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Here we are providing Pair of Online Education for Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Online Education Extra Questions for Class 10 Maths Arithmetic Progressions with Answers Solutions

Extra Questions for Class 10 Maths Chapter 5 Arithmetic Progressions with Solutions Answers

Arithmetic Progressions Class 10 Extra Questions Very Short Answer Type

Arithmetic Progression Class 10 Extra Questions Question 1.
Which of the following can be the n^th term of a_n AP?
4n + 3, 3n² + 5, n² + 1 give reason.
Solution:
4n + 3 because n^th term of a_n AP ca_n only be a linear relation in n as a_n = a + (n – 1)d.

Arithmetic Progression Extra Questions And Answers Class 10 Pdf Question 2.
Is 144 a term of the AP: 3, 7, 11, …? Justify your answer.
Solution:
No, because here a = 3 a_n odd number and d = 4 which is even. so, sum of odd and even must be odd whereas 144 is a_n even number.

Class 10 Arithmetic Progression Extra Questions Question 3.
The first term of a_n AP is p and its common difference is q. Find its 10^th term.
Solution:
210 = a + 9d = p + 99.

Ap Extra Questions Class 10 Question 4.
For what value of k: 2k, k + 10 and 3k + 2 are in AP?
Solution:
Given numbers are in AP
∴ (k + 10) – 2k = (3k + 2) – (k + 10)
⇒ -k + 10 = 2k – 8 or 3k = 18 or k = 6.

Ap Class 10 Extra Questions Question 5.
If a_n = 5 – 11n, find the common difference.
Solution:
We have a_n = 5 – 11n
Let d be the common difference
d = a_n+1 – a_n
= 5 – 11(n + 1) – (5 – 11n)
= 5 – 11n – 11 -5 + 11n = -11

Class 10 Ap Extra Questions Question 6.
If n^th term of a_n AP is \(\frac{3+n}{4}\) find its 8^th term.
Solution:

Arithmetic Progression Extra Questions Question 7.
For what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of AP?
Solution:
since 20 + 1, 13, 5p – 3 are in AP.
∴ second term – First term = Third term – second term
⇒ 13 – (2p + 1) = 5p – 3 – 13
⇒ 13 – 2p – 1 = 5p – 16
⇒ 12 – 2p = 5p – 16
⇒ -7p = – 28
⇒ p = 4

Extra Questions Of Ap Class 10 Question 8.
In a_n AP, if d = -4, n = 7, a, = 4 then find a.
Solution:
We know, a_n = a + (n – 1)d
Putting the values given, we get
⇒ 4 = a + (7 – 1)(-4) or a = 4 + 24
⇒ a = 28

Class 10 Maths Chapter 5 Extra Questions With Solutions Question 9.
Find the 25^th term of the AP: -5, \(\frac{-5}{2}\) , 0, \(\frac{-5}{2}\) ………
Solution:
Here, a = -5, b = –\(\frac{5}{2}\) – (-5) = \(\frac{5}{2}\)
We know,
a₂₅ = a + (25 – 1 )d
= (-5) + 24(\(\frac{5}{2}\)) = -5 + 60 = 55

Class 10 Maths Arithmetic Progression Extra Questions Question 10.
Find the common difference of an AP in which a₁₈ – a₁₄ = 32.
Solution:
Given, a₁₈ – a₁₄ = 32
⇒ (a + 17d) – (a + 13d) = 32
⇒ 17d – 13d = 32 or d = \(\frac{32}{4}\)

Important Questions Arithmetic Progression Class 10 With Solutions Pdf Question 11.
If 7 times the 7^th term of an AP is equal to 11 times its 11^th term, then find its 18^th term.
Solution:
Given, 7a₇= 11a₁₁
⇒ 7(a + 6d) = 11(a + 100) or 7a + 42d = 11a + 110d
⇒ 4a + 68d = 0 or a + 17d = 0
Now, a₁₈ = a + 17d = 0

Ap Questions Class 10 Question 12.
In an AP, if a = 1, a_n = 20, and s_n = 399, then find n.
Solution:
Given, An = 20
= 1 + (n – 1)d = 20
⇒ (n – 1) d = 19
s_n = \(\frac{n}{2}\) {2a + (n – 1)d}
⇒ 399 = \(\frac{n}{2}\){2 × 1 + 19}
⇒ \(\frac{399 \times 2}{21}\) = n
⇒ n = 38

Class 10 Maths Ch 5 Extra Questions Question 13.
Find the 9th term from the end (towards the first term) of the AP 5, 9, 13, …, 185.
Solution:
l = 185, d = 4
l₉ = l – (n – 1) d
= 185 – 8 × 4 = 153

Arithmetic Progressions Class 10 Extra Questions Short Answer Type 1

Ch 5 Maths Class 10 Extra Questions Question 1.
In which of the following situations, does the list of numbers involved to make an AP? If yes, give a reason.
(i) The cost of digging a well after every meter of digging, when it costs 150 for the first meter and rises by 50 for each subsequent meter.
(ii) The amount of money in the account every year, when 10,000 is deposited at simple interest at 8% per annum.
Solution:
(i) The numbers involved are 150, 200, 250, 300, …
Here 200 -150 = 250 – 200 = 300 – 250 and so on
∴ It forms a_n AP with a = 150, d = 50

(ii) The numbers involved are 10,800, 11,600, 12,400, …
which forms a_n AP with a = 10,800 and d = 800.

