CLASS 10 IMPORTANT QUESTIONS OF CHAPTER 5 (ARITHMETIC PROGRESSIONS) NCERT

 

Class 10 Important Questions Of  Chapter 5

Q.1: Write first four terms of the AP when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3

Solution:

(i) a = 10, d = 10

Let an AP be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20
a₃ = a₂ + d = 20 + 10 = 30
a₄ = a₃ + d = 30 + 10 = 40
a₅ = a₄ + d = 40 + 10 = 50
And so on…

Therefore, the AP will be 10, 20, 30, 40, 50 …

The first four terms of this AP will be 10, 20, 30, and 40.

(ii) a = -2, d = 0

Let an AP be a₁, a₂, a₃, a₄, a₅ …

a₁ = a = -2
a₂ = a₁ + d = -2 + 0 = -2
a₃ = a₂ + d = -2 + 0 = -2
a₄ = a₃ + d = -2 + 0 = -2

Therefore, the AP will be -2, -2, -2, -2 …

The first four terms of this AP will be -2, -2, -2 and -2.

(iii) a = 4, d = -3

Let an AP be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 4
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = -2
a₄ = a₃ + d = -2 – 3 = -5

Therefore, the AP will be 4, 1, -2, -5 …

And, the first four terms of this AP will be 4, 1, -2 and -5.

Q.2: Which term of the AP: 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.

Solution:

Given AP: 21, 18, 15,…

Here, a = 21,

d = 18 – 21 = –3

Let nth term of the given AP is -81.

So, a_n = –81
As we know,

a_n = a + ( n – 1) d
Thus,

– 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35

Therefore, the 35th term of the given AP is – 81.

Next, 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

⇒ 3(n – 1) = 21

⇒ n – 1 = 7

⇒ n = 8

Therefore, the eighth term is 0.

Q.3: Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .

Solution:

Given AP: 11, 8, 5, 2, …

First term, a = 11

Common difference, d = a₂ − a₁ = 8 − 11 = −3

Let −150 be the nth term of this AP.

As we know, for an AP,

a_n = a + (n − 1) d

-150 = 11 + (n – 1)(-3)

-150 = 11 – 3n + 3

⇒ -164 = -3n

⇒ n = 164/3

Clearly, n is not an integer but a fraction.

Therefore, – 150 is not a term of the given AP.

Q.4: If the 3rd and the 9th terms of an AP are 4 and  -8, respectively, then which term of this AP is zero.

Solution:

Given that,

3rd term, a₃ = 4
9th term, a₉ = −8

We know that, the nth term of AP is;
a_n = a + (n − 1) d

Therefore,
a₃ = a + (3 − 1) d
4 = a + 2d ……………………………………… (i)

a₉ = a + (9 − 1) d
−8 = a + 8d ………………………………………………… (ii)

On subtracting equation (i) from (ii), we get;
−12 = 6d
d = −2

Substituting d = -2 in equation (i), we get;
4 = a + 2 (−2)
4 = a − 4
a = 8

Let nth term of this AP be zero.
a_n = a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2n = 10
⇒ n = 5

Hence, 5th term of the given AP is 0.

Q.5: Which term of the AP 3, 15, 27, 39, … will be 132 more than its 54th term?

Solution:

Given AP is: 3, 15, 27, 39, …
First term, a = 3
Common difference, d = a₂ − a₁ = 15 − 3 = 12

We know that,
a_n = a + (n − 1) d

Therefore,

a₅₄ = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636
a₅₄ = 639

We have to find the term of this AP.which is 132 more than a₅₄, i.e. 771.
Let nth term be 771.
a_n = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
⇒ (n − 1) = 64
⇒ n = 65

Therefore, the 65th term is 132 more than the 54th term of the given AP.

Alternate Method:

Let nth term be 132 more than 54th term.
n = 54 + (132/12)

= 54 + 11

= 65th term

Q. 6: How many multiples of 4 lie between 10 and 250?

Solution:

The first multiple of 4 that is greater than 10 is 12.

The next multiple will be 16.
Therefore, the series formed as;

12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an AP with the first term as 12 and the common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows.
12, 16, 20, 24, …, 248

Let 248 be the nth term of this AP.
First term, a = 12
Common difference, d = 4
a_n = 248

As we know,
a_n = a + (n – 1) d
248 = 12 + (n – 1) × 4
⇒ 236/4 = n – 1
⇒ 59 = n – 1
⇒ n = 60

Therefore, there are 60 multiples of 4 between 10 and 250.

Q.7: The sum of 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

We know, the nth term of the AP is;
a_n = a + (n − 1) d
a₄ = a + (4 − 1) d
a₄ = a + 3d

Thus, we can write,
a₈ = a + 7d
a₆ = a + 5d
a₁₀ = a + 9d

Given in the question;

a₄ + a₈ = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 …………………………………………………… (i)
a₆ + a₁₀ = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 …………………………………….. (ii)

On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5

From equation (i), we get,

a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a₂ = a + d = −13 + 5 = −8
a₃ = a2 + d = −8 + 5 = −3

Therefore, the first three terms of this AP are −13, −8, and −3.

Q.8: Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

Solution:

Given that, Ramkali saved Rs.5 in the first week and then started increasing her savings each week by Rs.1.75.

Hence,

First term, a = 5
and common difference, d = 1.75

Also given,
a_n = 20.75
Find, n = ?

As we know, by the nth term formula,
a_n = a + (n − 1) d

Therefore,
20.75 = 5 + (n – 1) × 1.75

15.75 = (n – 1) × 1.75

(n – 1) = 15.75/1.75

= 1575/175

= 63/7

= 9
⇒ n = 10
Hence, n is 10.

Q.9: How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?

Solution:

Given AP: 24, 21, 18,…

Here, a = 24, d = 21 – 24 = –3, S_n = 78. We need to find n.
We know that;

S_n = n/2[2a + (n – 1)d]

So, 78 = n/2 [48 + (n – 1)(-3)]

78 = n/2 [51 – 3n]

156 = 51n – 3n²

3n² – 51n + 156 = 0

n² – 17n + 52 = 0

n² – 13n – 4n + 52 = 0

n(n – 13) – 4(n – 13) = 0

(n – 4) (n – 13) = 0

n = 4 or 13
Both values of n are admissible. So, the number of terms is either 4 or 13.

Q. 10: 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:

Given that,
first term, a = 5
last term, l = 45
Sum of the AP, S_n = 400

As we know, the sum of AP formula is;
S_n = n/2 (a + l)
400 = n/2 (5 + 45)

400 = n/2 (50)
Number of terms, n = 16

As we know, the last term of AP can be written as;
Last term, l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
Therefore, the Common difference is d = 40/15 = 8/3.