Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions

Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers

Extra Questions for Class 10 Maths Chapter 5 Arithmetic Progressions. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download NCERT Class 10 Maths Solutions to help you to revise complete syllabus and score more marks in your examinations.

Arithmetic Progressions Class 10 Extra Questions Very Short Answer Type

Question 1.
How many two digits numbers are divisible by 3? [CBSE 2019]
Answer:
Two digits numbers divisible by 3 are 12, 15, 18, …………, 99
Here, let a_n = 99
⇒ a + (n – 1 )d = 99
⇒ 12 + (n – 1)3 = 99
⇒ 3n + 9 = 99
⇒ n = 30
∴ 30 numbers of two digits are divisible by 3.

Question 2.
If the nth term of an A.P. is pn + q, find its common difference. [CBSE 2019 (C)]
Answer:
We know that d = t_n – t_{n – 1}
Here tn= pn + q
∴ t_{n – 1} = P(n – 1) + q = pn + q – p
∴ Common difference (d) = (pn + q) – (pn + q – 1) = P

Question 3.
For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P. ? [CBSE Outside Delhi 2016]
Answer:
(2k – 1) – (k + 9) = (2k + 7) – (2k -1) {∵ a₂ – a₁ = a₃ – a₂}
⇒ 2k – 1 – k – 9 = 2k + 7 – 2k + 1
⇒ k – 10 =8
⇒ k = 18

Question 4.
Find the 9th term from the end (towards the first term) of the A. P. 5, 9, 13, …………., 185. [CBSE 2016]
Answer:
Common difference, d, of the A.P. = 9 – 5 = 4
Last term ‘l’ of the A.P. = 185
nth term from the end of an A.P. = l – (n – 1 )d.
Thus, the 9th term from the end is
= 185 – (9 – 1)4 = 185 – 32 = 153

Question 5.
Find the Eleventh Term from the last of the A.P. 27, 23, 19, ……… – 65. [CBSE Sample Paper 2017]
Answer:
Here l (last term) = – 65
d = 23 – 27 = – 4
11^th term from last = l – (11 – 1) d
= – 65 – 10 × (- 4) = – 65 + 40 = – 25

Question 6.
In an A.P., if the common difference (d) = – 4 and the seventh term (a₇) is 4, then find the first term. [CBSE 2018]
Answer:
Å a_n = a + (n – 1)d
∴ a₇ = a + 6d
⇒ 4 = a + 6(- 4)
⇒ a = 4 + 24 = 28

Arithmetic Progressions Class 10 Extra Questions Short Answer Type-1

Question 1.
Find how many integers between 200 and 500 are divisible by 8. [CBSE 2017]
Answer:
Numbers between 200 and 500 which are divisible by 8 are 208, 216, 224 496, which is an A.P. with
a = 208, d = 8
Here a_n = 496
⇒ a + (n – 1)d = 496
⇒ 208 + (n – 1)8 = 496
⇒ 8n = 296
⇒ n = 37
∴ There are 37 such numbers.

Question 2.
Find the sum given below:
7 + 10 + 13 + …….. + 46
Or
If the 9^th term of an A.P. is zero, then show that its 29^th term is doable of its 19^th term. [CBSE 2019(C)]
Answer:
Given series is an A.P. with a = 7,d = 10 – 7 = 3
Let, t_n = 46
⇒ a + (n – 1 )d = 46
⇒ 7 + (n – 1)3 = 46
⇒ 3n + 4 = 46
⇒ 3n = 46 – 4
⇒ 3n = 42
⇒ n = 14
Required sum S₁₄ = \(\frac{14}{2}\) [7 + 46]
Using S_n = \(\frac{n}{2}\)(a + l)
= 7 × 53 = 371
Or
Given that t₉ = 0
⇒ a + 8d = 0 or a = – 8d …….. (i)
Now, t₂₉ = a + (29 – 1)d
= a + 28d
= – 8d + 28d [Using (i)]
= 20d ……… (ii)
Also, t₁₉ = a + (19 – 1)d
= a + 18d
= – 8d + 18d [Using (i)]
= 10d ………. (iii)
(ii) and (iii) ⇒ t₂₉ = 20d = 2(10d)
= 2t₁₉ Q.E.D.

