RBSE Class 10 Maths Important Questions Chapter 5 Arithmetic Progressions

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 5 Arithmetic Progressions Important Questions and Answers.

RBSE Class 10 Maths Chapter 5 Important Questions Arithmetic Progressions

Objective Type Questions—

Question 1.
The first term of an arithmetic progression is 2 and the common difference is 3, then the arithmetic progression will be—
(A) 2, 5, 8, 11, ….
(B) 2, 6, 18,54,….
(C) 2, 4, – 2, 3,….
(D) 2, 3, 3, 4,….
Answer:
(A) 2, 5, 8, 11, ….

Question 2.
The last term of the arithmetic progression 2, 5, 8, 11, …. of 16 terms will be—
(A) 47
(B) 48
(C) 49
(D) 46
Answer:
(A) 47

Question 3.
If first term of the arithmetic progression is 5 and the fifth term is 17, then the sum of the five terms will be—
(A) 55
(B) 56
(C) 57
(D) 58
Answer:
(A) 55

Question 4.
Which term of the series 72, 70, 68, 66, …..will be 40?
(A) 16th
(B) 15th
(C) 17th
(D)20th
Answer:
(C) 17th

Question 5.
The sum of 22 terms of the series a + b, a – b, a – 3b, will be—
(A) 22 (a – 20b)
(B) 22 (20b – a)
(C) 22 (20a – b)
(D) 22 (a – b)
Answer:
(A) 22 (a – 20b)

Question 6.
The 10th term of the series -4, -1, 2, 5, …… is—
(A) 23
(B) – 23
(C) 32
(D)-32
Answer:
(A) 23

Question 7.
The number of terms in the arithmetic pro¬gression 7, 10, 13,……, 43 is—
(A) 13
(B) 12
(C) 17
(D) 11
Answer:
(A) 13

Question 8.
If the first term of an arithmetic progres-sion is 5, last term is 45 and the sum of terms is 400, then the number of term will be—
(A) 8
(B) 10
(C) 16
(D) 20
Answer:
(C) 16

Very Short Answer Type Questions—

Question 1.
Which term of the arithmetic progression 3, 7, 11, 15,…. will be 499?
Solution:
a = 3, d = 7 – 3 = 4,
a_n = 499
Now, a_n = a + (n – 1) × d
⇒ 499 = 3 + (n – 1) × 4
⇒ z = n – 1
⇒ 124 = n – 1
∴ n = 124 + 1 = 125

Question 2.
How many terms of the arithmetic progression 18, 16, 14, …. be taken so that their sum be zero?
Solution:
Let n terms be taken
a = 18 , d = 16 – 18 = -2,
S_n = 0 (given)
S_n = \(\frac{n}{2}\)(2a + (n – 1)d)
0 = \(\frac{n}{2}\)[2 × 18 + (n – 1) × (- 2)]
⇒ 0 = 36 – 2n + 2
2n = 38 ∴ n = \(\frac{38}{2}\) = 19

Question 3.
Find the sum of 20 terms of the series
a + b, a – b, a – 3b
Solution:
Given
First term (A) = a + b
Common difference (d) = a – b – a – b
= -2 b
Now, S₂₀ = [2(a + b) + (20 – 1) × (-26)]
= 10(2a + 2b – 40b + 26)
= 10(2a- 36b)
= 20(a – 186)

Question 4.
Write the common difference and the next three terms of the A.P. 1,- 2,- 5, —8,….
Solution:
Common difference (d) = -2 – 1
= – 3
Next terms are : a₅ = – 11, a₆ = – 14,
a₇ = – 18

Question 5.
Find the sum to 9 terms of the arithmetic progression 8, 6, 4, …..
Solution:
First term (a) = 8
Common difference (d) = 6 – 8 = – 2
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₉ = \(\frac{9}{2}\)[2 × 8 + (9 – 1) × (- 2) ]
= \(\frac{9}{2}\)[16 – 16] = 0

Question 6.
If n arithmetic mean be inserted between two numbers a and b then find the value of the common difference d.
Solution:
d = \(\frac{b-a}{n+1}\)

Question 7.
Find those three numbers in arithmetic progression where sum is – 3 and product is 8.
Solution:
Three terms of the arithmetic progression will be
a + d, a, a – d
According to the question
a + d + a + a – d = – 3
a = – 1
(a + d) × a × (a – d) = 8
(- 1 + d) × (- 1)(- 1 – d) = 8
⇒ (1 + d)(d – 1) = 8
∴ d² – 1 = 8
d² = 9 or d= ±3
Hence more numbers will be 2, – 1,-4.