Extra Questions On Ap Class 10 Question 2.
Find the 20^th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
We have, last term = 1 = 253
And, common difference d = 2^nd term – 1^st term = 8 – 3 = 5
Therefore, 20^th term from end = 1 -(20 – 1) × d = 253 – 19 × 5 = 253 – 95 = 158.

Arithmetic Progression Class 10 Important Questions Question 3.
If the sum of the first p terms of an AP is ap² + bp, find its common difference.
Solution:
a_p = s_p – s_p-1 = (ap² + bp) -[a(p – 1)² + b(p – 1)]
= ap² + bp – (ap² + a – 2ap + bp – b)
= ap² + bp – ap² – a + 2ap – bp + b = 2ap + b-a .
= a₁ = 2a + b – a = a + b and a₂ = 4a + b – a = 3a + b
⇒ d = a₂ – a₁ = (3a + b) – (a + b) = 2a

Class 10 Maths Chapter 5 Extra Questions Question 4.
The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.
Solution:
Let the first term be ‘a’ and common difference be ‘d’.
Given, a = 5, T_n = 45, s_n = 400 .
Tn = a + (m – 1)d
⇒ 45 = 5 + (m – 1)d
⇒ (n – 1) d = 40 ………(i)
s_n = \(\frac{n}{2}\) (a + T_n)
⇒ 400 = \(\frac{n}{2}\) (5 + 45)
⇒ n = 2 × 8 = 16 substituting the value of n in (i)
⇒ (16 – 1)d = 40
⇒ d = \(\frac{40}{15}\) = \(\frac{8}{3}\)

Extra Questions Of Arithmetic Progression Class 10 Question 5.
Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Solution:
Natural numbers between 101 and 999 divisible by both 2 and 5 are 110, 120, … 990.
so, a1 = 110, d = 10, a_n = 990
We know, a_n = a₁ + (n – 1)d
990 = 110 + (n – 1) 10
(n – 1) = \(\frac{990-110}{10}\)
⇒ n = 88 + 1 = 89

Ap Important Questions Class 10 Question 6.
Find how many integers between 200 and 500 are divisible by 8.
Solution:
AP formed is 208, 216, 224, …, 496
Here, a_n = 496, a = 208, d = 8
a_n = a + (n – 1) d
⇒ 208 + (n – 1) x 8 = 496
⇒ 8 (n – 1) = 288
⇒ n – 1 = 36
⇒ n = 37

Extra Questions On Arithmetic Progression Class 10 Question 7.
The sum of the first n terms of an AP is 3n² + 6n. Find the nth term of this
Solution:
Given: s_n = 3n² +6n
s_n-1 = 3(n – 1)² + 6(n – 1)
⇒ 3(n² + 1 – 2n) + 6n – 6
⇒ 3m2 + 3 – 6n + 6n – 6 = 3n² – 3
The n^th term will be a_n
s_n = s_n-1 + a_n
a_n = s_n – s_n-1
⇒ 3n² + 6n – 3n² + 3
⇒ 6n + 3

Arithmetic Progression Class 10 Extra Questions With Solutions Question 8.
How many terms of the AP 18, 16, 14, …. be taken so that their sum is zero?
Solution:
Here, a = 18, d = -2, s_n = 0
Therefore, \(\frac{n}{2}\) [36 + (n – 1) (- 2)] = 0
⇒ n(36 – 2n + 2) = 0
⇒ n(38 – 2n) = 0
⇒ n = 19

Question 9.
The 4th term of an AP is zero. Prove that the 25^th term of the AP is three times its 11th term.
Solution:
∵ a₄ = 0 (Given)
⇒ a + 3d = 0
⇒ a = -3d
a₂₅ = a + 24d = – 3d + 24d = 21d
3a₁₁ = 3(a + 100) = 3(70) = 21d
∴ a₂₅ = 30₁₁
Hence proved.

Question 10.
If the ratio of sum of the first m and n terms of an AP is m² : n², show that the ratio of its mth and n^th terms is (2m – 1) : (2n – 1).
Solution:

Question 11.
What is the common difference of an AP in which a₂₁ – a₇ = 84?
Solution:
Given: a₂₁ – a₇ = 84
⇒ (a + 20d) – (a + 6d) = 84
⇒ 14d = 84
⇒ d = 6

Question 12.
For what value of n, are the n^th terms of two APs 63, 65, 67,… and 3, 10, 17,… equal?
Solution:
Let nth terms for two given series be a_n and a’_n
According to questions
a_n = a’_n
⇒ a + (n – 1)d = a + (n – 1)d’
⇒ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 5n = 65
⇒ n = 13.

