NCERT Exemplar Solutions: Arithmetic Progressions Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : NCERT Exemplar Solutions: Arithmetic Progressions Class 10 Notes | EduRev

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Exercise 5.1

Choose the correct answer from the given four options in the following questions:
1. In an AP, if d = –4, n = 7, a= 4, then a is
(a) 6 
(b) 7 
(c) 20 
(d) 28
Ans: (d)
Explanation:
We know that nth term of an AP is
an = a + (n – 1)d
where,
a = first term
an is nth term
d is the common difference
According to the question,
4 = a + (7 – 1)(- 4)
4 = a – 24
a = 24 + 4 = 28

2. In an AP, if a = 3.5, d = 0, n = 101, then an will be
(a) 0 
(b) 3.5 
(c) 103.5 
(d) 104.5
Ans: (b)
Explanation:
We know that nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
an = 3.5 + (101 – 1)0
= 3.5
(Since, d = 0, it’s a constant A.P)

3. The list of numbers – 10, – 6, – 2, 2,… is
(a) an AP with d = – 16
(b) an AP with d = 4
(c) an AP with d = – 4
(d) not an AP
Ans: (b)
Explanation:
According to the question,
a1 = – 10
a2 = – 6
a3 = – 2
a4 = 2
a2 – a1 = 4
a3 – a2 = 4
a4 – a3 = 4
a2 – a1 = a3 – a2 = a4 – a3 = 4
Therefore, it’s an A.P with d = 4

4. The 11th term of the AP: –5, (–5/2), 0, 5/2, …is
(a) –20
(b) 20
(c) –30
(d) 30

Ans: (b)
Explanation:
First term, a = – 5
Common difference,
d = 5 – (-5/2) = 5/2
n = 11
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a11 = – 5 + (11 – 1)(5/2)
a11 = – 5 + 25 = 20

5. The first four terms of an AP, whose first term is –2 and the common difference is –2, are
(a) – 2, 0, 2, 4
(b) – 2, 4, – 8, 16
(c) – 2, – 4, – 6, – 8
(d) – 2, – 4, – 8, –16

Ans: (c)
Explanation:
First term, a = – 2
Second Term, d = – 2
a1 = a = – 2
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
Hence, we have,
a2 = a + d = – 2 + (- 2) = – 4
Similarly,
a3 = – 6
a4 = – 8
So the A.P is
– 2, – 4, – 6, – 8

6. The 21st term of the AP whose first two terms are –3 and 4 is
(a) 17
(b) 137
(c) 143
(d) –143

Ans: (b)
Explanation:
First two terms of an AP are a = – 3 and a2 = 4.
We know, nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a2 = a + d
4 = – 3 + d
d = 7
Common difference, d = 7
a21 = a + 20d
= – 3 + (20)(7)
= 137

7. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30 
(b) 33 
(c) 37 
(d) 38

Ans: (b)
Explanation:
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a2 = a + d = 13 …..(1)
a5 = a + 4d = 25 …… (2)
From equation (1) we have,
a = 13 – d
Using this in equation (2), we have
13 – d + 4d = 25
13 + 3d = 25
3d = 12
d = 4
a = 13 – 4 = 9
a7 = a + 6d
= 9 + 6(4)
= 9 + 24 = 33

8. Which term of the AP: 21, 42, 63, 84… is 210?
(a) 9th 
(b) 10th 
(c) 11th 
(d) 12th
Ans: (b)
Explanation:
Let nth term of the given AP be 210.
According to question,
first term, a = 21
common difference, d = 42 – 21 = 21 and an = 210
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
210 = 21 + (n – 1)21
189 = (n – 1)21
n – 1 = 9
n = 10
So, 10th term of an AP is 210.