Question 3.
Which term of the A.P.: 3, 15, 27, 39, ……… will be 120 more than its 21st term?
Or
If S_n, the sum of first n terms of an AJP. is given by S_n = 3n² – 4n, find the nth term.
Answer:
Here, a = 3, d = 15 – 3 = 12
Let nth term be 120 more than a₂₁
i.e., a_n = a₂₁ + 120
⇒ a + (n – 1)d = a + 20d + 120
⇒ 3 + (n – 1)12 = 3 + 20 × 12 + 120
⇒ 12n – 12 = 360 ⇒ 12n = 372
⇒ n = 31
i. e., 31st term is the required term.
Or
Given, S_n = 3n² – 4n
∴ S_{n – 1} = 3(n – 1)² – 4(n – 1)
= 3n² – 10n + 7
[Replacing n by (n – 1) to get S_{n – 1})
Since, a_n = S_n – S_{n – 1}
⇒ a_n = (3n² – 4n) – (3n² – 10n + 7)
= 6n – 7

Question 4.
How many terms of the A.P. 18, 16, 14, ………… be taken so that their sum is zero? [CBSE Delhi 2016]
Answer:
The given A.P. is 18, 16, 14,….
Here ‘a’ = 18
‘d’ = 16 – 18 = – 2
Let the sum of the first x terms of the A.P. be 0.
Sum of the first x terms = \(\frac{x}{2}\) [2 × 18 + (x – 1)(- 2)] = 0
⇒ \(\frac{x}{2}\) [36 + (-2x + 2)] = 0
⇒ x [36 – 2x + 2)] = 0
⇒ x (38 – 2x) = 0
Rejecting x = 0 as the number of terms cannot be 0.
∴ 38 – 2x = 0
⇒ x = 19
Thus, the sum of the first 19 terms of the A.P. is 0.

Question 5.
Find the sum of first 8 multiples of 3. [CBSE 2018]
Answer:
First 8 multiples of 3 are 3, 6, 9, 12, ……, 24
Reqd. sum (S) = 3 + 6 + 9 + ……. + 24
= 3(1 + 2 + 3 + … + 8)
= 3(S₈ of A.P. with a = 1, 1 = 8)
= 3 × \(\frac{8}{2}\) [∵ S_n of A.P. = \(\frac{n}{2}\) (a + l)]
= 3 × 4 × 9 = 108

Question 6.
The first and last term of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.
Answer:
a = 5, a_n = 45
⇒ a + (n – 1)d = 45
⇒ 5 + (n – 1)d = 45
⇒ (n – 1)d = 45 – 5
⇒ (n – 1)d = 40 ……………. (1)
Now, S_n = 400

Question 7.
Which term of the sequence 114, 109, 104, ……, is first negative term?
Answer:
Here, a = 114, d = 109 – 114 = – 5
a_n = 114 + (n – 1) (- 5) = 119 – 5n
Let a_n < 0
⇒ 119 – 5n < 0 ⇒ 5n > 119
⇒ n > 23\(\frac{4}{5}\)
⇒ n = 24, 25, 26, ……..
∴ 24th is first negative term.

Question 8.
Angles of a triangle are in AP if the smallest angle is 45°. Find the largest angle.
Answer:
Let angles are (a – d)°, a°, (a + d)° [d> 0]
Sum of angles = 180°
a – d + a + a + d = 180°
⇒ 3a = 180°
⇒ a = 60°
∴ Angles are 60 – d, 60°, 60 + d.
Smallest angle = (60 – d)° = 45°
⇒ d = 15°
∴ Largest angle = (60° + d)° = 60° +15° = 75°

Arithmetic Progressions Class 10 Extra Questions Short Answer Type-2

Question 1.
Find the sum of n terms of the series
\(\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)\) + ………. [CBSE Delhi 2017]
Answer:

Question 2.
If m^th term of an A.P. is \(\frac{1}{n}\) and n^th term is \(\frac{1}{m}\), then find the sum of its first mn terms [CBSE Delhi 2017]
Answer:

Question 3.
If the m^th term of an A.P. is \(\frac{1}{n}\) and n^th term is \(\frac{1}{m}\) then show that its (mn)^th term 1 [CBSE Delhi 2017, 2019(C)].
Answer:
Let a be first term and d be common difference of A.P.

