NCERT Exemplar Solutions Class 10 Maths Chapter 5 – Free PDF Download
The NCERT Exemplar Class 10 Maths Chapter 5 Arithmetic Progressions is provided here in PDF format for students to download. They can practise these questions to prepare for the CBSE exam. These materials are prepared by our subject experts, keeping into consideration the CBSE Syllabus (2023-2024). Students can download exemplar problems with solutions and use them at their convenience to learn about the topics and also find answers to all questions given in the chapter. Furthermore, students can use this as a reference tool to do a complete revision of the entire chapter before the board exam.
The topic of arithmetic progression can sometimes be confusing for students. However, in Class 10, this topic has been discussed in detail in the Chapter 5 Maths textbook. The chapter introduces students to many core concepts, like the n^{th} term of an AP, common differences and more. There are also suitable examples provided to help students know the difference, as well as the right method to find the sum of arithmetic progressions. Thus, to make this chapter easy to learn, free NCERT Exemplar for Arithmetic Progressions is available here in PDF form for students. Click here to get exemplars for all chapters.
The topics covered in this material are related to
- Arithmetic progression – Finding the first term and common difference
- Finding the nth term of AP
- Finding the sum of the first n terms of AP
- Finding the sum of all terms of the AP
- Students can get a glimpse of the Class 10 Maths Chapter 5 NCERT Exemplar PDF below.
Download the PDF of NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression
Access Answers to the NCERT Exemplar Class 10 Maths Chapter 5
Exercise 5.1
Choose the correct answer from the given four options in the following questions:
1. In an AP, if d = â€“4, n = 7, a_{n }= 4, then a is
(A) 6 (B) 7 (C) 20 (D) 28
Solution:
(D) 28
Explanation:
We know that nth term of an AP is
a_{n =}Â a + (n – 1)d
where,
a = first term
a_{n}Â is nth term
d is the common difference
According to the question,
4 = a + (7 – 1)(- 4)
4 = a â€“ 24
a = 24 + 4 = 28
2. In an AP, if a = 3.5, d = 0, n = 101, then a_{n} will be
(A) 0 (B) 3.5 (C) 103.5 (D) 104.5
Solution:
(B) 3.5
Explanation:
We know that nth term of an AP is
a_{n =}Â a + (n – 1)d
Where,
a = first term
a_{n}Â is nth term
d is the common difference
a_{n}Â = 3.5 + (101 – 1)0
= 3.5
(Since, d = 0, itâ€™s a constant A.P)
3. The list of numbers â€“ 10, â€“ 6, â€“ 2, 2,… is
(A) an AP with d = â€“ 16
(B) an AP with d = 4
(C) an AP with d = â€“ 4
(D) not an AP
Solution:
(B) an AP with d = 4
Explanation:
According to the question,
a_{1 =}Â – 10
a_{2 =}Â – 6
a_{3 =}Â – 2
a_{4 =}Â 2
a_{2 –}Â a_{1 =}Â 4
a_{3 –}Â a_{2 =}Â 4
a_{4 –}Â a_{3 =}Â 4
a_{2}Â – a_{1 =}Â a_{3 –}Â a_{2 =}Â a_{4 –}Â a_{3 =}Â 4
Therefore, itâ€™s an A.P with d = 4
4. The 11th term of the AP: â€“5, (â€“5/2), 0, 5/2, …is
(A) â€“20 (B) 20 (C) â€“30 (D) 30
Solution:
(B) 20
Explanation:
First term, a = – 5
Common difference,
d = 5 â€“ (-5/2) = 5/2
n = 11
We know that the nth term of an AP is
a_{n =}Â a + (n – 1)d
Where,
a = first term
a_{n}Â is nth term
d is the common difference
a_{11 =}Â – 5 + (11 – 1)(5/2)
a_{11 =}Â – 5 + 25 = 20
5. The first four terms of an AP, whose first term is â€“2 and the common difference is â€“2, are
(A) â€“ 2, 0, 2, 4
(B) â€“ 2, 4, â€“ 8, 16
(C) â€“ 2, â€“ 4, â€“ 6, â€“ 8
(D) â€“ 2, â€“ 4, â€“ 8, â€“16
Solution:
(C) â€“ 2, â€“ 4, â€“ 6, â€“ 8
Explanation:
First term, a = – 2
Second Term, d = – 2
a_{1 =}Â a = – 2
We know that the nth term of an AP is
a_{n =}Â a + (n – 1)d
Where,
a = first term
a_{n}Â is nth term
d is the common difference
Hence, we have,
a_{2 =}Â a + d = – 2 + (- 2) = – 4
Similarly,
a_{3 =}Â – 6
a_{4 =}Â – 8
So the A.P is
– 2, – 4, – 6, – 8
6. The 21st term of the AP whose first two terms are â€“3 and 4 is
(A) 17 (B) 137 (C) 143 (D) â€“143
Solution:
(B) 137
Explanation:
First two terms of an AP are a = – 3 and a_{2 =}Â 4.
