In this post, you will find the NCERT solutions for class 10 maths ex 5.2. These solutions are based on the latest curriculum of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 5.2
1. Fill in the blanks in the following table, given that a is the first term, d is the common difference and a_{n} is
the nth term of AP.

a 
d 
n 
a_{n} 
(i) 
7 
3 
8 
… 
(ii) 
–18 
… 
10 
0 
(iii) 
… 
–3 
18 
–5 
(iv) 
–18.9 
2.5 
… 
3.6 
(v) 
3.5 
0 
105 
… 
Solution:
(i) a = 7, d = 3, n = 8
Here, we have to find a_{n}.
Using the formula, a_{n} = a + (n –1)d
Putting the values of a, d
and n, we get
a_{n} = 7 + (8 −1)3
= 7 + (7)3 = 7 + 21 = 28
(ii) a = –18, n = 10, a_{n} = 0
Here, we have to find d.
Using the formula, a_{n} = a + (n −1)d
Putting the values of a, n
and a_{n}, we get
0 = –18 + (10 – 1)d
⇒ 0 =
−18 + 9d
⇒ 18
= 9d
⇒ d = 2
(iii) d = –3, n = 18, a_{n} = −5
Here, we have to find a.
Using the formula, a_{n} = a + (n – 1)d
Putting the values of d, a_{n} and n, we
get
–5 = a + (18 –1) (–3)
⇒ −5
= a + (17) (−3)
⇒ −5
= a – 51
⇒ a = 46
(iv) a = –18.9, d = 2.5, a_{n} = 3.6
Here, we have to find n.
Using the formula, a_{n} = a + (n –1)d
Putting the values of d, a_{n} and n, we
get
3.6 = –18.9 + (n –1) (2.5)
⇒ 3.6
= −18.9 + 2.5n − 2.5
⇒ 2.5n = 25
⇒ n = 10
(v) a = 3.5, d = 0, n = 105
Here, we have to find a_{n}.
Using the formula, a_{n} = a + (n −1)d
Putting the values of a, d
and n, we get
a_{n} = 3.5 + (105 −1) (0)
⇒ a_{n} = 3.5
+ 104 × 0
⇒ a_{n} = 3.5
+ 0
⇒ a_{n} = 3.5
2. Choose the correct choice in the following and justify:
(i)
30^{th} term of the AP: 10, 7, 4 …, is
(A)
97 (B) 77 (C) –77 (D) –87
(ii)
11^{th} term of the AP: −3, −1/2, 2 …, is
(A)
28 (B) 22 (C) –38 (D) –48½
Solution:
(i) 10,
7, 4 …
Here, first term (a) = 10, common difference (d) = 7 – 10 = 4 –7 = –3
And n = 30 [Since
we have to find 30^{th} term]
Using the formula, a_{n} = a + (n – 1)d
⇒ a_{30 }= 10 + (30
−1) (−3) = 10 – 87 = −77
Therefore, the option (C) is
correct.
(ii) −3, −1/2, 2 …
Here, first term (a) = –3, common difference (d) = −1/2 −
(−3) = 2 − (−1/2) = 5/2
And n = 11 [Since we have to find 11^{th} term]
Using the formula, a_{n} = a + (n – 1)d
⇒
a_{11} = −3 + (11 –1) 5/2 = −3
+ 25 = 22
Therefore,
the option (B) is correct.
3.
In the following Aps, find the missing terms:
(i)
2, __, 26
(ii)
__, 13, __, 3
(iii)
5, __, __, 9½
(iv)
–4, __, __, __, __, 6
(v)
__, 38, __, __, __, –22
Solution:
(i) 2,
__, 26
We know that the difference
between consecutive terms in an A.P. is always equal.
Let
the missing term be x.
Then,
x – 2 = 26 – x
⇒ 2x = 28
⇒ x = 14
Therefore, the missing term
is 14.
(ii)
__,
13, __, 3
Let
the missing terms be x and y.
Then,
the sequence becomes x, 13, y, 3
We know that the difference between consecutive
terms in an A.P. is always equal.
Then, y – 13 = 3 – y
⇒ 2y = 16
⇒ y = 8
Again,
13 – x = y – 13
⇒ x + y = 26
Putting
the value of y, we get
⇒ x + 8 = 26
⇒ x = 18
Therefore, the missing terms
are 18 and 8.