Arithmetic Progressions Class 10 Extra Questions Short Answer Type 2

Question 1.
Which term of the AP: 3, 8, 13, 18, … , is 78?
Solution:
Let a_n be the required term and we have given AP
3, 8, 13, 18, …..
Here, a = 3, d = 8 – 3 = 5 and a_n = 78
Now, a_n = a + (n – 1)d
⇒ 78 = 3 + (n – 1) 5
⇒ 78 – 3 = (n – 1) × 5
⇒ 75 = (n – 1) × 5
⇒ \(\frac{75}{5}\) = n – 1
⇒ 15 = n – 1
⇒ n = 15 + 1 = 16
Hence, 16^th term of given AP is 78.

Question 2.
Find the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73.
Solution:
Let the first term be a and common difference be d.
Now, we have

Putting the value of d in equation (i), we have
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70
⇒ a = – 32
We have, a = -32 and d = 7
Therefore, a₃₁ = a + (31 – 1)d
⇒ a₃₁ = a + 30d
⇒ (-32) + 30 × 7
⇒ – 32 + 210
= a₃₁ = 178

Question 3.
An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term.
Solution:
Let a be the first term and d be the common difference.
since, given AP consists of 50 terms, so n = 50
a₃ = 12
⇒ a + 2d = 12 …(i)
Also, a₅₀ = 106
⇒ a + 490 = 106 … (ii)
subtracting (i) from (ii), we have
47d = 94
⇒ d = \(\frac{94}{47}\) = 2
Putting the value of d in equation (i), we have
a + 2 × 2 = 12
⇒ a = 12 – 4 = 8
Here, a = 8, d = 2
∴ 29th term is given by
a₂₉ = a + (29 – 1)d = 8 + 28 × 2
⇒ a₂₉ = 8 + 56
⇒ a₂₉ = 64

Question 4.
If the 8th term of an AP is 31 and the 15^th term is 16 more than the 11^th term, find the AP.
Solution:
Let a be the first term and d be the common difference of the AP.
We have, a₈ = 31 and a₁₅ = 16 + a₁₁
⇒ a + 7d = 31 and a + 14d = 16 + a + 10d
⇒ a + 7d = 31 and 4d = 16
⇒ a + 7d = 31 and d = 4
⇒ a + 7 × 4 = 31
⇒ a + 28 = 31
⇒ a= 3
Hence, the AP is a, a + d, a + 2d, a + 3d…..
i.e., 3, 7, 11, 15, 19, …

Question 5.
Which term of the arithmetic progression 5, 15, 25, …. will be 130 more than its 31^st term?
Solution:
We have, a = 5 and d = 10
∴ a₃₁ = a + 30d = 5 + 30 × 10 = 305
Let n^th term of the given AP be 130 more than its 31^st term. Then,
a_n = 130 + a₃₁
∴ a + (n – 1)d = 130 + 305
⇒ 5 + 10(n – 1) = 435
⇒ 10(n – 1) = 430
⇒ n – 1 = 43
⇒ n = 44
Hence, 44^th term of the given AP is 130 more than its 31^st term.

Question 6.
Which term of the progression \(20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}\), is the first negative term?
Solution:

Question 7.
Find the sum given below:
7 + 10 \(\frac{1}{2}\) + 14 +…+ 84
Solution:
Let a be the first term, d be the common difference and a_n be the last term of given AP.

Question 8.
In an AP: given l = 28, s = 144, and there are total 9 terms. Find a.
Solution:
We have, l = 28, s = 144 and n = 9
Now, l = a_n = 28
28 = a + (n – 1) d 28 = a + (9 – 1)d
⇒ 28 = a + 8d ……(i)
and S = 144
⇒ 141 = \(\frac{1}{2}\) [2a + (n – 1)d]
⇒ 144 = \(\frac{9}{2}\) [12a +(9 – 1) d]
\(\frac{144 \times 2}{9}\) = 2a + 8d
⇒ 32 = 2a + 8d
⇒ 16 = a + 4d … (ii)
Now, subtracting equation (ii) from (i), we get
4d = 12 or d = 3
Putting the value of d in equation (i), we have
a + 8 × 3 = 28
⇒ a + 24 = 28
⇒ a = 28 – 24
∴ a = 4.

Question 9.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Let sum of n terms be 636.
s_n = 636, a = 9, d = 17 – 9 = 8
⇒ \(\frac{n}{2}\)[2a + (n – 1) d] = 636
⇒ \(\frac{n}{2}\)[2 x 9 + (n – 1) × 8] = 636
⇒ \(\frac{n}{2}\) × 2[9+ (n – 1) 4] = 636
⇒ n[9 + 4n – 4] = 636
⇒ n[5 + 4n] = 636
⇒ 5n + 4n² = 636
⇒ 4n² + 5n – 636 = 0

Thus, the sum of 12 terms of given AP is 636.

Question 10.
How many terms of the series 54, 51, 48 ……. be taken so that, their sum is 513? Explain the
double answer.
Solution:
Clearly, the given sequence is an AP with first term a = 54 and common difference d = -3. Let
the sum of n terms be 513. Then,
s_n = 513
⇒ \(\frac{n}{2}\) {2a + (n – 1) d} = 513
⇒ \(\frac{n}{2}\) (108 + (n – 1) – 3) = 513
⇒ n[108 – 3n + 3) = 1026
⇒ -3n² + 111n = 1026
⇒ n² – 37n + 342 = 0
⇒ (n – 18) (n – 19) = 0
⇒ n = 18 or 19
Here, the common difference is negative. so, 19th term is given by
a₁₉ = 54 + (19 – 1) × – 3 = 0
Thus, the sum of 18 terms as well as that of 19 terms is 513.

