Class10 NCRT Arithmetic Progressions Exercise – 5.2 pdf  UP Board
Arithmetic
Progressions
Exercise
– 5.2
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} the nth term of the A.P:

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 
… 
(i) Here , a = 7, d = 3 and n = 8 Therefore, substituting the value in a_{n} = a
+ (n – 1)d,
a_{n} = 7 + (8 – 1)3
a_{n} = 7 + 21 = 28
(ii) Here ,
a = 18, n = 10 and a_{n} = 0 Therefore, substituting the value in a_{n} = a + (n – 1)d,
0 = 18 + (10 – 1)d
18 = 9d
d = 2
(iii) Here ,
d = 3, n = 18 and a_{n} = 5 Therefore, substituting the value in a_{n} = a + (n – 1)d,
5 = a + (18 – 1)×3
5 = a + 17×3
5 = a – 51
a = 51 – 5 = 46
(iv) Here ,
a = 18.9, d = 2.5 and a_{n} = 3.6 Therefore, substituting the value in a_{n} = a + (n – 1)d,
3.6 = 18.9 + (n – 1)(2.5)
3.6 = 18.9 + 2.5n – 2.5
2.5n = 3.6 + 21.4
2.5n = 25.0
n = 10
(v) Here ,
a = 3.5, d = 0 and n = 105 Therefore, substituting the value in a_{n} = a + (n – 1)d,
a_{n} = 3.5 + (105 – 1)0
a_{n} = 3.5 + 0 = 3.5
2. Choose
the correct choice in the following and justify:
(i) 30^{th} term of the A.P.:
10, 7, 4, ……, is:
(A) 97 (B) 77 (C) 77 (D) 87
Here a = 10, d = 7 – 10 =  3 and n = 30,
Therefore, substituting the value in a_{n} = a + (n – 1)d,
a_{30} = 10 + (30 – 1)×3
a_{30} = 10 – 87 = 77
Hence, option (C) is correct.
(ii) 11^{th} term of the A.P.: 3, ½, 2, ……, is:
(A) 28
(B) 22 (C)
38 (D) 48½
Here a = 3, d =  ½ – (3) = 5/2 and n = 11,
Therefore, substituting the value in a_{n} = a + (n – 1)d,
a_{11} = 3 + (11 – 1)×5/2
a_{11} = 3 + 25 = 22
Hence, option (B) is correct.
3. In
the following APs, find the missing terms in the boxes:
(i) 2, _, 26
Given, a =
2 and a_{3} = 26,
To find a_{2}
a_{3}
= a + (3 – 1)d = 26 [given that]
2 + 2d =
26
d = 12
Therefore,
a_{2} = a + (2 – 1)d = 2 + 12 = 14
(ii) _,
13, _, 3
Given, a_{2}
= 13 and
a_{4} = 3,
To find a_{1}
and
a_{3}
a_{2}
= a + (2 – 1)d = 13
[given that]
a + d = 13
a = 13 –
d …………(1)
and
a_{4} = 3
a + (4 –
1)d = 3
a + 3d = 3
Substituting
the value of a in equation (1),
13 – d +
3d = 3
13 + 2d =
3
d = 5
Substituting
the value of d in equation (1),
a = 13 –
(5)
a = 13 + 5
a = 18
Therefore
a_{1} = 18
and
a_{3} = a + (3 –
1)d = 18 + 2(5) = 8
(iii) 5,
_, _, 9 ½
Given, a =
5 and
a_{4} = 9 ½,
To find a_{2}
and
a_{3}
a_{4}
= a + (4 – 1)d = 9 ½ [given that]
5 + 3d =
19/2
3d = 19/2
– 5
3d = 9/2
d = 3/2
Therefore
a_{2} = a + d = 5
+ 3/2 = 6 ½ and a_{3} = a + 2d =
5 + 2(3/2) = 8
(iv) 4,_,
_, _, _, 6
Given, a =
 4 and a_{6}
= 6,
To find a_{2},
a_{3}, a_{4} and
a_{5}
a_{6}
= a + (6 – 1)d = 6
[given that]
4 + 5d =
6
5d = 10
d = 2
Therefore
a_{2} = a + d =
4 + 2 = 2
a_{3}
= a + 2d = 4 + 2×2 = 0
a_{4}
= a + 3d = 4 + 3×2 = 2
a_{5}
= a + 4d = 4 + 4×2 = 4
(v) _, 38,
_, _, _, 22
Given, a_{2}
= 38 and a_{6}
= 22,
To find a_{1},
a_{3}, a_{4}
and
a_{5}
a_{2}
= a + (2 – 1)d = 38
[given that]
a + d = 38
a = 38 –
d …………(1)
and
a_{6} = 22
a + 5d =
22
Substituting
the value of a in equation (1),
38  d + 5d = 22
38 + 4d =
22
d = 15
Substituting
the value of d in equation (1),
a = 38 –
(15) = 53
Therefore,
a_{1} = a = 53
a_{3}
= a + 2d = 53 + 2×15 = 23
a_{4}
= a + 3d = 53 + 2×15 = 8
a_{5}
= a + 4d = 53 + 2×15 = 7
4. Which term of the A.P.: 3, 8, 13, 18,
… is
78?
Given
a = 3
and
d = 8 – 3 = 5,
Let, nth
term of the A.P.
is
78.
Therefore, a_{n} = 78
a + (n –
1)d = 78
3 + (n –
1)5 = 78
(n – 1)5 =
75
n – 1 = 15
n = 16
Hence,
16th term of the A.P.: 3, 8, 13, 18, … is
78
5. Find the number of terms in each of
the following APs?