(iii)
5, __, __, 9½
Here,
first term (a) =
5, n = 4 and fourth term (a_{4}) = 9½
Using the formula, a_{n} = a + (n – 1)d, we get
a_{4 }= 5
+ (4 − 1)d
⇒ 19/2 = 5 + 3d
⇒ 19 = 2(5 + 3d)
⇒ 19 = 10 + 6d
⇒ 6d = 19 – 10
⇒ 6d = 9
⇒ d = 3/2
Therefore, we get the common
difference (d) = 3/2
Second term = a + d
= 5 + 3/2 = 13/2
Third term = second term + d = 13/2 + 3/2 = 16/2 = 8
Therefore, the missing terms
are 13/2 and 8.
(iv)
–4,
__, __, __, __, 6
Here,
first term (a) = –4, n = 6 and sixth term (a_{6}) = 6
Using the formula, a_{n} = a + (n – 1)d, we get
a_{6 }= −4
+ (6 − 1)d
⇒ 6 =
−4 + 5d
⇒ 5d = 10
⇒ d = 2
Therefore, the common
difference (d) = 2
Second term = first term + d = a
+ d = –4 + 2 = –2
Third term = second term + d = –2 + 2 = 0
Fourth term = third term + d = 0 + 2 = 2
Fifth term = fourth term + d = 2 + 2 = 4
Therefore, the missing terms
are –2, 0, 2 and 4.
(v) __, 38, __, __, __,
–22
Here,
the 2^{nd} and the 6^{th} terms are
given.
Using the formula, a_{n} = a + (n – 1)d
Second term (a_{2}) = a + (2 − 1)d
⇒ 38 = a + d …. (1)
And sixth term (a_{6}) = a + (6 − 1)d
⇒ −22
= a + 5d …. (2)
Equations (1) and (2) are
the linear equations in two variables.
Let us solve it using
substitution method.
Using
equation (1), we get a = 38 – d …. (3)
Putting
the value of a in equation (2), we get
−22
= 38 – d + 5d
⇒ 4d = −60
⇒ d = −15
Putting
this value of d in equation (1), we
get
38
= a – 15
⇒ a = 53
Therefore,
we get a = 53 and d = −15
First term (a) = 53
Third term = second term + d = 38 – 15 = 23
Fourth term = third term + d = 23 – 15 = 8
Fifth term = fourth term + d = 8 – 15 = –7
Therefore, the missing terms
are 53, 23, 8 and –7.
4. Which term of the AP: 3, 8, 13, 18, …, is 78?
Solution:
Here, first term (a) = 3, common
difference (d) = 8 – 3 = 13 – 8 = 5
Let
the n^{th} term of the given
AP is 78, then a_{n} = 78
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{n} = 3 + (n − 1)5
⇒ 78
= 3 + (n − 1) 5
⇒ 75
= 5n − 5
⇒ 80
= 5n
⇒ n = 16
It means
the 16^{th} term of the given AP is 78.
5.
Find the number of terms in each of the following APs:
(i)
7, 13, 19, …, 205
(ii)
18, 15½, 13, …, −47
Solution:
(i) 7,
13, 19, …, 205
Here, first term (a) = 7, common difference (d) = 13 – 7 = 19 – 13 = 6
Let the number of terms in
the given AP is n, then a_{n} = 205
Using
the formula, a_{n} = a + (n
– 1)d, we get
205
= 7 + (n − 1)6
⇒ 205 =
7 + 6n – 6
⇒ 205
= 6n + 1
⇒ 204
= 6n
⇒ n = 34
Therefore, there are 34
terms in the given AP.
(ii)
18, 15½, 13, …, −47
Let the number of terms in
the given AP is n, then a_{n} = −47
Using the formula, a_{n} = a + (n – 1)d, we get
−47
= 18 + (n − 1)(−5/2)
⇒ −47
= 18 – (5/2)n + 5/2
⇒ −94
= 36 − 5n + 5
⇒ 5n = 135
⇒ n = 27
Therefore, there are 27
terms in the given AP.
6. Check whether –150 is a term of the AP: 11, 8, 5, 2…
Solution:
Let −150 be the n^{th} term of AP: 11, 8, 5, 2 …,
which means that a_{n} =
−150
Here, first term (a) = 11, common difference (d) = 8 – 11 = –3
Using
the formula, a_{n} = a + (n
– 1)d, we get
−150
= 11 + (n − 1) (−3)
⇒
−150 = 11 − 3n + 3
⇒ 3n = 164
⇒ n = 164/3
But, n cannot be in fraction.