Question 11.
The first term, common difference and last term of an AP are 12, 6 and 252 respectively. Find the sum of all terms of this AP.
Solution:
We have, a = 12, d = 6 and l = 252
Now, l = 252
⇒ a_n = 252
= l = a + (n – 1)d
⇒ 252 = 12 + (n – 1) × 6
⇒ 240 = (n – 1) × 6
⇒ n – 1 = 40 or n = 41
Thus, S_n = \(\frac{n}{2}\)(a + l)
⇒ S₄₁ = \(\frac{41}{2}\)(12 + 252) = \(\frac{41}{2}\) (264) = 41 × 132 = 5412

Question 12.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
We have, S₇ = 49
⇒ 49 = \(\frac{7}{2}\)[2a + (7 – 1) × d]
⇒ 49 × \(\frac{2}{7}\) = 2a + 6d
⇒ 14 = 2a + 6d
⇒ a + 3d = 7
and S₁₇ = 289
⇒ 289 = \(\frac{17}{2}\) [2a + (17 – 1)d]
⇒ 2a + 16d = \(\frac{289 \times 2}{17}\) = 34
⇒ a + 8d = 17
Now, subtracting equation (i) from (ii), we have
5d = 10 ⇒ d = 2
Putting the value of d in equation (i), we have
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
Here, a = 1 and d = 2
Now, s_n = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 × 1 + (n – 1) × 2]
= \(\frac{n}{2}\)[2 + 2n – 2]
= \(\frac{n}{2}\) × 2n = n²

Question 13.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Question 14.
If the seventh term of an AP is a and its ninth term is, find its 63^rd term.
Solution:

Question 15.
The sum of the 5^th and the 9^th terms of an AP is 30. If its 25^th term is three times its 8^th term, find the AP.
Solution:
a₅ + a₉ = 30
⇒ (a + 4d) + (a + 8d) = 30 = 2a + 12d = 30
⇒ a + 6d = 15
a = 15 – 6d …(i)
a₂₅ = 3a₈
⇒ a + 24d = 3(a + 7d)
á + 24d = 3a + 21d
⇒ 2a = 3d
Putting the value of a form (i), we have
2(15 – 6d) = 3d
⇒ 30 – 12d = 3d
⇒ 15d = 30
⇒ d = 2
So, a = 15 – 6 × 2 = 15 – 12 [From equation (i)]
⇒ a = 3
The AP will be 3, 5, 7, 9….

Question 16.
The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28^th term of this AP.
Solution:
sum of first seven terms,

Question 17.
If the ratio of the sum of first n terms of two AP’s is (7n + 1): (4n + 27), find the ratio of their mth terms.
Solution:

Question 18.
Find the sum of the following series :
5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + … + (-5) + 81 + (-3)
Solution:
The series can be rewritten as,
(5 + 9 + 13 + … +81) + (41 + (-39) + (-37) + … + (-5) + (-3))
For the series 5 + 9 + 13 + … 81
a = 5, d = 4 and a_n = 81
nth term = a + (n – 1)d = a_n
⇒ 5+ (n – 1)4 = 81
⇒ 4n = 80
⇒ n =
20 sum of 20 terms for this series.
s_n = \(\frac{20}{2}\) (5 + 81) = 860 …(i)
[∵ s_n = \(\frac{n}{2}\) (a + a_n)]
For the series (41) + (-39) + (-37)… + (-5) + (-3)
a = 41, d = 2 and and = -3
nth term = a + (n – 1)d = a_n.
⇒ -41 + (n – 1)2 = -3
⇒ 2n = 40
⇒ n = 20
sum of 20 terms for this series
s_n = \(\frac{20}{2}\) (-41 – 3) = – 440 …(ii)
By adding (i) and (ii), we get
sum of series = 860 – 440
= 420

Question 19.
Find the sum of all two digit natural numbers which are divisible by 4.
Solution:
Here a = 12, d = 4, a_n = 96
The formula is a_n = a + (n – 1)d
Therefore 96 = 12 + (n – 1) × 4
⇒ 96 = 8 + 4n
⇒ n = \(\frac{88}{4}\)
⇒ n = 22
Apply the formula for sum,
s_n = \(\frac{n}{2}\) [2a + (n – 1)d]
Hence, s₂₂ = 11/24 + 21 × 4] = 11[24 + 84]
= 11 × 108 = 1188.