We know, nth term of an AP is
a_{n =}Â a + (n – 1)d
Where,
a = first term
a_{n}Â is nth term
d is the common difference
a_{2 =}Â a + d
4 = – 3 + d
d = 7
Common difference, d = 7
a_{21 =}Â 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 7^{th} term?
(A) 30 (B) 33 (C) 37 (D) 38
Solution:
(B) 33
Explanation:
We know that the nth term of an AP is
a_{n =}Â a + (n – 1)d
Where,
a = first term
a_{n}Â is nth term
d is the common difference
a_{2 =}Â a + d = 13 â€¦..(1)
a_{5 =}Â a + 4d = 25 â€¦â€¦ (2)
From equation (1) we have,
a = 13 â€“ d
Using this in equation (2), we have
13 – d + 4d = 25
13 + 3d = 25
3d = 12
d = 4
a = 13 – 4 = 9
a_{7 =}Â a + 6d
= 9 + 6(4)
= 9 + 24 = 33
8. Which term of the AP: 21, 42, 63, 84… is 210?
(A) 9^{th} (B) 10^{th} (C) 11^{th} (D) 12^{th}
Solution:
(B) 10^{th}
Explanation:
Let nth term of the given AP be 210.
According to question,
first term, a = 21
common difference, d = 42 â€“ 21 = 21 and a_{n}Â = 210
We know that the nth term of an AP is
a_{n =}Â a + (n – 1)d
Where,
a = first term
a_{n}Â 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 a_{18} â€“ a_{13}?
(A) 5 (B) 20 (C) 25 (D) 30
Solution:
(C) 25
Explanation:
Given, the common difference of AP i.e., d = 5
Now,
As we know, nth term of an AP is
a_{n =}Â a + (n – 1)d
where a = first term
a_{n}Â is nth term
d is the common difference
a_{18}Â -a_{13 =}Â a + 17d â€“ (a + 12d)
= 5d
= 5(5)
= 25
Exercise 5.2
1. Which of the following form an AP? Justify your answer.
(i) â€“1, â€“1, â€“1, â€“1,…
Solution:
We have a_{1}Â = – 1 , a_{2}Â = – 1, a_{3}Â = – 1 and a_{4}Â = – 1
a_{2}Â – a_{1}Â = 0
a_{3}Â – a_{2}Â = 0
a_{4}Â – a_{3}Â = 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 a_{1}Â = 0, a_{2}Â = 2, a_{3}Â = 0 and a_{4}Â = 2
a_{2}Â – a_{1}Â = 2
a_{3}Â – a_{2}Â = – 2
a_{4}Â – a_{3}Â = 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 a_{1}Â = 1 , a_{2}Â = 1, a_{3}Â = 2 and a_{4}Â = 2
a_{2}Â – a_{1}Â = 0
a_{3}Â – a_{2}Â = 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 a_{1}Â = 11, a_{2}Â = 22 and a_{3}Â = 33
a_{2}Â – a_{1}Â = 11
a_{3}Â – a_{2}Â = 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 a_{1}Â = Â½Â , a_{2}Â = 1/3Â and a_{3}Â =Â Â¼
a_{2}Â – a_{1}Â = -1/6
a_{3}Â – a_{2}Â = -1/12
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.