Question 8.
Write the formula of finding the general term of an arithmetic progression.
Solution:
a_n = a + (n – 1 )d

Question 9.
If an arithmetic progression fourth term is 64 and 54th term is – 61. Then what will be its common difference?
Solution:
According to the question
a₄ = a + 3d
⇒ 64 = a + 3d … (1)
Similarly a₅₄ = a + 53d
⇒ -61 =a + 53d …(2)
Operating equation (1) – equation (2)
125 = -50d
⇒ d = \(\frac{-125}{50}\) = \(\frac{-5}{2}\)

Question 10.
Write the formula of finding out the sum of n term of an arithmetic progression.
Solution:
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]

Question 11.
If the third term of an arithmetic progression is 18 and the seventh term is 30, then write its 15th term.
Solution:
According to the question
18 = a₃ = a + 2d …(1)
30 = a₇ = a + 6d … (2)
Equation (1) – Equation (2)
– 12 = – 4d ∴ d = 3
From equation (1)
18 = a + 2 × 3
⇒ 18 = a + 6
⇒ a = 12
a₁₅ = a + 14d
= 12 + 14 × 3
= 12 + 42 = 54

Question 12.
Write the arithmetic mean of a and b.
Solution:
Arithmetic mean = \(\frac{a+b}{2}\)

Question 13.
For what value of m will 10, m, – 2 be in arithmetic progression?
Solution:
We know that if three numbers a, b and c are in arithmetic progression,
then b = \(\frac{a+c}{2}\)
⇒ m = \(\frac{10+(-2)}{2}\)
= \(\frac{10-2}{2}\)
= \(\frac{8}{2}\) = 4
∴ m = 4

Question 14.
If a_n = 9 – 5n is the nth term of an A.P., then write the common difference.
Solution:
Given:
a_n = 9 – 5n
Putting n = 1, 2, 3, …..
a₁ = 9 – 5(1) = 4
a₂ = 9 – 5(2) = 9 – 10 = – 1
a₃ = 9 – 5(3) = 9 – 15 = – 6
Common difference for A.P.
(d) = a₂ – a₁ = a₃ – a₂
⇒ d = -1 – 4 = -5

Question 15.
Write next two terms of the A.P. 4, 1, -2, -5, …….
Solution:
The given progression is 4, 1, -2, -5, …..
Common difference of the progression
d = 1 – 4 = – 3
So 5th term (a₅) = – 5 + (-3)
= -5 – 3 = -8
6th term (a₆) = – 8 + (-3)
= – 8 – 3 =- 11
i.e., a₅ = – 8 and a₆ = – 11

Question 16.
Write the common difference of the A.P. 7, 5, 3, 1, – 1, – 3, ……..
Solution:
Common difference is – 2.
[∵ 5 – 7 = 3 – 5 = -2]

Short Answer Type Questions—

Question 1.
Find the 10th term of the A.P.: 2, 7, 12,….
Solution:
Here a = 2, d = 7 – 2 = 5 and n= 10
∵ a_n = a + (n – 1) d so
a₁₀ = 2+ (10 – 1) × 5
= 2 + 45 = 47
Hence the 10th term of the given A.P. is 47.

Question 2.
Determine the A.P. whose third term is 5 and 7th term is 9.
Solution:
According to the question
a₃ = a + (3 – 1) d = a + 2d = 5 … (1)
and a₇ = a + (7 – 1)d = a + 6d = 9 … (2)
Solving the pair of equation (1) and (2)
a = 3, d= 1
Hence the required A.P. is : 3, 4, 5, 6, 7,….

Question 3.
Find the last term of the Arithmetic Progression 7, 16, 25, 34,…. of 28 terms.
Solution:
Given : First term (a) = 7
Common difference (d) = 16 – 7 = 9
To determine—Last term i.e ., 28th term = a₂₈
Formula— a_n = a + (n – 1) d
a₂₈ = 7 + (28 – 1)9
= 7 + (27 × 9)
= 7 + 243 = 250
Hence, last term = 250.