Arithmetic Progressions Class 10 Extra Questions Long Answer Type

Question 1.
The sum of the 4^th and 8^th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.
Solution:
We have, a₄ + a₈ = 24
⇒ a + (4 – 1)d + a + (8 – 1) d = 24
⇒ 2a + 3d + 7d = 24
⇒ 2a + 10d = 24
⇒ 2(a + 5d) = 24
∴ a + 5d = 12
and, a₆ + a₁₀ = 44
⇒ a + (6 – 1)d + a + (10 – 1) d = 44
⇒ 2a + 5d + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22
subtracting (i) from (ii), we have
2d = 10
∴d = \(\frac{10}{2}\) = 5
Putting the value of d in equation (i), we have
a + 5 × 5 = 12
⇒ a = 12 – 25 = -13
Here, a = -13, d = 5
Hence, first three terms are
-13, -13, + 5, -13 + 2 × 5 i.e., -13, -8, -3

Question 2.
The sum of the first n terms of an AP is given by s_n = 3n² – 4n. Determine the AP and the 12^th term.
Solution:
We have, s_n = 3n² – 4n …(i)
Replacing n by (n – 1), we get
s_n-1 = 3(n – 1)² – 4(n – 1) ….(ii)
We know, .
a_n = s_n – s_n-1 = {3n² – 4n} – {3(n – 1)² – 4(n – 1)}.
= {3n² – 4n} – {3n² + 3 – 6n – 4n + 4}
= 3n² – 4n – 3n² – 3 + 6n + 4n – 4 = 6n – 7
so, nth term a_n = 6n – 7
To get the AP, substituting n = 1, 2, 3… respectively in (iii), we get
a₁ = 6 × 1-7 = -1,
a₂ = 6 × 2 – 7 = 5
a₃ = 6 × 3 – 7 = 11,…
Hence, AP is – 1,5, 1:1, …
Also, to get 12th term, substituting n = 12 in (iii), we get
a₁₂ = 6 × 12 – 7 = 72 – 7 = 65

Question 3.
Divide 56 into four parts which are in AP such that the ratio of product of extremes to the product of means is 5 : 6.
Solution:
Let the four parts be a – 3d, a-d, a + d, a + 3d.
Given, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56

Question 4.
In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.
Solution:
Let ‘a’ be the first term and ‘d be the common difference.

Question 5.
If s, denotes the sum of the first n terms of an AP, prove that s₃₀ = 3 (s₂₀ – s₁₀).
Solution:

Question 6.
A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief?
Solution:
Let total time be n minutes
Total distance covered by thief = 100 n metres
Total distance covered by policeman = 100 + 110 + 120 + … + (n – 1) terms
∴ 100m = \(\frac{n-1}{2}\) [100(2) + (n – 2)10]
⇒ 200n = (n – 1)(180 + 10n)
⇒ 10²– 30n – 180 = 0
⇒ n² – 3n – 18 = 0
⇒ (n-6) (n + 3) = 0
⇒ n = 6
Policeman took (n – 1) = (6 – 1) = 5 minutes to catch the thief.

Question 7.
The houses in a row are numbered consecutively from 1 to 49. show that there exists a value of X such that sum of numbers of houses preceeding the house numbered X is equal to sum of the numbers of houses following X. Find value of X.
Solution:
The numbers of houses are 1, 2, 3, 4……….49.
The numbers of the houses are in AP, where a = 1 and d = 1
sum of n terms of an AP = \(\frac{n}{2}\)[2a + (n – 1)d]
Let X^th number house be the required house.
sum of number of houses preceding X^th house is equal to s_x-1 i.e.,

since number of houses is positive integer,
∴ X = 35

Question 8.
If the ratio of the 11^th term of an AP to its 18th term is 2:3, find the ratio of the sum of the first five terms to the sum of its first 10 terms.
Solution:

Arithmetic Progressions Class 10 Extra Questions HOTS

Question 1.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are
8, 16, 24, … 120
Clearly, these numbers are in AP with first term a = 8 and common difference, d = 16 – 8 = 8

Question 2.
Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder.
Solution:
Two digit natural numbers which when divided by 3 yield 1 as remainder are:
10, 13, 16, 19, …, 97, which forms an AP.
with a = 10, d = 3, a_n = 97
a_n = 97 = a + (n – 1) d = 97
or 10 + (n – 1)3 = 97
⇒ (n – 1) = \(\frac{87}{3}\) = 29
⇒ n = 30
Now, s₃₀ = [2 × 10 + 29 × 3) = 15(20 + 87) = 15 × 107 = 1605

Question 3.
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the prizes be a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
Therefore, the sum of prizes is
a + 60 + a + 40 + a + 20 + a ta – 20 + a – 40 +a – 60 = 700
= 7a = 700
⇒ a = \(\frac{700}{7}\) = 100
Thus, the value of seven prizes are
100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60
i.e., ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹340

Question 4.
If the mth term of an AP is \(\frac{1}{n}\) and nth term is \(\frac{1}{m}\), then show that its (mn)th term is 1.
Solution:
Let a and d be the first term and common difference respectively of the given AP. Then
a_m = a + (m – 1) d
⇒ a + (m – 1)d = \(\frac{1}{n}\) …… (i)

Question 5.
If the sum of m terms of a_n AP is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero.
Solution:
Let a be the first term and d be the common difference of the given AP.
Then, s_m = s_n

Question 6.
The ratio of the sums of m and n terms of a_n AP is m² : n². show that the ratio of the mth and nth terms is (2m – 1): (2n – 1).
Solution:
Let a be the first term and d the common difference of the given AP. Then, the sums of m and n terms are given by

Question 7.
The sum of n, 2_n, 3_n terms of a_n AP are s₁, s₂, and s₃ respectively. Prove that s₃ = 3(s₂ – s₁).
Solution:
Let a be the first term and d be the common difference of the AP

Question 8.
If a², b², c², are in AP, prove that \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in AP.
Solution:
since a², b², c², are in AP …(i)

Here we are providing Online Education for A Visit to Cambridge Extra Questions and Answers Class 8 English Honeydew, Extra Questions for Class 8 English was designed by subject expert teachers.