Hence Proved

Question 4.
If the sum of first m terms of an A.P. is the same as the sum of its n terms, show that the sum of its first (m + n) terms is zero. [CBSE Delhi 2017]
Answer:
Let a, d be first term and common difference of A.P. respectively
a_p = q ⇒ a + (p – 1) d = q …………. (i)
a_q = p ⇒ a + (q – 1)d = p ………… (ii)
(i) – (ii) ⇒ (p – q)d = q – p ⇒ d = – 1
Put for d in (i) we get a = p + q – 1
a_n = a + (n – 1)d
= (p + q – 1) + (n – 1) (- 1) = p + q – n

Question 5.
If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P. [CBSE 2016]
Answer:
Let the first term and the common difference of the given AP be a and d, respectively.
Sum of the first 7 terms, S₇ = 49

⇒ a + 8d = 17
Subtracting (1) from (2), we get
5d = 10
⇒ d = 2
Substituting the value of d in (1),
We get, a = 1
Now, S_n = \(\frac{n}{2}\) [2a + (n – 1)d] = \(\frac{n}{2}\) [2 × 1 + 2(n – 1)]
= n(1 + n – 1) = n²
Therefore, the sum of the first n terms of the A.P. is n².

Question 6.
If the ratio of the sum of first n terms of two A.P/s is (7n + 1):(4n + 27), find the ratio of their mth terms. [CBSE Outside Delhi 2016] [All India 2017]
Answer:
Let a₁, d₁; a₂ d₂ be first term and common difference of two A.P.’s respectively.

∴ Required rate is (14n – 6) : (8m + 23)
Hence, in 10th week her savings will be ₹ 20.75.

Question 7.
Find the sum of all three digit natural numbers which leave the remainder 3 when divided by 5.
Answer:
The three digit numbers when divided by 5 leave the remainder 3 are 103, 108, 113, ….., 998.
This forms an AP with
a = 103, d = 108 – 103 = 5, last term = 998
Now, a_n = 998
⇒ 103 + (n – 1) 5 = 998
[∵ a_n = a + (n – 1)d]
⇒ (n – 1)5 = 998 – 103 = 895
5n = 895 + 5 = 900
⇒ n = 180
∴ The required sum = S₁₈₀
= \(\frac{180}{2}\) [103 + 998] [∵ S_n = \(\frac{n}{2}\) [a + a_n]]
= 90 (1101)
= 99090.

Question 8.
The sum of first 16 terms of an AP is 112 and sum of its next fourteen terms is 518. Find theAP.
Answer:
Let first term = a and common difference is = d.
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
⇒ S₁₆ = \(\frac{16}{2}\) [2a + (16 – 1 )d] = 112
⇒ 8 [2a + 15 d] = 112 ……………… (i)
⇒ 2a + 15d = 14
S₃₀ = S₁₆ + S₁₄
∴ S₃₀ = 112 + 518 = 630
⇒ \(\frac{30}{2}\) [2a + (30 – 1)d] = 630
⇒ 2a + 29d = 42 ……………… (ii)
Now, (i) – (ii) gives

⇒ d = 2
∴ From (i) we get
2a + 15 × 2 = 14
⇒ 2a = 14 – 30 = -16
⇒ a = – 8
∴ AP is – 8, -6, -4, -2, 0, ……..

Arithmetic Progressions Class 10 Extra Questions Long Answer Type 1

Question 1.
If the sum of the first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the stun of its first M terms. [CBSE 2019]
Answer:
Given, S₄ = 40 and S₁₄ = 280
If a, d be the first term and common difference of given A.P. Then
⇒ \(\frac{4}{2}\) [2a + (4 – l)d]
= 40, \(\frac{14}{2}\) [2a + (14 – 1 )d] = 280
⇒ 4a + 6d = 40, 14a + 91 d = 280
⇒ 2a + 3d = 20, ……….. (1)
2a + 13d = 40 ……….. (2)
Subtracting (2) from (1)
⇒ – 10d = – 20
⇒ d = 2
Put d = 2 in (1)
2a + 3(2) = 20 ⇒ a = 7
∴ S_n = \(\frac{n}{2}\) [2a + (n – 1)d] = \(\frac{n}{2}\) (2 × 7 + (n – 1)2]
= n[7 + (n – 1)]
= n(n + 6)
= n² + 6n