Question 4.
Find the last term of the Arithmetic Progression 18, 16, 14, 12, …. of 17 terms.
Solution:
Given :
First term (a) = 18
Common difference (d) = 16 – 18 = -2
To determine—Last term, i.e., 17th term = a₁₇
Formula— a_n = a + (n – 1) d
a₁₇ = 18 + (17 – 1) (-2)
= 18 + [16 × (-2)]
= 18 – 32 = – 14
Hence, last term = (- 14)

Question 5.
How many numbers of two digits are divisible by 3?
Solution:
The list of two digit numbers divisible by 3 is : 12, 15, 18,. .. ., 99
It is an A.P. ? Yes, it is true. Here a = 12, d = 3 and a_n = 99.
∵ a_n = a + (n – 1) d
So 99 = 12 + (n – 1) × 3
or 87 = (n – 1) × 3
or n – 1 = \(\frac{87}{3}\) = 29
or n = 29 + 1 = 30
Hence, there are 30 two digit numbers divisible by 3.

Question 6.
Find the sum of 11 terms of an arithmetic progression when the sum of its first and the last term is 40.
Solution:
Given—
(1) Sum of first and last term (a + l) = 40
(2) Number of terms = 11
To determine—Sum of 11 terms (S₁₁)
S_n = \(\frac{n(a+l)}{2}\)
S₁₁ = \(\frac{11(40)}{2}\) = 11 × 20 = 220

Question 7.
Find the sum of the first 22 terms of A.P.: 8, 3, – 2,….
Solution:
Here a = 8, d = 3 – 8 = -5 and n = 22
We khow that
S = \(\frac{n}{2}\)[2a + (n – 1)d]
So, S = \(\frac{22}{2}\)[16 + 21 (-5)]
= 11 (16 – 105)
= 11 (-89)
= -979
So the sum of first 22 terms of the A.P. is – 979

Question 8.
If the nth term of an arithmetic progression is 2n + 5 then find the sum of seven terms of the progression.
Solution:
Since it is given that
a_n = 2n + 5
Putting n = 1, 2, 3 …….
a₁ = 2 × 1 + 5 = 2 + 5 = 7
a₂ = 2 × 2 + 5= 4 + 5 = 9
a₃ = 2 × 3 + 5 = 6 + 5 = 11
∴ The progression will be
7, 9, 11 …
∴ d = a₂ – a₁ = 9 – 7 = 2
∴ S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{7}{2}\)[2 × 7 + (7 – 1) × 21]
= \(\frac{7}{2}\)[14 + 12= \(\frac{7}{2}\) × 26
= 7 × 13 = 91

Question 9.
Find the 11th term of the arithmetic progression 2, 7, 12….
Solution:
Here a = 2, d = 7 – 2 = 5 and n= 10
∵ a_n = a + (n – 1 )d
So, a₁₁ = 2 + (11 – 1) × 5
= 2 + 50 = 52
Hence the 11th term of the given arithmetic progression is 52.

Question 10.
If sum of the first 12 terms of an A.P. is 468 and its common difference is 6 then find its 10th term.
Solution:
According to the question here S₁₂ = 468, n = 12 and common difference (d) = 6
∵ S_n = \(\frac{n}{2}\)[2a + (n – 1 )d]
∴ 468 = \(\frac{12}{2}\)[2a + (12- 1)6]
or 468 = 6 [2a + 66]
or 2a + 66 = \(\frac{468}{6}\) = 78
or 2a = 78 – 66 = 12
∴ a = \(\frac{12}{2}\) = 6
Now 10th term
a₁₀ = a + (n – 1) d
= 6 + (10 – 1)6
= 6 + 9 x 6
= 6 + 54 = 60
i.e. 10th term is 60.