Therefore, our supposition
is wrong.
Hence, −150 cannot be a term
of the given AP.
7.
Find the 31^{st} term of an AP whose 11^{th} term is
38 and the 16^{th} term is 73.
Solution:
Here, we are given that a_{11} = 38 and a_{16} = 73
Using
the formula, a_{n} = a + (n
– 1)d, we have
38
= a + (11 − 1)(d)
⇒ 38
= a + 10d ….. (1)
And 73
= a + (16 − 1)(d)
⇒ 73
= a + 15d ….. (2)
These are linear equations in
two variables.
Let
us solve these equations using substitution method.
From
equation (1), we get
a = 38 − 10d …..
(3)
Putting
the value of a in equation (2), we
get
73 =
38 − 10d + 15d
⇒ 35
= 5d
⇒ d = 7
Putting
the value of d in equation (1), we get
38
= a + 10 × 7
⇒ a = −32
Therefore, common difference (d) = 7 and first term (a) = –32
Using
the formula, a_{n} = a + (n
− 1)d, we get
a_{31 }= −32 + (31 − 1)(7)
=
−32 + 210 = 178
Therefore,
the 31^{st} term of the given AP is 178.
8.
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:
The AP consists of 50 terms and a_{50} = 106
It
is also given that a_{3} = 12
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{50 }= a + (50 − 1)d
⇒ 106
= a + 49d ….. (1)
And a_{3 }= a + (3 − 1)d
⇒ 12
= a + 2d ….. (2)
These are linear equations in
two variables.
Let
us solve these equations using substitution method.
Using
equation (1), we get
a = 106 − 49d …..
(3)
Putting
the value of a in the equation (2), we get
12 =
106 − 49d + 2d
⇒ 47d = 94
⇒ d = 2
Putting
the value of d in equation (3), we get
a = 106 – 49(2)
= 106 – 98
a = 8
Therefore,
first term (a) =
8 and common difference (d) = 2
Let us use the formula, a_{n} = a + (n – 1)d to find the 29^{th} term.
a_{29 }= 8 + (29 − 1)2
= 8 + 56
= 64
Therefore, the 29th term of
the given AP is 64.
9.
If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of
this AP is zero?
Solution: It is given that the 3^{rd} and the 9^{th} terms of an AP are 4 and –8 respectively.
It means that, a_{3} = 4 and a_{9} = −8
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{3} = a + (3 – 1)d
4 = a + 2d ….. (1)
And, a_{9} = a + (9 –
1)d
–8 = a + 8d ….. (2)
These are linear equations in
two variables.
Let
us solve these equations using substitution method.
From
equation (1), we get
a = 4 − 2d …..
(3)
Putting
the value of a in equation (2), we get
−8 =
4 − 2d + 8d
⇒ −12
= 6d
⇒ d = −2
Putting
the value of d in equation (2), we get
−8
= a + 8(−2)
⇒ −8
= a – 16
⇒ a = 8
Therefore,
first term (a) =
8 and common difference (d) = −2
We want to know which term
of the given AP is equal to zero.
Let the n^{th} term of the given AP is 0.
Using
the formula, a_{n} = a + (n
− 1)d, we get
0 =
8 + (n − 1) (−2)
⇒ 0 =
8 − 2n + 2
⇒ 0 =
10 − 2n
⇒ 2n = 10
⇒ n = 5
Therefore,
the 5^{th} term of the given AP is 0.
10.
The 17^{th} term of an AP exceeds its 10^{th} term by
7. Find the common difference.
Solution:
It is given that a_{17} = a_{10} + 7 ….. (1)
Using
the formula, a_{n} = a + (n
– 1)d, we have
a_{17} = a + 16d ….. (2)
a_{10} = a + 9d ….. (3)
Putting the values of a_{17 }and_{ }a_{10} from equations (2) and
(3) in equation (1), we get
a + 16d = a + 9d + 7
⇒ 7d = 7
⇒ d = 1
Therefore, the common difference is 1.
11.
Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54^{th} term?
Solution:
Let us first calculate the 54^{th} term of the given AP.