Question 2.
The first term of an A.P. is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the A.P. [CBSE 2019]
Answer:
Here, a = 3, a_n = 83 and S_n = 903
∵ a_n = a + (n – 1 )d
⇒ 83 = 3 + (n – 1)d
⇒ (n – 1 )d = 80 …….. (1)
Also, S_n = \(\frac{n}{2}\) [2a + (n – 1)d] (Using (1))
⇒ 903 = \(\frac{n}{2}\) [2 × 3 + (n – 1 )d
⇒ 903 = \(\frac{n}{2}\) (6 + 80)
⇒ 903 = 43n
⇒ n = 21
Putting n = 21 in (1), we get
⇒ (21 – 1 )d = 80
20 d = 80
⇒ d = 4

Question 3.
The ratio of the sums of first m and first terms of an A. P. is m² : n². Show that the ratio of its m”1 and nth terms is (2m – 1) : (2n – 1) [CBSE Delhi 2017]
Answer:
Let a, d be first term & common difference of A. P. respectively.

⇒ 2an + (mn – n)d = 2am + (mn – m)d
⇒ 2a(m – n) = (m – n)d
⇒ 2a = d (∵ m ≠ n)

Thus a_m : a_n = (2m – 1) : (2n – 1).
[Hence Proved]

Question 4.
If the sum of first m terms of an A.P. is the same as the sum of its n terms, show that the sum of its first (m + n) terms is zero. [CBSE Delhi 2017]
Answer:
Let a, d be first term and common difference of A.P. respectively
S_m = S_n = \(\frac{m}{2}\)[2a + (m – 1)d] = \(\frac{n}{2}\)[2a + (n – 1)d]
⇒ 2a(m – n) = – d(m + n – 1) (m – n)
⇒ 2a + d(m + n – 1) = 0 ……. (i)
S_{m + n} = \(\frac{m+n}{2}\) [2a + d(m + n – 1)] = 0
= \(\frac{m+n}{2}\) (0) [By (i)]
= 0
Proved

Question 5.
If the p^th term of an A.P. is \(\frac{1}{q}\) and q^th term is \(\frac{1}{p}\), prove that the sum of first pq terms of the A.P. is \(\left(\frac{p q+1}{2}\right)\). [CBSE Delhi 2017]
Answer:
Let a, d be first term and common difference of A.P. respectively.

Proved

Question 6.
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers. [CBSE 2018]
Answer:
Let the reqd. numbers be α – 3β, α – β, α + β, α + 3β, α – 3β + α – β + α + β + α + 3β = 32
4α = 32
α = 8
∴ Numbers are 8 – 3β, 8 – β, 8 + β, 8 + 3β

⇒ 64 × 15 – 9 × 15β² = 7 × 64 – 7β²
⇒ 960 – 135β² = 448 – 7β²
⇒ 128β² = 512
⇒ β² = \(\frac{512}{128}=\frac{32}{8}\) = 4
⇒ β = ± 2
When β = 2
Numbers are 2, 6, 10, 14
When β = – 2
Numbers are 14, 10, 6, 2.

Question 7.
If the sum of first 14 terms of an A.P. is 1050 and its first term is 10, find the 20th term.
Or
The first term of an A.P. is 5, the last term is 45 and sum is 400. Find the number of terms and the common difference. [CBSE SQP 2019-20 Basic]
Answer:
Let common difference bed;
∴ S₁₄ = 1050
⇒ \(\frac{14}{2}\) [2(10) + (n -1 )d] = 1050
[S_n = \(\frac{n}{2}\) (2a + (n – 1)d]
⇒ 7(20 + (14 – 1)d) = 1050
⇒ 20 + (13)d = 150
⇒ (13)d = 130
⇒ d = 10
a₂₀ = a + 19
⇒ 10 + 19(10) = 200
Or
a = 5
a_n = 45
⇒ a + (n – 1)d = 45
⇒ 5 + (n – 1)d = 45
⇒ (n – 1)d = 40 …………… (i)
⇒ S_n = 400
⇒ \(\frac{n}{2}\) (a + a_n) = 400
⇒ \(\frac{n}{2}\) (5 + 45) = 400
⇒ 50n = 800
⇒ n = 16 ……………… (ii)
(i) and (ii)
⇒ (16 – 1 )d = 40
⇒ d = \(\frac{40}{15}=\frac{8}{3}\)
∴ Number of terms = 16 and common difference = \(\frac{8}{3}\)

