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KSEEB SSLC Solutions for Class 10 Maths – Progressions

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

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.1 Q1

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.1 Q2

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.1 Q3

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q4

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q5

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q6

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q7

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q8

Question 4:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q9

Question 5:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q10

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q11

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q12

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q13

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q14

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q15

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q16

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.2 Q17

Exercise 3.3:

Question 1:

If Tn = 2n + 3, find S3

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q18;

Question 2:

If Tn = n2 – 1 find
i. S5
ii. S2

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q19

Question 3(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q20

Question 3(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q21

Question 3(iii):

Find the sum of 3a, a, -a …… to a terms.
KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q22

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q23

Question 3(iv):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q24

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q25

Question 3(v):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q26

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q27

Question 3(vi):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q28

Question 4:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q29

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q30

Question 5(a):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q31

Question 5(b):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q32

Question 6:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q33

Question 7:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q34

Question 8:

Find the sum of all natural numbers between 1 and 201 which are divisible by 5.
KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q35
If the smallest angle, ∠AOB = 20°, find ∠BOC and ∠COD.

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q36

Question 9:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q37

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q38

Question 10:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q39

Question 11(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q40

Question 11(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q41

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q42

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q43

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q44

Question 15:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q45

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q46

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.3 Q47

Exercise 3.4:

Question 1(i):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q48

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q49

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q50

Question 1(iii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q51

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q52

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q53

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q54

Question 1(vi):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q55

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q56

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q57

Question 2(ii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q58

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q59

Question 3:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q60

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q61

Question 4:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q62

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.4 Q63

Exercise 3.5:

Question 1:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q64

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q65

Question 2(i):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q66

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q67

Question 2(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q68

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q69

Question 4(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q70

Question 4(iii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q71

Question 4(iv):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q72

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q73

Question 4(v):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q74

Question 5(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q75

Question 5(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q76

Question 5(iii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q77

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q78

Question 5(iv):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q79

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q80

Question 6:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q81

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q82

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q83

Question 8:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q84

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q85

Question 10(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q86

Question 10(ii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q87

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q88

Question 10(iii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q89

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q90

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.5 Q91
∴ 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.

Exercise 3.6:

Question 1(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q92

Question 1(ii):

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

Solution :

&KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q93

Question 1(iii):
KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q94

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q95

Question 1(iv):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q96

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q97

Question 2:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q98

Question 3(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q99

Question 3(ii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q100

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q101

Question 4:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q102

Question 5(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q103

Question 5(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q104

Question 6(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q105

Question 6(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q106

Question 7:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q107

Question 8(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q108

Question 8(ii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q109

Question 8(iii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q110

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q111

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q112

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.6 Q113

Exercise 3.7:

Question 1(i):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q114

Question 1(ii):

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q115

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q116

Question 1(iii):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q117

Question 1(iv):

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q118

Question 2:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q119

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q120

Question 4:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q121

Question 5:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q122

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q123

Question 6:

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

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q124

Question 7:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q125

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q126

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q127

Question 9:

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q128

Solution :

KSEEB Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q129

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 Solutions for Class 10 Maths Chapter 3 Progressions Ex 3.7 Q130

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