Here, first term (a) = 3 and common difference (d) = 15 – 3 = 12
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{54 }= a + (54 − 1)d
= 3
+ 53(12)
= 3 + 636 = 639
We have
to find that which term is 132 more than its 54^{th} term.
Let
us suppose it is n^{th} term which is 132 more than the 54^{th} term.
⇒ a_{n} = a_{54} + 132
⇒ 3 +
(n − 1)12 = 639 + 132
⇒ 3 +
12n – 12 = 771
⇒ 12n – 9 = 771
⇒ 12n = 780
⇒ n = 65
Therefore,
the 65^{th} term is 132 more than its 54^{th} term.
12.
Two APs have the same common difference. The difference between their 100^{th} terms
is 100, what is the difference between their 1000^{th} terms.
Solution:
Let the first term of 1^{st} AP = a and the first term
of 2^{nd} AP = a′
It is given that their
common difference is the same.
Let
their common difference be d.
It
is also given that the difference between their 100^{th} terms
is 100.
Using
the formula, a_{n} = a + (n – 1)d, we get
a + (100 − 1)d – [a′ + (100 − 1)d] = 100
⇒ a + 99d − a′ − 99d = 100
⇒ a − a′ = 100 ….. (1)
We have to find the
difference between their 1000^{th} terms,
which means we have to calculate:
a + (1000 − 1)d – [a′ + (1000 − 1)d] = a + 999d − a′ − 999d
= a – a′
Putting the value of a − a′ from equation (1)
in the above equation, we get
a + (1000 − 1)d – [a′ + (1000 − 1)d] = 100
Therefore,
the difference between their 1000^{th} terms is
equal to 100.
13.
How many three digit numbers are divisible by 7?
Solution: The first threedigit number divisible by 7 is 105 and the last threedigit
number divisible by 7 is 994.
Therefore, we have AP of the
form 105, 112, 119, …, 994
Let there
are total n terms in the above AP.
Then, a_{n} = 994
We have to find n here.
First term (a) = 105 and common difference (d) = 112 – 105 = 7
Using
the formula, a_{n} = a + (n
– 1)d, we get
994
= 105 + (n − 1) (7)
⇒ 994
= 105 + 7n − 7
⇒ 896
= 7n
⇒ n = 128
It
means 994 is the 128^{th} term of the above AP.
Therefore, there are 128
terms in the above AP.
Hence, there are 128 threedigit numbers which
are divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Solution: The first multiple of 4 which lies between 10 and 250 is 12 and the
last multiple of 4 which lies between 10 and 250 is 248.
Therefore, the AP is of the
form: 12, 16, 20, …, 248
Let there
are total n terms in the above AP.
Then, a_{n} = 248
We have to find n here.
Here, first term (a) = 12 and common difference (d) = 4
Using
the formula, a_{n} = a + (n – 1)d, we get
248
= 12 + (n − 1) (4)
⇒ 248
= 12 + 4n − 4
⇒ 240
= 4n
⇒ n = 60
It
means that 248 is the 60^{th} term of
the above AP.
Hence, we can say that there are 60 multiples
of 4 which lie between 10 and 250.
15.
For what value of n, are the n^{th} terms of two APs: 63, 65,
67, … and 3, 10, 17, … equal?
Solution:
Let us consider the first AP: 63, 65, 67, …
Here, first term (a) = 63 and common difference (d) = 65 – 63 = 2
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{n} = 63 + (n − 1)(2) …..
(1)
Now, let us consider the second AP: 3, 10, 17,
…
Here, first term (a) = 3 and common difference (d) = 10 – 3 = 7
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{n} = 3 + (n − 1)(7) …..
(2)
According to the question,
we have
Equation (1) = Equation (2)
⇒ 63
+ (n − 1)(2) = 3 + (n − 1)(7)
⇒ 63
+ 2n – 2 = 3 + 7n − 7
⇒ 65
= 5n
⇒ n = 13
Therefore,
the 13^{th} terms of both the APs are equal.
16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term
by 12.
Solution:
Let the first term of the given AP be a and the common difference be d.
It
is given that its 3^{rd} term is equal to 16.
Using
the formula, a_{n} = a + (n
– 1)d, we get
16
= a + (3 − 1)d
⇒ 16
= a + 2d ….. (1)
It
is also given that the 7^{th} term
exceeds the 5^{th} term by 12.