(vi) 2, 2^{2}, 2^{3}, 2^{4}, …
Solution:
We have a_{1}Â = 2 , a_{2}Â = 2^{2}, a_{3}Â = 2^{3}Â and a_{4}Â = 2^{4}
a_{2}Â – a_{1}Â = 2^{2}Â – 2 = 4 – 2 = 2
a_{3}Â – a_{2}Â = 2^{3}Â – 2^{2}Â = 8 – 4 = 4
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.
(vii) âˆš3, âˆš12, âˆš27, âˆš48, …
Solution:
We have,
a_{1 }= âˆš3, a_{2} = âˆš12, a_{3} = âˆš27 and a_{4} = âˆš48
a_{2} â€“ a_{1 }= âˆš12 â€“ âˆš3 = 2âˆš3 â€“ âˆš3 = âˆš3
a_{3} â€“ a_{2 }= âˆš27 â€“ âˆš12 = 3âˆš3 â€“ 2âˆš3 = âˆš3
a_{4} â€“ a_{3 }= âˆš48 â€“ âˆš27 = 4âˆš3 â€“ 3âˆš3 = âˆš3
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 â€“1, -3/2, â€“2, 5/2,… forms an AP as
a_{2} â€“ a_{1} = a_{3} â€“ a_{2}.
Solution:
False
a_{1} = -1, a_{2} = -3/2, a_{3} = -2 and a_{4} = 5/2
a_{2} â€“ a_{1 }= -3/2 â€“ (-1) = – Â½
a_{3} â€“ a_{2 }= – 2 â€“ (- 3/2) = – Â½
a_{4} â€“ a_{3 }= 5/2 â€“ (-2) = 9/2
Clearly, the difference of successive terms in not same, all though, a_{2}Â – a_{1}Â = a_{3}Â – a_{2}Â but a_{4}Â – a_{3}Â â‰ a_{3}Â – a_{2}Â therefore it does not form an AP.
3. For the AP: â€“3, â€“7, â€“11, …, can we find directly a_{30} â€“ a_{20} without actually finding a_{30} and a_{20}? Give reasons for your answer.
Solution:
True
Given
First term, a = – 3
Common difference, d = a_{2}Â – a_{1}Â = – 7 – (- 3) = – 4
a_{30}Â – a_{20}Â = 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
a_{n}Â = a + (n – 1)d
A_{n}Â = A + (n – 1)D
As common difference is equal for both AP’s
We have D = d
Using this we have
A_{n}Â – a_{n}Â = a + (n – – 1)d – [ A + (n – 1)D]
= a + (n – 1)d – A – (n – 1)d
= a â€“ 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 A | Column B |
(A_{1}) 2, â€“ 2, â€“ 6, â€“10,… | (B_{1}) 2/3 |
(A_{2}) a = â€“18, n = 10, a_{n} = 0 | (B_{2}) â€“ 5 |
(A_{3}) a = 0, a_{10} = 6 | (B_{3}) 4 |
(A_{4}) a_{2 }= 13, a_{4} =3 | (B_{4}) â€“ 4 |
(B_{5}) 2 | |
(B_{6}) 1/2 | |
(B_{7}) 5 |
Solution:
(A_{1}) AP is 2, – 2, – 6, – 10, â€¦.
So common difference is simply
a_{2}Â – a_{1}Â = – 2 – 2 = – 4 = (B_{3})
(A_{2}) Given
First term, a = – 18
No of terms, n = 10
Last term, a_{n}Â = 0
By using the nth term formula
a_{n =}Â a + (n – 1)d
0 = – 18 + (10 – 1)d
18 = 9d
d = 2 = (B_{5})
(A_{3}) Given
First term, a = 0
Tenth term, a_{10}Â = 6
By using the nth term formula
a_{n =}Â a + (n – 1)d
a_{10}Â = a + 9d
6 = 0 + 9d
d = 2/3 = (B_{6})
(A_{4}) Let the first term be a and common difference be d
Given that
a_{2}Â = 13
a_{4}Â = 3
a_{2}Â – a_{4}Â = 10
a + d – (a + 3d) = 10
d – 3d = 10
– 2d = 10
d = – 5= (B_{1})
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_{1 }= 0
a_{2} = Â¼
a_{3} = Â½
a_{4} = Â¾
a_{2} â€“ a_{1} = Â¼ – 0 = Â¼
a_{3} â€“ a_{2} = Â½ – Â¼ = Â¼
a_{4} â€“ a_{3} = Â¾ – Â½ = Â¼
Since, difference of successive terms are equal,
Hence, 0, 1/4, 1/2, 3/4… is an AP with common differenceÂ Â¼.
