# 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 3Exercise 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 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 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):

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

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

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 :

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 :

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 3rd term is 12 and last term is 106. Find the 29th term.

Solution :

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 :

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 :

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 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 :

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 Chapter 3Exercise 3.3

Question 1:

If Tn = 2n + 3, find S3

Solution :

;

Question 2:

If Tn = 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. Sn = 55
ii. Sn = 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 = 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 Chapter 3Exercise 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 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 :

Question 4(ii):

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

Solution :

Question 4(iii):

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

Solution :

Question 4(iv):

Solution :

Question 4(v):

Find the following 4th and 8th 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 5th hour.

Solution :

Question 8:

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

Solution :

Question 9:

If the 4th and 8th 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 3Exercise 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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