Arithmetic Progressions Class 10 Extra Questions HOTS

Question 1.
Find the sum of first 24 terms of the AP, a₁, a₂, a₃, ………, if it is known that a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225.
Answer:
We know that in an AP, the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.
∴ In an AP of 24 terms,
a₁ + a₂₄ = a₅ + a₂₀ = a₁₀ + a₁₅ ………… (i)
Now a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225
⇒ (a₁ + a₂₄) + (a₅ + a₂₀) + (a₁₀ + a₁₅) = 225
⇒ (a₁ + a₂₄) + (a₁ + a₂₄) + (a₁ + a₂₄) = 225 [Using (i)]
⇒ 3 (a₁ + a₂₄) = 225
⇒ a₁ + a₂₄ = 75 …….. (ii)
Now S₂₄ = \(\frac{24}{2}\) (a₁ + a₂₄)
[∵ S_n = \(\frac{n}{2}\) (a + l)]
= 12 (75)
= 900.

Question 2.
Find the sum of 2n terms of the series:
1² – 2² + 3² – 4² + 5² – 6² + ……..
Answer:
(1² – 2²) + (3² – 4²) + (5² – 6²) + ………….. to n terms
= – 3 – 7 – 11 ……. n terms
= – (3 + 7 + 11 + …… n terms)
= {\(\frac{n}{2}\) (2 × 3 + (n – 1) × 4)} = {\(\frac{n}{2}\) [4n + 2]}
= – n (2n + 1)

Question 3.
Find the sum of the series 3 + 5 + 7 + 6 + 9 + 12 + 9 +13 +17 + … to 3n terms.
Answer:

= (3 + 6 + 9 + ……. to n terms) + (5 + 9 + 13 + … n terms) + (7 + 12 + 17 + …….. n terms)
= (S_n of AP with a = 3, d = 3) + (S_n of AP with a = 5, d = 4) + (S_nof AP with a = 7, d = 5)
= \(\frac{n}{2}\) [2 × 3 + (n – 1) × 3] + \(\frac{n}{2}\) [2 × 5 + (n – 1) × 4] + \(\frac{n}{2}\) [2 × 7 + (n – 1) × 5]
= \(\frac{n}{2}\) [3n + 3] + \(\frac{n}{2}\) [4n + 6] + \(\frac{n}{2}\) [5n + 9]
= \(\frac{n}{2}\) [(3n + 3) + (4n + 6) + (5n + 9)]
= \(\frac{n}{2}\) [12n + 18] = n(6n + 9) = 6n² + 9n

Question 4.
Balls are arranged in rows to form an equilateral triangle. First row consists of one ball, the second two balls and so on. If 669 more balls are added, then all the balls can be arranged in the shape of a square and each of its side then contains 8 balls less than each side of the triangle. Find initial number of balls.
Answer:
Let there be n balls in each side of the triangle.
∴ Total numbers of balls in the triangle = Sum of no. of balls in n rows of triangle
= 1 + 2 + 3 + ….+ M
= S_n of AP with a = 1, d = 1

We are also given that,
Number of balls in each square = (n – 8)
Total balls in the square = (n – 8)²
According as given
\(\frac{n(n+1)}{2}\) + 669 = (n – 8)²
⇒ n2² + n + 1338 = 2(n² + 64 – 16n)
⇒ n² – 33n – 1210 = 0
⇒ n² – 55n + 22n – 1210 = 0
⇒ (n – 55) (n + 22) = 0
⇒ n = 55, – 22
[Rejecting – 22 as number of balls cannot be negative.]
∴ n = 55
∴ Initial number of balls
= \(\frac{n(n+1)}{2}\) = \(\frac{55 \times(55+1)}{2}\) = \(\frac{55 \times 56}{2}\) = 1540

Question 5.
If a₁, a₂, ……, a_n are in AP where a_i > 0 for all i, show that:

Answer:
Let d be the common difference of the given AP Then
a₂ – a₁ = a₃ – a₂ = …………. = a_n – a_{n – 1} = d
Rationalising the denominator of each term, we have
LHS

Question 6.
Which term of the sequence: 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), ……………. is the first negative term? [All India 2017]
Answer:

or 83 – 3n < 0
or 83 < 3n or n > \(\frac{83}{3}\) = 27\(\frac{2}{3}\)
or n ≥ 28
Hence, 28th term of the given sequence is the first negative term.

