Class 10 Maths Arithmetic Progressions questions with solutions

Q 11 :

In Mathematics, relations can be expressed in various ways. Matchstick patterns are an example of linear relations. Different strategies can be used to calculate the number of matchsticks used in different figures.

Based on the above information, answer the following questions:

(iii) (a) If 2 more matchsticks are added in each figure, then will the number of matchsticks in each figure still form an AP? If yes, find the number of matchsticks used in the 10th figure.

77
74
73
70

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(1)

If two more matchsticks are added to each figure, the new sequence is:
12 + 2, 19 + 2, 26 + 2, … i.e., 14, 21, 28, …

$24 - 14 = 7 = 28 - 21$

Hence, the sequence is an AP where a=14,??d=7

We need to find $a_{10} : a_{n} = a + (n - 1) d a_{10} = 14 + 9 \times 7 = 14 + 63 = 77$

The number of matchsticks used in the 10th figure is 77

Q 12 :

In Mathematics, relations can be expressed in various ways. Matchstick patterns are an example of linear relations. Different strategies can be used to calculate the number of matchsticks used in different figures.

Based on the above information, answer the following questions:

(iii) (b) If from each figure 5 matchsticks are removed, then will the number of matchsticks in each figure still form an AP? If yes, find which figure will have 91 matchsticks.

13th figure
14th figure
12th figure
11th figure

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(1)

If 5 matchsticks are removed from each figure, the new sequence formed is:
12 – 5, 19 – 5, 26 – 5, … i.e., 7, 14, 21, …

$14 - 7 = 7, 21 - 14 = 7$

Hence, the sequence is an AP where $a = 7, d = 7$

Let the nth figure have 91 matchsticks:

$a_{n} = 91 7 + (n - 1) 7 = 91 (n - 1) 7 = 84 \Rightarrow n - 1 = 12 \Rightarrow n = 13$

Q 13 :

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to complete the race in 31 seconds.

Based on the above information answer the following question:

(i) Write AP for the given situation

51, 49, 47
52, 49, 47
51, 53, 47
51, 50, 74

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(1)

The general form of an AP is a, a + d, a + 2d, ...., where a is the first term and d is the common difference.

Here a = 51, d = –2 ∴ The required AP is: 51, (51 – 2), (51 – 4), .... $\Rightarrow$ 51, 49, 47, ....

Q 14 :

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to complete the race in 31 seconds.

Based on the above information answer the following question:

(ii) What is the minimum number of days he needs to practice till his goal is achieved?

11
12
13
15

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(1)

Let he has to practice for n days

$a ₙ = 31 \Rightarrow a + (n - 1) d = 31, a = 51, d = - 2 \Rightarrow 51 + (n - 1) (- 2) = 31 \Rightarrow - 2 (n - 1) = 31 - 51 = - 20 \Rightarrow n - 1 = 10 \Rightarrow n = 11$

Minimum number of days = 11

Q 15 :

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to complete the race in 31 seconds.

Based on the above information answer the following question:

(iii) (a) If nth term of an AP is given by $a ₙ = 2 n + 3$ , then find the common difference of the AP.

1
2
5
4

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(2)

$G i v e n, a ₙ = 2 n + 3 \Rightarrow a ₙ ₋ ₁ = 2 (n - 1) + 3 = 2 n + 1 ∴ d = a ₙ - a ₙ ₋ ₁ = (2 n + 3) - (2 n + 1) = 2$

Q 16 :

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to complete the race in 31 seconds.

Based on the above information answer the following question:

(iii) (b) Find the value of n, for which 2n, n + 10, 3n + 2 are three consecutive terms of an AP.

5
4
6
3

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(3)

$S i n c e, 2 n, n + 10, 3 n + 2 a r e i n A P . 2 (n + 10) = 2 n + 3 n + 2 [∵ i f a, b, c a r e i n A P t h e n 2 b = a + c] \Rightarrow 2 n + 20 = 5 n + 2 \Rightarrow 18 = 3 n \Rightarrow n = 6$

Q 17 :

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following:

Based on the above information answer the following questions:

(i) Find the amount paid by him in the 30th instalment

3900
3700
3600
3500

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(1)

We have an AP of monthly instalments 1000, 1100, 1200, ...
Here a=1000 ,d=100

(i) Amount paid in 30th instalment

$a_{30} = a + (30 - 1) d \Rightarrow a_{30} = a + 29 d = 1000 + 29 \times 100 = 1000 + 2900 = ₹ 3900$

Q 18 :

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following:

Based on the above information answer the following questions:

(ii) What amount does he still have to pay after the 30th instalment?

45500
44500
42500
43000

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(2)

Now, total amount paid in 30 instalment = $S_{30}$

$S_{30} = 30 / 2 (2 \times 1000 + (30 - 1) \times 100) [∵ S_{n} = n / 2 (2 a + (n - 1) d)] \Rightarrow S_{30} = 30 (1000 + 29 \times 50) = 30 \times 2450 = ₹ 73, 500 ∴ A m o u n t s t i l l h e h a s t o p a y = ₹ 1, 18, 000 - ₹ 73, 500 = ₹ 44, 500$

Q 19 :

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following

Based on the above information answer the following questions:

(iii) (a) If the total number of installments is 40, what is the amount paid in the last installment?

4400
4700
4900
4100

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(3)

(a) As total number of instalment is 40, therefore amount paid in last instalment $= a_{40}$

$a_{40} = a + (40 - 1) d = a + 39 d = 1000 + 39 \times 100 = 1000 + 3900 = ₹ 4900$

Q 20 :

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following

Based on the above information answer the following questions:

(iii) (b) Find the ratio of the 1st instalment to the last instalment.

10:49
10:45
11:49
10:19

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(1)

(b) We have, first instalment = Rs 1000
and last instalment = Rs 4900

∴ Required ratio $= 1000 / 4900 = 10 / 49 = 10 : 49$

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