9. If the common difference of an AP is 5, then what is a18 – a13?
(a) 5 
(b) 20 
(c) 25 
(d) 30
Ans: (c)
Explanation:
Given, the common difference of AP i.e., d = 5
Now,
As we know, nth term of an AP is
an = a + (n – 1)d
where a = first term
an is nth term
d is the common difference
a18 -a13 = a + 17d – (a + 12d)
= 5d
= 5(5)
= 25


Exercise 5.2

1. Which of the following form an AP? Justify your answer.
(i) –1, –1, –1, –1,…

Solution:
We have a1 = – 1 , a2 = – 1, a3 = – 1 and a4 = – 1
a2 – a1 = 0
a3 – a2 = 0
a4 – a3 = 0
Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

(ii) 0, 2, 0, 2,…
Solution:
We have a1 = 0, a2 = 2, a3 = 0 and a4 = 2
a2 – a1 = 2
a3 – a2 = – 2
a4 – a3 = 2
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iii) 1, 1, 2, 2, 3, 3…
Solution:
We have a1 = 1 , a2 = 1, a3 = 2 and a4 = 2
a2 – a1 = 0
a3 – a2 = 1
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iv) 11, 22, 33…
Solution:
We have a1 = 11, a2 = 22 and a3 = 33
a2 – a1 = 11
a3 – a2 = 11
Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

(v) 1/2,1/3,1/4, …
Solution:
We have a1 = ½ , a2 = 1/3 and a3 = ¼
a2 – a1 = -1/6
a3 – a2 = -1/12
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vi) 2, 22, 23, 24, …
Solution:
We have a1 = 2 , a2 = 22, a3 = 23 and a4 = 24
a2 – a1 = 22 – 2 = 4 – 2 = 2
a3 – a2 = 23 – 22 = 8 – 4 = 4
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vii) √3, √12, √27, √48, …
Solution:
We have,
a= √3, a2 = √12, a3 = √27 and a4 = √48
a2 – a= √12 – √3 = 2√3 – √3 = √3
a3 – a= √27 – √12 = 3√3 – 2√3 = √3
a4 – a= √48 – √27 = 4√3 – 3√3 = √3
Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

2. Justify whether it is true to say that –1, -3/2, –2, 5/2,… forms an AP as

a2 – a1 = a3 – a2.
Solution:
False
a1 = -1, a2 = -3/2, a3 = -2 and a4 = 5/2
a2 – a= -3/2 – (-1) = – ½
a3 – a= – 2 – (- 3/2) = – ½
a4 – a= 5/2 – (-2) = 9/2
Clearly, the difference of successive terms in not same, all though, a2 – a1 = a3 – a2 but a4 – a3 ≠ a3 – a2 therefore it does not form an AP.

3. For the AP: –3, –7, –11, …, can we find directly a30a20 without actually finding a30 and a20? Give reasons for your answer.
Solution:
True
Given
First term, a = – 3
Common difference, d = a2 – a1 = – 7 – (- 3) = – 4
a30 – a20 = a + 29d – (a + 19d)
= 10d
= – 40
It is so because difference between any two terms of an AP is proportional to common difference of that AP

4. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?
Solution:
Suppose there are two AP’s with first terms a and A
And their common differences are d and D respectively
Suppose n be any term
an = a + (n – 1)d
An = A + (n – 1)D
As common difference is equal for both AP’s
We have D = d
Using this we have
An – an = a + (n – 1)d – [ A + (n – 1)D]
= a + (n – 1)d – A – (n – 1)d
= a – A
As a – A is a constant value
Therefore, difference between any corresponding terms will be equal to a – A.


Exercise 5.3

1. Match the APs given in column A with suitable common differences given in column B.

Column AColumn B
(A1) 2, – 2, – 6, –10,…(B1) 2/3
(A2) = –18, = 10, an = 0(B2) – 5
(A3) = 0, a10 = 6(B3) 4
(A4) a= 13, a4 =3(B4) – 4

(B5) 2

(B6) 1/2

(B7) 5

Solution:
(A1) AP is 2, – 2, – 6, – 10, ….
So common difference is simply
a2 – a1 = – 2 – 2 = – 4 = (B3)
(A2) Given
First term, a = – 18
No of terms, n = 10
Last term, an = 0
By using the nth term formula
an = a + (n – 1)d
0 = – 18 + (10 – 1)d
18 = 9d
d = 2 = (B5)
(A3) Given
First term, a = 0
Tenth term, a10 = 6
By using the nth term formula
an = a + (n – 1)d
a10 = a + 9d
6 = 0 + 9d
d = 2/3 = (B6)
(A4) Let the first term be a and common difference be d
Given that
a2 = 13
a4 = 3
a2 – a4 = 10
a + d – (a + 3d) = 10
d – 3d = 10
– 2d = 10
d = – 5= (B1)