You can refer to NCERT Solutions for Class 8 English honeydew Chapter 7 A Visit to Cambridge to revise the Questions and Answers in the syllabus effectively and improve your chances of securing high marks in your board exams.

Online Education for A Visit to Cambridge Extra Questions and Answers Class 8 English Honeydew

A Visit to Cambridge Extra Questions and Answers Short Answer Type

A Visit To Cambridge Extra Questions Class 8 Question 1.
Who was Stephen Hawking?
Answer:
Stephen Hawking was one of the greatest scientists of our time. He suffered from a form of paralysis that confined him to a wheelchair, and allowed him to ‘speak’ only by punching buttons on a computer, which spoke for him in a machine-like voice.

A Visit To Cambridge Class 8 Extra Questions Answers Question 2.
Who is Firdaus Kanga?
Answer:
Firdaus Kanga is a writer and journalist who lives and works in Mumbai. Kanga was born with ‘brittle bones’ that tended to break easily when he was a child.

Visit To Cambridge Extra Questions Class 8 Question 3.
Who received the phone at Stephen Hawking’s house and what were the conversations?
Answer:
The author called to Stephen Hawking’s house. There was his assistant on the line and the author told him that he wants to meet Mr. Hawking. Even ten minutes would do “Half an hour”. Then he said “from three-thirty to four”.

A Visit To Cambridge Extra Question Answer Class 8 Question 4.
What was the thing that makes author feel stronger?
Answer:
The only thing that makes you stronger is somebody like you. It seems like achieving something huge. This makes him feel stronger.

Extra Questions Of A Visit To Cambridge Class 8 Question 5.
What did Mr. Hawking reply when he asked “lot of people seem to think that disabled people are chronically unhappy”?
Answer:
When he asked “A lot of people seem to think that disabled people are chronically unhappy”. About after three minutes later, he responded, that “I find it amusing when people patronize me”.

A Visit To Cambridge Class 8 Extra Questions  Question 6.
“A first glimpse of Hawking is shocking”. Why?
Answer:
A first glimpse of Hawking is shocking, because he was like a still photograph – as if all those pictures of him in magazines and newspapers have turned 3 – Dimensional.

A Visit To Cambridge Short Question Answer Class 8 Question 7.
Why was author watching his wrist?
Answer:
He shifted his chair or turned his wrist to watch the time. He wanted to make every one of thirty minutes count.

A Visit To Cambridge Question Answer Class 8 Question 8.
What did Hawking reply when author asked “Is there any advice you can give disabled people, something that might help make life better”?
Ans.
Hawking replied that they should concentrate on what they are good at and according to him things I like the disabled Olympics are a waste of time.

A Visit To Cambridge Extracts Class 8 Question 9.
The author asked Hawking that he had been an inspiration beyond clinched for me, what Hawking replied to it?
Answer:
Hawking said “no”, according to him if his body is like claustrophobic room & the walls are growing narrower day by day, if doesn’t do much good to know that there are people outside smiling with admiration to see him breathing still.

Extra Questions From A Visit To Cambridge Class 8 Question 10.
The author had got appointment only for half an hour then why did he stay there for some more time?
Answer:
He had got appointment only for half an hour but when he was going to back his way Stephen stopped him and said, “Have some tea and he can see his garden”. So he stayed there for some more time.

A Visit To Cambridge Extra Questions And Answers Class 8 Question 11.
Describe the environment of Stephen’s Garden.
Answer:
The garden was as big as a park, but Stephen Hawking covered every inch, rumbling along motorized wheelchair. The environment of garden was entirely different.

Extra Questions On A Visit To Cambridge Class 8 Question 12.
Why was there silence in the garden?
Answer:
They cannot talk very much in the garden, because the sun made them silent, the letters on his screen disappeared because of the sunlight. So there were silence in the garden.

A Visit To Cambridge Extract Based Questions Class 8 Question 13.
How did the author say good bye to Mr. Hawking?
Answer:
The author didn’t know what to do at the time of leaving him. He couldn’t kiss him or cry. So he touched * his shoulder and wheeled out in the summer evening.

Class 8 English A Visit To Cambridge Extra Questions Question 14.
Describe the emotions of the author at the end of the chapter.
Answer:
He was in the feeling of embodiment of his bravest self. He thought Stephen Hawking might be waving at him though he wasn’t. At the end, he said his journey is over and he had a great interview of Mr. Hawking.

