s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.4-4s

KSEEB SSLC Solutions for Class 10 Maths Chapter 3 Exercise 3.5

Question 1:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-1
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-1

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-1s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-1s

Question 2(i):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-2(i)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-2(i)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-2(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-2(i)s

Question 2(ii):

Do as directed In the G.P 729, 243, 81, ……. Find T7
Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-2(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-2(ii)s

Question 3:

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

Solution :

T5 = 64 = ar4
 r = 2
T12= ?
Tn = arn-1
T12 = ar12-1
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 5th and 8th terms of the G.P. 3, 6, 12, …………

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(i)s

Question 4(ii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(ii)s

Question 4(iii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(iii)s

Question 4(iv):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(iv)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(iv)s

Question 4(v):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(v)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-4(v)s

Question 5(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(i)s

Question 5(ii):

Find the last term of the following series : 4, 42, 43 …… to 2n terms.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(ii)s

Question 5(iii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iii)s

Question 5(iv):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iv)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-5(iv)s

Question 6:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-6
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-6

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-6s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-6s

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 5th hour.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-7s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-7s

Question 8:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-8s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-8s

Question 9:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-9s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-9s

Question 10(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(i)s

Question 10(ii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(ii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(ii)s

Question 10(iii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(iii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(iii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-10(iii)s

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
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.5-11(iii)s
∴ 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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(i)s

Question 1(ii):

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

Solution :

&KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(ii)s

Question 1(iii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iii)s

Question 1(iv):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iv)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iv)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-1(iv)s

Question 2:

Find the first term of a G.P in which S8 = 510 and r = 2
Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-2s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-2s

Question 3(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-3(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-3(i)s

Question 3(ii):

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-3(ii)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-3(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-3(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-3(ii)s

Question 4:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-4s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-4s

Question 5(i):

Find  S2: S4 for the series 5 + 10 + 20 + ………

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-5(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-5(i)s

Question 5(ii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-5(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-5(ii)s

Question 6(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-6(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-6(i)s

Question 6(ii):

Find the G.P if S10 : S5 = 33 : 32 and T5 = 64

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-6(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-6(ii)s

Question 7:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-7s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-7s

Question 8(i):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-8(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-8(i)s

Question 8(ii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-8(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-8(ii)s

Question 8(iii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-8(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-8(iii)s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-9s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-9s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-10(a)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-10(a)

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 – Progressions -Ex-3.6-10(b)
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.6-10(b)

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(i)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(i)s

Question 1(ii):
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(ii)s

Question 1(iii):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(iii)s

Question 1(iv):

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-1(iv)s

Question 2:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-2s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-2s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-3s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-3s

Question 4:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-4s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-4s

Question 5:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-5
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-5

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-5s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-5s

Question 6:

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

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-6s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-6s

Question 7:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-7
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-7

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-7s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-7s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-8s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-8s

Question 9:

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-9
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-9

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-9s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-9s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-10s
KSEEB SSLC Solutions for Class 10 Maths – Progressions -Ex-3.7-10s

All Chapter KSEEB Solutions For Class 10 Maths

—————————————————————————–

All Subject KSEEB Solutions For Class 10

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