Multiple Choice Questions

Choose the correct option for each of the following:

Question 1.
The n^th term of the sequence -15, -5, 5, 15, ………….. is
(a) 10n + 25
(b) 10n – 25
(c) 10n – 5
(d) 2n – 5
Answer:
(b) 10n – 25

Question 2.
The 23rd term of an A.P. whose first two terms are – 2 and 5 is
(a) 154
(b) 165
(c) 156
(d) 152
Answer:
(d) 152

Question 3.
For an A.P. if d = – 3, n = 8, an = -14, then a is
(a) 5
(b) 6
(c) 8
(d) 7
Answer:
(d) 7

Question 4.
For an A.P. {a_n}, a₁₈ – a₁₀ = 72 difference of A.P. is
(a) 7
(b) 9
(c) 8
(d) none of these
Answer:
(b) 9

Question 5.
n^th term of an A.P. is given by t_n = 2 – 3n then S₂₅ is
(a) – 925
(b) 925
(c) 1125
(d) 525
Answer:
(a) – 925

Question 6.
If 8 times the 8^th term of m A.P. is equal to 11 times the 11^th term then a₁₉
(a) can’t be determined
(b) zero
(c) 27
(d) \(\frac{19}{5}\)
Answer:
(b) zero

Question 7.
The 10^th term from end of an A.P. 4, 9, 14, 19, ……………., 254 is
(a) 109
(b) 152
(c) 209
(d) 178
Answer:
(c) 209

Question 8.
The first negative term of the sequence 114, 109, 104, ……………… is
(a) a₂₃
(b) a₂₄
(c) a₂₁
(d) a₂₅
Answer:
(b) a₂₄

Question 9.
If the n^th term of A.P. is 3n – 5 then its common difference is
(a) – 3
(b) 3
(c) 6
(d) 5
Answer:
(b) 3

Question 10.
Which term of the A.P. 4, 9,14, …….. is 254?
(a) 219
(b) 291
(c) 229
(d) 209
Answer:
(d) 209

Question 11.
The next two terms of sequence √12, √27, √48, √75 are
(a) √108, √147
(b) √96, √105
(c) √108, √221
(d) √87, √147
Answer:
(a) √108, √147

Question 12.
The number of terms in the A.P. 3, 8, 13, 18, ………… 78 are
(a) 15
(b) 14
(c) 16
(d) 18
Answer:
(c) 16

Question 13.
The middle term of the sequence 2, 5, 8, 11, ……. 92 is
(a) 49
(b) 48
(c) 47
(d) 46
Answer:
(c) 47

Question 14.
If the sum of n terms of an A.P. is given by S_n = An + Bn², where A and B are constants, then the common difference of A.P. is
(a) 2B
(b) 2A
(c) A
(d) B
Answer:
(a) 2B

Question 15.
For an A.P. S_n = 3n² + 5. The number of term that equals 159 is
(a) 9
(b) 18
(c) 13
(d) 27
Answer:
(d) 27

Fill in the blanks

Question 1.
The famous mathematicians associated with finding the sum of the first 100 natural numbers is __________ .
Answer:
Gauss

Question 2.
Next term of sequence √8, √18, √32, ……… is __________ .
Answer:
√50

Question 3.
The value of k for which x, 2x + k, 3x + 6 are any three consecutive terms of A.P. is __________ .
Answer:
3

Question 4.
Middle term of A.P. 1, 8, 15 , ……………, 505 is __________ .
Answer:
253

Question 5.
0, _______ , – 8, – 12, _______ are in A.?.
Answer:
– 4, – 16

Question 6.
The total number of two digit numbers divisible by 8 is __________ .
Answer:
11

Question 7.
If the sum of first n even natural numbers is k times the sum of first n odd natural numbers then k equals __________ .
Answer:
\(\frac{(n+1)}{n}\)

Question 8.
The common difference of A.P. with 18 terms whose first term is a and last term is c is __________ .
Answer:
\(\frac{(c-a)}{17}\)

Question 9.
For an A.P. {a_n} the quantity a_n – a_{n – 1} is called __________ .
Answer:
common difference

Question 10.
For an A.P. (a_n) the quantity S_n – S_{n – 1} is __________ .
Answer:
n^th term

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths

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