Therefore, the next three term will be,
Â¾ + Â¼ , Â¾ + 2(Â¼), Â¾ + 3(Â¼)
1, 5/4 , 3/2
(ii) 5, 14/3, 13/3, 4…
Solution:
Here,
a_{1 }= 5
a_{2} = 14/3
a_{3} = 13/3
a_{4} = 4
a_{2} â€“ a_{1} = 14/3 â€“ 5 = -1/3
a_{3} â€“ a_{2} = 13/3 â€“ 14/3 = -1/3
a_{4} â€“ a_{3} = 4 â€“ 13/3 = -1/3
Since, difference of successive terms are equal,
Hence, 5, 14/3, 13/3, 4… is an AP with common differenceÂ -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) âˆš3 , 2âˆš3, 3âˆš3,…
Solution:
Here,
a_{1 }= âˆš3
a_{2} = 2âˆš3
a_{3} = 3âˆš3
a_{4} = 4âˆš3
a_{2} â€“ a_{1} = 2âˆš3 â€“ âˆš3 = âˆš3
a_{3} â€“ a_{2} = 3âˆš3 â€“ 2âˆš3= âˆš3
a_{4} â€“ a_{3} = 4âˆš3 â€“ 3âˆš3= âˆš3
Since, difference of successive terms are equal,
Hence, âˆš3 , 2âˆš3, 3âˆš3,… is an AP with common differenceÂ âˆš3.
Therefore, the next three term will be,
4âˆš3 + âˆš3, 4âˆš3 + 2âˆš3, 4âˆš3 + 3âˆš3
5âˆš3, 6âˆš3, 7âˆš3
(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …
Solution:
Here
a_{1}Â = a + b
a_{2}Â = (a + 1) + b
a_{3}Â = (a + 1) + (b + 1)
a_{2}Â – a_{1}Â = (a + 1) + b – (a + b) = 1
a_{3}Â – a_{2}Â = (a + 1) + (b + 1) – (a + 1) – b = 1
Since, difference of successive terms are equal,
Hence, a + b, (a + 1) + b, (a + 1) + (b + 1), … is an AP with common differenceÂ 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 a_{1}Â = a
a_{2}Â = 2a + 1
a_{3}Â = 3a + 2
a_{4}Â =Â 4a + 3
a_{2}Â – a_{1}Â = (2a + 1) â€“ (a) = a + 1
a_{3}Â – a_{2}Â = (3a + 2) â€“ (2a + 1) = a + 1
a_{4}Â – a_{3}Â = (4a + 3) â€“ (3a+2) = a + 1
Since, difference of successive terms are equal,
Hence, a, 2a + 1, 3a + 2, 4a + 3,… is an AP with common differenceÂ 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 a and d are as given below:
- a =1/2, d = -1/6
- a = â€“5, d = â€“3
- a = 2 , d = 1/âˆš2
Solution:
(i) a =1/2, d = -1/6
We know that,
First three terms of AP are :
a, a + d, a + 2d
Â½, Â½ + (-1/6), Â½ + 2 (-1/6)
Â½, 1/3, 1/6
(ii) a = â€“5, d = â€“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 = âˆš2 , d = 1/âˆš2
We know that,
First three terms of AP are :
a, a + d, a + 2d
âˆš2, âˆš2+1/âˆš2, âˆš2+2/âˆš2
âˆš2, 3/âˆš2, 4/âˆš2
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, câ€¦ to be in AP
it has to satisfy the condition,
a_{5}Â – a_{4}Â = a_{4}Â – a_{3}Â = a_{3}Â – a_{2}Â = a_{2}Â – a_{1}Â = d
Where d is thecommon difference
7 – a = b – 7 = 23 – b = c â€“ 23 â€¦(1)
Let us equate,
b – 7 = 23 â€“ b
2b = 30
b = 15 (eqn 1)
And,
7 – a = b â€“ 7
From eqn 1
7 – a = 15 â€“ 7
a = – 1
And,
c – 23 = 23 â€“ b
c – 23 = 23 â€“ 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,
5^{th}Â term, a_{5}Â = 19
Using the n^{th} term formula,
a_{n}Â = a + (n – 1)d
We get,
a + 4d = 19
a = 19 – 4d â€¦(1)
Also,
13^{th}Â term – 8^{th}Â term = 20
a + 12d – (a + 7d) = 20
5d = 20
d = 4
Substituting d = 4 in equation 1,
We get,
a = 19 â€“ 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,
S_{5}Â + S_{7}Â = 167
Using the formula for sum of n terms,
S_{n} = (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 â€¦(1)
We have,
S_{10}Â = 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 â€“ 167
23d = 115
d = 5
Substituting the value of d = 5 in equation (1)
12a = 167 – 31(5)
12a = 167 â€“ 155
12a = 12
a = 1
We know that,
S_{20 }= (n/2) [2a + (20 â€“ 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 â€“ 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, 30â€¦, 490.