Extra Questions A Visit To Cambridge Class 8 Question 15.
(i) Did the prospect of meeting Stephen Hawking make the writer nervous?
(ii) Did he at the same time feel very excited? If so, why?
Answer:
(i) Yes, the prospect of meeting Stephen Hawking made the writer nervous. Stephen Hawking was the greatest scientist on the earth. He was the author of the world-renowned book “A brief History of Time”. He was doubtful whether he would be allowed to meet him even for ten minutes or not. Also, he was fed up with people asking him to be brave.

(ii) Yes, he felt very excited at the same time because Stephen Hawking was totally paralysed, still, he had made great achievement. This got him the strength to do still better.

A Visit To Cambridge Class 8 Question Answers  Question 16.
Guess the first question put to the scientist by the writer.
Answer:
The writer’s first question might be about Hawking’s disability and how he had accepted it.

Question 17.
Stephen Hawking said, “I’ve had no choice.” Does the writer think was a choice? What was it?
Answer:
The writer thought that there was a choice. Stephen Hawking could have chosen to leave everything, be sad and depressed. He could have sulked. However, he chose to live creatively, knowing the reality of his disintegrating body.

Question 18.
“I could feel his anguish”. What could be the anguish?
Answer:
The auguish was that he found it very difficult to express himself with the right words on his computer. He felt frustrated. He could not express himself freely although ideas were floating in his mind.

Question 19.
What endeared the scientist to the writer so that he said he was looking at one of the most beautiful men in the world?
Answer:
When the writer asked the scientist if he felt annoyed when someone like the writer went and disturbed him in his work, then scientist answered ‘yes’ with his one-way smile. This endeared the writer as he knew without being sentimental or silly he was looking at one of the most beautiful man in the world.

Question 20.
Read aloud the description of the beautiful’ man. Which is the most beautiful sentence in the description?
Answer:
The sentences describing the inner glow of Hawking’s personality which makes his physical looks irrelevant is probably one of the most beautiful descriptions of beauty.

Question 21.
(i) If ‘the lantern’ is the man, what would its ‘walls’ be?
(ii) What is housed within the thin walls?
(iii) What general conclusion does the writer draw from this comparison?
Answer:
(i) The walls of the lantern in Hawking’s case were his skeleton like physical structure.
(ii) The glow, the external soul was housed within the thin walls of his body.
(iii) The writer draws conclusion that the eternal soul is more important than the body.

Question 22.
What is the scientist’s message for the disabled?
Answer:
The scientist Stephen Hawking’s message for the disabled is they should concentrate on what they are good at. They should make the best use of their talent and thank God.

Question 23.
Why does the writer refer to the guitar incident? Which idea does it support?
Answer:
The scientist Stephen Hawking said that Olympics for the handicapped or disabled people are waste of time. The writer agreed with him. He remembered the years which he tried to play a big guitar. He felt defeated. So he destroyed it one night. He supports Stephen Hawking’s idea that the disabled should only concentrate on what they are good at.

Question 24.
The writer expresses his great gratitude to Stephen Hawking. What is the gratitude for?
Answer:
The writer expressed his gratitude to Stephen Hawking because he had been an inspiration for him. He saw Stephen as the embodiment of his bravest self. He felt that if he had been as brave as Stephen, he would have achieved a lot. He felt he was moving towards the embodiment that he had believed in for many years. That is why he expressed his great gratitude to him as he had made him realised what great heights he could reach.

A Visit to Cambridge Extra Questions and Answers Long Answer Type

Question 1.
The narrator called him ‘a beautiful man’. How does he define beauty?
Answer:
The narrator was inspired by the achievements of Stephen Hawking whose creativity is beyond measures. His never ‘giving up’ attitude attracted him the most. He could be associated with the difficulties faced by Stephen as he was also confined to a wheel chair. So it is rightly said that beautiful is the one, who do things beautifully and Stephen Hawkings contributed so much that people will also be inspired and awestruck by his achievements.

Question 2.
What is Stephen’s idea behind saying- ‘They should concentrate on what they are good at’?
Answer:
Stephen Hawking was a differently able person who overcame him deformity. He was one of the greatest scientists of the present era. He had the idea to achieve what one is good at. Rather than justifying or proving to others, he wanted challenged person to accept the way they are. Because they do have potential and focus to achieve much more than any other normal people. He did not appreciate the things like the disable Olympics as they are a wastage of time to him.

A Visit to Cambridge Extra Questions and Answers Reference to Context

Passage 1

It was on a walking tour through Cambridge that the guide mentioned Stephen Hawking, ‘poor man, who is quite disabled now, though he is a worthy successor to Issac Newton, whose chair he has at the university.’ And I started, because I had quite forgotten that this most brilliant and completely paralysed astrophysicist, the author of ‘A Brief History of Time’, one of the biggest best-sellers ever, lived here.

When the walking tour was done, I rushed to a phone booth and, almost tearing the cord so it could reach me outside, phoned Stephen Hawking’s house. There was his assistant on the line and I told him I had come in a wheelchair from India (perhaps he thought I had propelled myself all the way) to write about my travels in Britain. I had to see Professor Hawking – even ten minutes would do. “Half an hour,” he said. “From three-thirty to four.”