According to this condition,
we have
a_{7} = a_{5} + 12
⇒ a + (7 – 1)d = a + (5 – 1)d + 12 [Using the formula, a_{n} = a + (n – 1)d]
⇒ 6d = 4d + 12
⇒ 2d = 12
⇒ d = 6
Putting
the value of d in equation (1), we
get
16
= a + 2(6)
⇒ a = 4
Therefore, the first term (a) = 4 and common difference (d) = 6
Thus, AP is 4, 4 + 6, 4 + 12,
4 + 18, …
Hence, AP is 4, 10, 16, 22, …
17.
Find the 20^{th} term from the last term of the AP: 3, 8, 13, …,
253.
Solution: We have to find the 20^{th} term of the
given AP from the last.
Let us write the given AP from
the last: 253, …, 13, 8, 3
Here, first term (a) = 253 and common difference (d) = 8 – 13 = –5
Using
the formula, a_{n} = a + (n
– 1)d, we get
a_{20} = 253 + (20 − 1)(−5)
⇒ a_{20} = 253 + 19(−5) = 253 – 95 = 158
Therefore,
the 20^{th} term of the given AP from the last is 158.
18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of 6th and
10th terms is 44. Find the first three terms of the AP.
Solution: It is given that the sum of the 4^{th} and 8^{th} terms of an AP is 24
Then, a_{4 }+_{ }a_{8
}= 24 ….. (1)
It is
also given that the sum of the 6^{th} and 10^{th} terms is
44.
Then, a_{6} + a_{10} = 44 ….. (2)
Using
the formula, a_{n} = a + (n − 1)d
in equation (1), we get
a + (4 − 1)d + [a + (8 − 1)d] = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ….. (3)
Using
the formula, a_{n} = a + (n − 1)d
in equation (2), we get
a + (6 − 1)d + [a + (10 − 1)d] = 44
a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d =22 ….. (4)
Equations (3) and (4) are linear equations in
two variables.
Let us solve these equations using
substitution method.
From
equation (3), we have a = 12 − 5d ….. (5)
Putting
the value of a in equation (4), we get
12 −
5d + 7d = 22
⇒ 12
+ 2d = 22
⇒ 2d = 10
⇒ d = 5
Putting
the value of d in equation (5), we get
a = 12 – 5(5) = 12 – 25 = −13
Therefore,
first term (a) =
−13 and common difference (d) = 5
Therefore, the AP is –13,
–8, –3, 2 …
Its first three terms are
–13, –8 and –3.
19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an
increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
Subba Rao’s starting salary = ₹ 5000
It means that the first term
(a) = 5000
He gets an increment of ₹ 200 after every year.
Therefore, the common
difference (d) = 200
His salary after one year = 5000 + 200 = ₹ 5200
His salary after two years = 5200 + 200 = ₹ 5400
Therefore, it is an AP of
the form: 5000, 5200, 5400, 5600, …, 7000
We have to find in which
year his income reaches ₹ 7000.
Let
after n years, his salary reaches ₹ 7000.
Using
the formula, a_{n} = a + (n − 1)d,
we get
7000
= 5000 + (n − 1) (200)
⇒
7000 = 5000 + 200n − 200
⇒
7000 – 5000 + 200 = 200n
⇒
2200 = 200n
⇒ n = 11
It means after 11 years,
Subba Rao’s income would be ₹ 7000.
Hence, his income reached ₹ 7000 in the year 2006.
20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings
by ₹ 1.75. If in the nth week, her
weekly savings become ₹ 20.75, find n.
Solution:
Ramkali saved ₹ 5 in the first week of a
year. It means first term (a) = 5
Ramkali increased her weekly
savings by ₹ 1.75.
Therefore, the common
difference (d) = 1.75
Money saved by Ramkali in
the second week = a + d = 5 + 1.75 = ₹ 6.75
Money saved by Ramkali in
the third week = 6.75 + 1.75 = ₹ 8.5
Therefore, it is an AP of
the form: 5, 6.75, 8.5, …, 20.75
We have to find in which week
her weekly savings become ₹ 20.75.
Let in nth week, her weekly savings become ₹ 20.75.
Using
the formula, a_{n} = a + (n − 1)d,
we get
20.75
= 5 + (n − 1) (1.75)
⇒
20.75 = 5 + 1.75n − 1.75
⇒
17.5 = 1.75n
⇒ n = 10
It means that in the 10^{th} week, her weekly savings become ₹ 20.75.