Hence,
We can conclude that 10, 20, 30â€¦, 490 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using n^{th} term formula,
a_{n}Â = a + (n – 1)d
490 = 10 + (n – 1)10
480 = (n – 1)10
n – 1 = 48
n = 49
Sum of an AP,
S_{n} = (n/2) [a + a_{n}], here a_{n} is the last term, which is given]
= (49/2) Ã— [10 + 490]
= (49/2) Ã— [500]
= 49 Ã— 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, 30â€¦, 500.
Hence,
We can conclude that 10, 20, 30â€¦, 500 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using n^{th} term formula,
a_{n}Â = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP,
S_{n} = (n/2) [ a + a_{n}], here a_{n} is the last term, which is given]
= (50/2) Ã—[10+500]
= 25Ã— [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 â€“ Multiple of LCM (2, 5)
Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 â€“ 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, 6â€¦ 500) + (5, 10, 15â€¦ 500) – (10, 20, 30â€¦ 500)
Required sum = sum(2, 4, 6,â€¦, 500) + sum(5, 10, 15,â€¦, 500) – sum(10, 20, 30,., 500)
Consider the first series,
2, 4, 6, â€¦., 500
First term, a = 2
Common difference, d = 2
Let n be no of terms
a_{n}Â = a + (n – 1)d
500 = 2 + (n – 1)^{2}
498 = (n – 1)^{2}
n – 1 = 249
n = 250
Sum of an AP, S_{n }= (n/2) [ a + a_{n}]
Let the sum of this AP be S_{1,}
S_{1 }= S_{250} = (250/2) Ã—[2+500]
S_{1} = 125(502)
S_{1} = 62750 â€¦ (1)
Consider the second series,
5, 10, 15, â€¦., 500
First term, a = 5
Common difference, d = 5
Let n be no of terms
By nth term formula
a_{n}Â = a + (n – 1)d
500 = 5 + (n – 1)
495 = (n – 1)5
n – 1 = 99
n = 100
Sum of an AP, S_{n }= (n/2) [ a + a_{n}]
Let the sum of this AP be S_{2,}
S_{2 }= S_{100} = (100/2) Ã—[5+500]
S_{2} = 50(505)
S_{2} = 25250 â€¦ (2)
Consider the third series,
10, 20, 30, â€¦., 500
First term, a = 10
Common difference, d = 10
Let n be no of terms
a_{n}Â = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP, S_{n }= (n/2) [ a + a_{n}]
Let the sum of this AP be S_{3,}
S_{3 }= S_{50} = (50/2) Ã— [2+510]
S_{3} = 25(510)
S_{3} = 12750 â€¦ (3)
Therefore, the required Sum, S = S_{1} + S_{2} – S_{3}
S = 62750 + 25250 â€“ 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
n^{th} term of an AP, a_{n}Â = a + (n – 1)d
According to the question,
a_{s} = Â½ a_{2}
2a_{8}Â = a_{2}
2(a + 7d) = a + d
2a + 14d = a + d
a = – 13d â€¦(1)
Also,
a_{11} = 1/3 a_{4} + 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,
a_{15}Â = a + 14d
= – 39 + 14(3)
= – 39 + 42
= 3
So 15^{th}Â 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
n^{th} term of an AP, a_{n}Â = a + (n – 1)d
Since, n = 37 (odd),
Middle term will be (n+1)/2 = 19^{th }term
Thus, the three middle most terms will be,
18^{th}, 19^{th}Â and 20^{th}Â terms
According to the question,
a_{18}Â + a_{19}Â + a_{20}Â = 225
Using a_{n}Â = a + (n – 1)d
a + 17d + a + 18d + a + 19d = 225
3a + 54d = 225
3a = 225 – 54d
a = 75 – 18d â€¦ (1)
Now, we know that last three terms will be 35^{th}, 36^{th}Â and 37^{th}Â terms.