2. Verify that each of the following is an AP, and then write its next three terms.
(i) 0, 1/4, 1/2, 3/4,…
Solution:
Here,
a= 0
a2 = ¼
a3 = ½
a4 = ¾
a2 – a1 = ¼ – 0 = ¼
a3 – a2 = ½ – ¼ = ¼
a4 – a3 = ¾ – ½ = ¼
Since, difference of successive terms are equal,
Hence, 0, 1/4, 1/2, 3/4… is an AP with common difference ¼.
Therefore, the next three term will be,
¾ + ¼ , ¾ + 2(¼), ¾ + 3(¼)
1, 5/4 , 3/2

(ii) 5, 14/3, 13/3, 4…
Solution:
Here,
a= 5
a2 = 14/3
a3 = 13/3
a4 = 4
a2 – a1 = 14/3 – 5 = -1/3
a3 – a2 = 13/3 – 14/3 = -1/3
a4 – a3 = 4 – 13/3 = -1/3
Since, difference of successive terms are equal,
Hence, 5, 14/3, 13/3, 4… is an AP with common difference -1/3.
Therefore, the next three term will be,
4 + (-1/3), 4 + 2(-1/3), 4 + 3(-1/3)
11/3 , 10/3, 3

(iii)3 , 23, 33,…
Solution:
Here,
a= √3
a2 = 2√3
a3 = 3√3
a4 = 4√3
a2 – a1 = 2√3 – √3 = √3
a3 – a2 = 3√3 – 2√3= √3
a4 – a3 = 4√3 – 3√3= √3
Since, difference of successive terms are equal,
Hence, √3 , 2√3, 3√3,… is an AP with common difference √3.
Therefore, the next three term will be,
4√3 + √3, 4√3 + 2√3, 4√3 + 3√3
5√3, 6√3, 7√3

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …
Solution:
Here
a1 = a + b
a2 = (a + 1) + b
a3 = (a + 1) + (b + 1)
a2 – a1 = (a + 1) + b – (a + b) = 1
a3 – a2 = (a + 1) + (b + 1) – (a + 1) – b = 1
Since, difference of successive terms are equal,
Hence, + b, (+ 1) + b, (+ 1) + (+ 1), … is an AP with common difference 1.
Therefore, the next three term will be,
(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1(2), (a + 1) + (b + 1) + 1(3)
(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)

(v) a, 2a + 1, 3a + 2, 4a + 3,…
Solution:
Here a1 = a
a2 = 2a + 1
a3 = 3a + 2
a4 = 4a + 3
a2 – a1 = (2a + 1) – (a) = a + 1
a3 – a2 = (3a + 2) – (2a + 1) = a + 1
a4 – a3 = (4a + 3) – (3a+2) = a + 1
Since, difference of successive terms are equal,
Hence, a, 2+ 1, 3+ 2, 4+ 3,… is an AP with common difference a + 1.
Therefore, the next three term will be,
4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)
5a + 4, 6a + 5, 7a + 6

3. Write the first three terms of the APs when and are as given below:
(i) a = 1/2, d = -1/6
(ii) a = –5, d = –3
(iii) a = 2 , d = 1/2
Solution: 
(i) a =1/2, d = -1/6
We know that,
First three terms of AP are:
a, a + d, a + 2d
½, ½ + (-1/6), ½ + 2 (-1/6)
½, 1/3, 1/6
(ii) a = –5, d = –3
We know that,
First three terms of AP are:
a, a + d, a + 2d
-5, – 5 + 1 (- 3), – 5 + 2 (- 3)
– 5, – 8, – 11
(iii) a = √2 , d = 1/√2
We know that,
First three terms of AP are:
a, a + d, a + 2d
√2, √2+1/√2, √2+2/√2
√2, 3/√2, 4/√2

4. Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c.
Solution:
For a, 7, b, 23, c… to be in AP
it has to satisfy the condition,
a5 – a4 = a4 – a3 = a3 – a2 = a2 – a1 = d
Where d is the common difference
7 – a = b – 7 = 23 – b = c – 23 …(1)
Let us equate,
b – 7 = 23 – b
2b = 30
b = 15 (eqn 1)
And,
7 – a = b – 7
From eqn 1
7 – a = 15 – 7
a = – 1
And,
c – 23 = 23 – b
c – 23 = 23 – 15
c – 23 = 8
c = 31
So a = – 1
b = 15
c = 31
Then, we can say that, the sequence – 1, 7, 15, 23, 31 is an AP

5. Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Solution:
We know that,
The first term of an AP = a
And, the common difference = d.
According to the question,
5th term, a5 = 19
Using the nth term formula,
an = a + (n – 1)d
We get,
a + 4d = 19
a = 19 – 4d …(1)
Also,
13th term – 8th term = 20
a + 12d – (a + 7d) = 20
5d = 20
d = 4
Substituting d = 4 in equation 1,
We get,
a = 19 – 4(4)
a = 3
Then, the AP becomes,
3, 3 + 4 , 3 + 2(4),…
3, 7, 11,…


Exercise 5.4

1. The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.
Solution:
We know that, in an A.P.,
First term = a
Common difference = d
Number of terms of an AP = n
According to the question,
We have,
S5 + S7 = 167
Using the formula for sum of n terms,
Sn = (n/2) [2a + (n-1)d]
So, we get,
(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167
5(2a + 4d) + 7(2a + 6d) = 334
10a + 20d + 14a + 42d = 334
24a + 62d = 334
12a + 31d = 167
12a = 167 – 31d …(1)
We have,
S10 = 235
(10/2) [2a + (10-1)d] = 235
5[ 2a + 9d] = 235
2a + 9d = 47
Multiplying L.H.S and R.H.S by 6,
We get,
12a + 54d = 282
From equation (1)
167 – 31d + 54d = 282
23d = 282 – 167
23d = 115
d = 5
Substituting the value of d = 5 in equation (1)
12a = 167 – 31(5)
12a = 167 – 155
12a = 12
a = 1
We know that,
S20 = (n/2) [2a + (20 – 1)d]
= 20/(2[2(1) + 19 (5)])
= 10[ 2 + 95]
= 970
Therefore, the sum of first 20 terms is 970.

2. Find the
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.
[Hint (iii): These numbers will be: multiples of 2 + multiples of 5 – multiples of 2 as well as of 5]
Solution:
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
We know that,
Multiples of 2 as well as of 5 = LCM of (2, 5) = 10
Multiples of 2 as well as of 5 between 1 and 500 = 10, 20, 30…, 490.
Hence,
We can conclude that 10, 20, 30…, 490 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using nth term formula,
an = a + (n – 1)d
490 = 10 + (n – 1)10
480 = (n – 1)10
n – 1 = 48
n = 49
Sum of an AP,
Sn = (n/2) [a + an], here an is the last term, which is given]
= (49/2) × [10 + 490]
= (49/2) × [500]
= 49 × 250
= 12250
Therefore, sum of those integers between 1 and 500 which are multiples of 2 as well as of 5 = 12250

(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

We know that,
Multiples of 2 as well as of 5 = LCM of (2, 5) = 10
Multiples of 2 as well as of 5 from 1 and 500 = 10, 20, 30…, 500.
Hence,
We can conclude that 10, 20, 30…, 500 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using nth term formula,
an = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP,
Sn = (n/2) [ a + an], here an is the last term, which is given]
= (50/2) ×[10 + 500]
= 25× [10 + 500]
= 25(510)
= 12750
Therefore, sum of those integers from 1 to 500 which are multiples of 2 as well as of 5= 12750

(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.
We know that,
Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5)
Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (10)
Multiples of 2 or 5 from 1 to 500 = List of multiple of 2 from 1 to 500 + List of multiple
of 5 from 1 to 500 – List of multiple of 10 from 1 to 500
= (2, 4, 6… 500) + (5, 10, 15… 500) – (10, 20, 30… 500)
Required sum = sum(2, 4, 6,…, 500) + sum(5, 10, 15,…, 500) – sum(10, 20, 30,., 500)
Consider the first series,
2, 4, 6, …., 500
First term, a = 2
Common difference, d = 2
Let n be no of terms
an = a + (n – 1)d
500 = 2 + (n – 1)2
498 = (n – 1)2
n – 1 = 249
n = 250
Sum of an AP, S= (n/2) [ a + an]
Let the sum of this AP be S1,
S= S250 = (250/2) ×[2 + 500]
S1 = 125(502)
S1 = 62750 … (1)
Consider the second series,
5, 10, 15, …., 500
First term, a = 5
Common difference, d = 5
Let n be no of terms
By nth term formula
an = a + (n – 1)d
500 = 5 + (n – 1)
495 = (n – 1)5
n – 1 = 99
n = 100
Sum of an AP, S= (n/2) [ a + an]
Let the sum of this AP be S2,
S= S100 = (100/2) ×[5 + 500]
S2 = 50(505)
S2 = 25250 … (2)
Consider the third series,
10, 20, 30, …., 500
First term, a = 10
Common difference, d = 10
Let n be no of terms
an = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP, S= (n/2) [ a + an]
Let the sum of this AP be S3,
S= S50 = (50/2) × [2 + 510]
S3 = 25(510)
S3 = 12750 … (3)
Therefore, the required Sum, S = S1 + S2 – S3
S = 62750 + 25250 – 12750
= 75250