Question 11.
How many terms of the A.P.: 17, 15, 13, must be taken, so that their sum is 81?
Solution:
Given A.P.,
17, 15, 13, …..
Here a = 17,d = 15 – 17 = – 2 and S_n = 81
∴ \(\frac{n}{2}\)[2a + (n – 1)d] = 81
or \(\frac{n}{2}\)[2(17) + (n – 1)(-2)] = 81
or \(\frac{n}{2}\)[34 – 2n + 2] = 81
or \(\frac{n}{2}\)[-2n + 36] = 81
or n[-n + 18] = 81
or -n² + 18n – 81 = 0
or n² – 18n + 81 = 0
or n² – 9n – 9n + 81 = 0
or n(n – 9) – 9(n – 9) =0
or (n – 9) (n – 9) = 0
or n – 9)² = 0
or n = 0

Long Answer Type Questions—

Question 1.
Which term of the A.P. : 21, 18, 15, ……… is – 81? Also, is any term of this A.P. zero? Give reason for your answer.
Solution:
Here, a = 21, d = 18 – 21 = – 3 and a_n = – 81.
We have to find n.
∵ a_n = a + (n – 1)d
so -81 = 21 + (n – 1) (-3)
or -81 = 24 – 3n
or – 105 = – 3n
so n = 35
Therefore the 35th term of the given A .P. is -81.
Now, we want to know if there is any n for which a_n = 0. If such an n is there. Then
21 + (n – 1) (- 3) = 0
or 3(n – 1) = 21
or n = 8
Hence, 8th term is 0.

Question 2.
Check whether 301 is a term of the list of numbers 5, 11, 17, 23, ……..
Solution:
We have :
a₂ – a₁ = 11 – 5 = 6
a₃ – a₂ = 17 – 11 = 6
a₄ – a₃ = 23 – 17 = 6
Since for k = 1, 2, 3, etc. a_k+1 – a_k is the same number, therefore, the given list of numbers is an A.P.
Here a = 5 and d = 6
Let nth term of this A.P. be 301.
We know that
a_n = a + (n – 1)d
So 301 = 5 + (n – 1) × 6
i.e. 301 = 6n – 1
Hence n = \(\frac{302}{6}\) = \(\frac{151}{3}\)
But n should be a positive integer. So, 301 is not a term of the given list of numbers.

Question 3.
If pth term of an arithmetic progression is \(\frac{1}{q}\) and qth term is \(\frac{1}{p}\), then prove that the sum of first pq terms should be \(\frac{1}{2}\)(pq + 1)
Solution:
Let the first term of the arithmetic progression be A and the common difference be d.
∴ According to the question
a_p = A + (p – 1 )d = \(\frac{1}{q}\) … (i)

Question 4.
If the sum of the first 14 terms of an AP 1050 and its first term is 10, find the 20th term.
Solution:
Here S₁₄ = 1050, n = 14 and a = 10
∵ S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
so 1050 = \(\frac{14}{2}\)[20+ 13d] = 140 + 91d
i.e. 910 = 91d
or d = 10
so a₂₀ = 10 +(20 – 1) × 10 = 200
i.e., 20th term is 200.

Question 5.
How many terms of the A.P.: 24, 21, 18,…. must be taken 20 that their sum is 78?
Solution:
Here a = 24, d = 21 – 24 = – 3 and S_n = 78.
We need to find n.
We know that
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
So, 78 = \(\frac{n}{2}\)[48 +(n – 1)(-3)
= \(\frac{n}{2}\)[51 – 3_n]
or 3n² – 51n + 156 = 0
or n² – 17n + 52 = 0
or n² – 4n – 13n + 52 = 0
or n(n- 4) – 13 (n – 4) = 0
or (n – 4) (n – 13) = 0
So, n = 4 or 13
Both these values of n are possible and can be accepted. So, the required number of terms is either 4 or 13.

Question 6.
Find the sum of:
(i) The first 1000 positive integers
(ii) The first n positive integers
Solution:
(i) Let S = 1 + 2 + 3 + ……. + 1000
Using the formula S_n = \(\frac{n}{2}\)(a + l) for the sum of the first n terms of an A.P.
S₁₀₀₀ = \(\frac{1000}{2}\)(1 + 1000)
= 500 × 1001
= 500500
So, the sum of the fi rst 1000 positive integers is 500500.