Question 1.
Why did the guide call Stephen Hawking a ‘poor man’?
Answer:
The guide called Stephen Hawking a ‘poor man’ because he was a differently-abled man.

Question 2.
What did ‘I’ speak about Stephen Hawking?
Answer:
He said that Stephen Hawking is the most brilliant and completely paralysed astrophysicist, author of a best seller ‘A brief History of Time’ lived in Cambridge.

Question 3.
Why had he ‘almost tearing the cord’?
Answer:
He rushed to a phone booth to take an appointment with Stephen Hawking and as he was on a wheel chair, he found it difficult to reach to the telephone.

Question 4.
What time slot was given to him?
Answer:
He was given half an hour from three thirty to four to meet Stephen Hawking.

Question 5.
Who is astrophysics?
Answer:
Astrophysics is the branch of astronomy that employs the principles of physics and chemistry “to ascertain the nature of the heavenly bodies, rather than their positions or motions in space.”

Passage 2

“I haven’t been brave,” said his disembodied computer-voice, the next afternoon. “I’ve had no choice.” Surely, I wanted to say, living creatively with the reality of his disintegrating body was a choice? But I kept quiet, because I felt guilty every time I spoke to him, forcing him to respond. There he was, tapping at the little switch in his hand, trying to find the words on his computer with the only bit of movement left to him, his long, pale fingers. Every so often, his eyes would shut in frustrated exhaustion. And sitting opposite him I could feel his anguish, the mind buoyant with thoughts that came out in frozen phrases and sentences stiff as corpses. “A lot of people seem to think that disabled people are chronically unhappy,” I said. “I know that’s not true myself. Are you often laughing inside?”

Question 1.
What do you get about Stephen when he spoke ‘I have had no choice’?
Answer:
Stephen Hawking accepted his disability. He tried to be brave.

Question 2.
How did he manage to express himself?
Answer:
Stephen Hawking was tapping at the little switch in his hand, trying to find the words on his computer with the only bit of movement left to him.

Question 3.
What is the general opinion about disabled?
Answer:
The general opinion about the disabled people are that they chronically unhappy.

Question 4.
How did Stephen take his visitors?
Answer:
Stephen found it amusing when others patronize him and get annoyed when someone comes and disturb him.

Question 5.
Change the word ‘patron’ into a verb.
Answer:
Patronise.

Passage 3

“Yes,” he said; it was a disadvantage of his voice synthesiser that it could convey no inflection, no shades or tone. And I could not tell how enthusiastically he agreed with me. Every time I shifted in my chair or turned my wrist to watch the time – I wanted to make every one of our thirty minutes count — I felt a huge relief and exhilaration in the possibilities of my body. How little it mattered then that I would never walk, or even stand. I told him how he had been an inspiration beyond cliche’ for me, and, surely, for others — did that thought help him? “No,” he said; and I thought how foolish I was to ask. When your body is a claustrophobic room and the walls are growing narrower day by day, it doesn’t do much good to know that there are people outside smiling with admiration to see you breathing still.

Question 1.
What was the disadvantage of his voice synthesizer?
Answer:
The disadvantage of his voice synthesizer was that it could convey no infection, no shade or tone.

Question 2.
Why was he worried about time?
Answer:
He wanted to make every minute of meeting count.

Question 3.
How was his body a ‘Claustrophobic Room’?
Answer:
As he was dependent on others and confined to a wheel chair, he called his body to be a suffocating place.

Question 4.
How did he feel after seeing admiring people?
Answer:
He didn’t get to know that there are people outside smiling with admiration to see him breathing still.

Question 5.
Choose a word that means the same as follows.
‘Phrase or idea used so often that it loses it meaning”
Ans.
cliche.

Passage 4

“Stay.” I waited. “Have some tea. I can show you the garden.” The garden was as big ks a park, but Stephen Hawking covered every inch, rumbling along in his motorised wheelchair while I dodged to keep out of the way. We couldn’t talk very much; the sun made him silent, the letters on his screen disappearing in the glare. An hour later, we were ready to leave. I didn’t know what to do. I could not kiss him or cry. I touched his shoulder and wheeled out into the summer evening. I looked back; and I knew he was waving, though he wasn’t. Watching him, an embodiment of my bravest self, the one I was moving towards, the one I had believed in for so many years, alone, I knew that my journey was over. For now.

Question 1.
Why did Stephen say ‘stay’?
Answer:
Stephen wanted him to stay with him and to show him the garden.

Question 2.
Why did ‘the sun made him silent’?
Answer:
The letters on his screen couldn’t be displayed because of the sunlight. So he couldn’t communicate. It made him silent.

Question 3.
What did the narrator So to did bade him goodbye?
Answer:
The narrator touched his shoulder and wheeled out into the summer evening.

Question 4.
How was the narrator’s journey?
Answer:
The narrator’s journey was inspiring and thought provoking.

Question5.
Find the superlative adjective of‘brave’ from the above lines?
Answer:
Bravest.