According to the question,
a_{35}Â + a_{36}Â + a_{37}Â = 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 + 2dâ€¦.
i.e. 3, 7, 11â€¦.
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 â€“ Total numbers divisible by 9]
Solution:
(i) The number between 100 and 200 which is divisible by 9 = 108, 117, 126, â€¦198
Let the number of terms between 100 and 200 which is divisible by 9 = n
a_{n}Â = a + (n – 1)d
198 = 108 + (n – 1)9
90 = (n – 1)9
n – 1 = 10
n = 11
Sum of an AP = S_{n }= (n/2) [ a + a_{n}]
S_{n} = (11/2) Ã— [108 + 198]
= (11/2) Ã— 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) â€“ (sum of total numbers between 100 and 200 which is divisible by 9)
Sum, S = S_{1} – S_{2}
Here,
S_{1} = sum of AP 101, 102, 103, – – – , 199
S_{2} = 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,
a_{n}Â = a + (n – 1)d
199 = 101 + (n – 1)1
98 = (n – 1)
n = 99
Sum of an AP = S_{n }= (n/2) [ a + a_{n}]
Sum of this AP,
S_{1 }= (99/2) Ã— [199 + 101]
= (99/2) Ã— 300
= 99(150)
= 14850
For AP 108, 117, 126, – – – – , 198
First term, a = 108
Common difference, d = 9
Last term, a_{n}Â = 198
Number of terms = n
Then,
a_{n}Â = a + (n – 1)d
198 = 108 + (n – 1)9
10 = (n – 1)
n = 11
Sum of an AP = S_{n }= (n/2) [ a + a_{n}]
Sum of this AP,
S_{2 }= (11/2) Ã— [108 + 198]
= (11/2) Ã— (306)
= 11(153)
= 1683
Substituting the value of S_{1} and S_{2} in the equation, S = S_{1} – S_{2}
S = S_{1} + S_{2}
= 14850 â€“ 1683
= 13167
At BYJUâ€™S, students get free online learning materials such as notes, exemplar books, question papers and Class 10 Maths NCERT Solutions to help them prepare for first- and second-term CBSE exams in an effective way and score good marks. Sample papers and previous years’ question papers will help students to get an idea of the question pattern of the chapter, Arithmetic Progression and its marks weightage.
Download BYJUâ€™S – The Learning App for advanced learning through video lessons based on different topics and improving conceptual knowledge related to Maths and Science.
Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Maths Chapter 5
What are the topics covered in Chapter 5 of NCERT Exemplar Solutions for Class 10 Maths?
1. Arithmetic progression â€“ Finding the first term and common difference
2. Finding the nth term of AP
3. Finding the sum of the first n terms of AP
4. Finding the sum of all terms of the AP
How many exercises are present in Chapter 5 of NCERT Exemplar Solutions for Class 10 Maths?
Exercise 5.1 â€“ 9 questions
Exercise 5.2 â€“ 4 questions
Exercise 5.3 â€“ 5 questions
Exercise 5.4 â€“ 5 questions
Why should I choose NCERT Exemplar Solutions for Class 10 Maths Chapter 5?
Also Read
Also AccessÂ |
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic ProgressionsÂ |
CBSE Notes for Class 10 Maths Chapter 5 Arithmetic ProgressionsÂ |
Comments