(ii) Let S_n= 1 + 2 + 3 + …….+ n
Here a = 1 and the last term l = n
Therefore, S_n = \(\frac{n(1+n)}{2}\)
or S_n = \(\frac{n(n+1)}{2}\)
Thus, the sum of first n positive integers is obtained by the formula
S_n = \(\frac{n(n+1)}{2}\)

Question 7.
A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. So these interests form an A.P.? If so find the interest at the end of 30 years making use of this fact.
Solution:
We know that the formula to calculate simple interest is given by
Simple Interest = \(\frac{P \times R \times T}{100}\)
So, the interest at the end of the first year
= ₹ \(\frac{1000 \times 8 \times 1}{100}\)
= ₹ 80
The interest at the end of the second year
= ₹ \(\frac{1000 \times 8 \times 2}{100}\)
= ₹ 160
The interest at the end of the second year
= ₹ \(\frac{1000 \times 8 \times 3}{100}\)
= ₹ 240
Similarly, we can calculate the interest at the end of the fourth year, fifth year, and so on so, the interest (in Rs.) at the end of first, second, third, years, respectively are :
80, 160, 240, …….
It is an AP on the difference between any two consecutive term is 80, i.e., d = 80. Also here a = 80.
So, to find the interest at the end of 30 years, we shall find a₃₀.
Now, a₃₀ = a + (30 – 1)d
= 80 + 29 × 80
= 2400
So, the interest at the end of 30 years will be ₹ 2400.

Question 8.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Solution:
The number of rose plants in the first, second, third, rows are respectively : 23, 21, 19, ……. 5. Therefore an A.P. Let the
number of rose be n.
So, a = 23, d = 21 – 23 = – 2 and a_n = 5
∵ a_n = a + (n – 1 )d
∴ 5 = 23 + (n – 1) (-2)
or – 18 = (n – 1) (-2)
or n = 10
So, there are 10 rose in the flower bed.

Question 9.
Find the sum of first 24 terms of the list of numbers where nth term is given by a_n = 3 + 2n.
Solution:
Since
a_n = 3 + 2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 × 2 = 7
a₃ = 3 + 2 × 3 = 9
The list of numbers thus obtained is 5, 7, 9, 11, ….
Here 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on so find. S₂₄ here n = 24, a = 5, d = 2
Therefore, S₂₄ = [2 × 5 + (24 – 1) × 2]
= 12 [10 + 46]
= 672
So, sum of first 24 terms of the list of numbers is 672.

Question 10.
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed numbers every year, find :
(i) the production in the 1st year
(ii) the production in the 10th year
(iii) the total production in first 7 years.
Solution:
(i) Since the production increase uniformly by a fixed number every year, therefore the number of TV sets manufactured in first, second, third, …….. year will form an A.P.
Let us denote the number of TV sets manufactured in the nth year by a_n.
Then, a₃ = 600
and a₇ = 700
or a + 2d = 600
and a + 6d = 700
Solving these, we get
d = 25 and a = 550.
Therefore, the number of TV sets produced in the first year in 550.

(ii) Now a₁₀ = a + 9 d
= 550 + 9 × 25
= 550 + 225
= 775
So, the number of TV sets produced in the 10th year is 775.

(iii) Also, S₇ = \(\frac{7}{2}\)[2 × 550 + (7 – 1) 25]
= \(\frac{7}{2}\)[1100+150]
= 4375
So, the total production of TV sets in first 7 years is 4375.

Question 11.
Find the sum of numbers divisible by 6 from 1 to 100.
Solution:
The list of positive integers divisible by 6 from 1 to 6 is 6, 12, 18, 24, 30, ….., 96
The above numbers are in AP sequence.
First term a = 6 and
Common difference d = 12 – 6 = 6
Last term a_n = 96
Sum of terms S_n = ?
Now a_n = a + (n – 1) × 6
96 = 6 + (n – 1) × 6
⇒ 96 – 6 = (n – 1) × 6
⇒ 90 = (n – 1) × 6
⇒ (n – 1) = \(\frac{90}{6}\) = 15
⇒ n = 15 + 1 = 16
We know that
S_n = \(\frac{n}{2}\)(a + a_n)
∴ S₁₆ = \(\frac{16}{2}\)(6 + 96)
= \(\frac{16 \times 102}{2}\)
= 8 × 102 = 816