3. The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
Solution:
We know that,
First term of an AP = a
Common difference of AP = d
nth term of an AP, an = a + (n – 1)d
According to the question,
as = ½ a2
2a8 = a2
2(a + 7d) = a + d
2a + 14d = a + d
a = – 13d …(1)
Also,
a11 = 1/3 a4 + 1
3(a + 10d) = a + 3d + 3
3a + 30d = a + 3d + 3
2a + 27d = 3
Substituting a = -13d in the equation,
2 (- 13d) + 27d = 3
d = 3
Then,
a = – 13(3)= – 39
Now,
a15  = a + 14d
= – 39 + 14(3)
= – 39 + 42
= 3
So 15th term is 3.

4. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
Solution:
We know that,
First term of an AP = a
Common difference of AP = d
nth term of an AP, an = a + (n – 1)d
Since, n = 37 (odd),
Middle term will be (n+1)/2 = 19th term
Thus, the three middle most terms will be,
18th, 19th and 20th terms
According to the question,
a18 + a19 + a20 = 225
Using an = a + (n – 1)d
a + 17d + a + 18d + a + 19d = 225
3a + 54d = 225
3a = 225 – 54d
a = 75 – 18d … (1)
Now, we know that last three terms will be 35th, 36th and 37th terms.
According to the question,
a35 + a36 + a37 = 429
a + 34d + a + 35d + a + 36d = 429
3a + 105d = 429
a + 35d = 143
Substituting a = 75 – 18d from equation 1,
75 – 18d + 35d = 143 [ using eqn1]
17d = 68
d = 4
Then,
a = 75 – 18(4)
a = 3
Therefore, the AP is a, a + d, a + 2d….
i.e. 3, 7, 11….

5. Find the sum of the integers between 100 and 200 that are
(i) divisible by 9
(ii) not divisible by 9
[Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9]
Solution:
(i) The number between 100 and 200 which is divisible by 9 = 108, 117, 126, …198
Let the number of terms between 100 and 200 which is divisible by 9 = n
an = a + (n – 1)d
198 = 108 + (n – 1)9
90 = (n – 1)9
n – 1 = 10
n = 11
Sum of an AP = S= (n/2) [ a + an]
Sn = (11/2) × [108 + 198]
= (11/2) × 306
= 11(153)
= 1683
(ii) Sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9)
Sum, S = S1 – S2
Here,
S1 = sum of AP 101, 102, 103, – – – , 199
S2 = sum of AP 108, 117, 126, – – – , 198
For AP 101, 102, 103, – – – , 199
First term, a = 101
Common difference, d = 199
Number of terms = n
Then,
an = a + (n – 1)d
199 = 101 + (n – 1)1
98 = (n – 1)
n = 99
Sum of an AP = S= (n/2) [ a + an]
Sum of this AP,
S= (99/2) × [199 + 101]
= (99/2) × 300
= 99(150)
= 14850
For AP 108, 117, 126, – – – – , 198
First term, a = 108
Common difference, d = 9
Last term, an = 198
Number of terms = n
Then,
an = a + (n – 1)d
198 = 108 + (n – 1)9
10 = (n – 1)
n = 11
Sum of an AP = S= (n/2) [ a + an]
Sum of this AP,
S= (11/2) × [108 + 198]
= (11/2) × (306)
= 11(153)
= 1683
Substituting the value of S1 and S2 in the equation, S = S1 – S2
S = S1 + S2
= 14850 – 1683
= 13167

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