(i) 7, 13,
19, …, 205
Here
a = 7
and
d = 13 – 7 = 6
Let there
be n terms in the A.P.
Therefore,
a_{n} = 205
a + (n –
1)d = 205
7 + (n –
1)6 = 205
(n – 1)6 =
198
n – 1 = 33
n = 34
Hence, there
are 34 terms
in A.P.
(ii) 18,
15 ½ , 13, …… , 47
Here
a = 18
and
d = 15 ½  18 = 5/2
Let there
be n terms in the A.P.
Therefore,
a_{n }= 47
a + (n –
1)d = 47
18 + (n –
1)(5/2) = 47
(n –
1)(5/2) = 65
n – 1 = 26
n = 27
Hence,
there are 27 terms
in A.P.
6. Check whether 150 is a term of the A.P.:
11, 8, 5, 2, …?
Here
a = 11
and
d = 8 – 11 = 3
Let, the
nth term of the A.P. is 150.
Therefore
a_{n} = 150
a + (n –
1)d = 150
11 + (n –
1)×3 = 150
11 – 3n +
3 = 150
3n = 164
Here n is
not an integer. Hence, 150 is not the term of A.P.: 11, 8, 5, 2, ….
7. Find the 31^{st} term of an AP
whose 11^{th} term is 38 and the 16^{th} term is 73.
Here, a_{11}
= 38 and
a_{16} = 73,
To find a_{31}
a_{11}
= a + (11 – 1)d = 38
[given that]
a + 10d =
38
a = 38 –
10d ……………(1)
and
a_{16} = 73
a + 15d =
73
Substituting
the value of a in equation (1),
38 – 10d +
15d = 73
5d = 35
d = 7
Substituting
the value of d in equation (1),
a = 38 –
10×7 = 32
Therefore
a_{31} = a + 30d = 32 + 30×7 = 178
Hence, the
31st
term of A.P 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.
Given, a_{3}
= 12 and a_{50}
= 106,
To find a_{29}
a_{3}
= a + (3 – 1)d = 12
[given that]
a + 2d =
12
a = 12 –
2d …………(1)
and
a_{50} = 106
a + 49d =
106
Substituting
the value of a in equation (1),
12 – 2d +
49d = 106
47d = 94
d = 2
Substituting
the value of d in equation (1),
a = 12 –
2×2 = 8
Therefore
a_{29} = a + 28d = 8 + 28×2 = 64
Hence, 29th term of the A.P. is 64.
9. If the 3^{rd} and the 9^{th}
terms of an AP are 4 and 8 respectively, which term of this AP is zero?
Given, a_{3}
= 4 and a_{9}
= 8,
To find n:
where a_{n} = 0
a_{3}
= a + (3 – 1)d = 4
[given that]
a + 2d = 4
a = 4 –
2d …………(1)
and
a_{9} = 8
a + 8d =
8
Substituting
the value of a in equation (1),
4 – 2d +
8d = 8
6d =  12
d = 2
Substituting
the value of d in equation (1),
a = 4 –
2×2 = 8
So putting
the value in a_{n} = 0,
a_{n}
= a + (n – 1)d = 0
8 + (n –
1)×2 = 0
n – 1 = 4
n = 5
So, A.P. 5th term of is zero.
10. The 17^{th} term of an AP
exceeds its 10^{th} term by 7. Find the common difference.
Let the
first term = a
and common difference
= d
According
to question,
a_{17}
= a_{10} + 7
a + 16d =
a + 9d + 7
7d = 7
d = a
So this
A.P. The common difference of is 1.
11. Which term of the A.P.: 3, 15, 27, 39, …
will be 132 more
than its 54^{th} term?
First term
= 3
and common difference
= 15 – 3 = 12
Let the
nth term of this A.P be 132 more than its 54th term.
According
to question,
a_{n}
= a_{54} + 132
a + (n –
1)d = a + 53d + 132
(n – 1)12
= 53×12 + 132
(n – 1)12
= 768
n – 1 =
768/12 = 64
n = 65
Hence, 65^{th}
term of the A.P.:
3,
15,
27,
39,
… will be 132 more
than its 54th
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?
Let the
first term of the first AP = A
and the common difference
= d
and the
first term of the second AP = a
and the common difference
= d
Difference
between their 100^{th} term = A_{100} – a_{100} =
100
(A + 99d)
– (a + 99d) = 100
A – a = 100
Difference
between their 1000^{th} term = A_{1000} – a_{1000}
(A + 999d)
– (a + 999d)
A – a
⇒ 100 [ A – a = 100]
Hence,
these A.P. The difference of 1000th terms will be 100.
13. How many threedigit numbers are
divisible by 7?
Three
digit numbers divisible by 7: 105, 112, 119, … , 994
Let the
total number of three digit numbers divisible by 7 be n.
Given that,
a = 105 and d
= 112 – 105 = 7,
To find n:
where a_{n}
= 994
a_{n}
= a + (n – 1)d = 994
[given that]
105 + (n –
1)7 = 994
7(n – 1) =
889
n – 1 =
889/7
n = 128
Hence, 128
three digit numbers are
divisible by 7.
14. How many multiples of 4 lie between 10
and 250?
Multiples
of 4 between
10 and
250:
12, 16, 20, … , 248
Let there
be total n multiples of 4 between 10
and 250.
Given that,
a = 12 and
d = 16 – 12 = 4,
To find n:
where a_{n}
= 248
a_{n}
= a + (n – 1)d = 248
[given that]
12 + (n –
1)4 = 248
4(n – 1) =
236
n – 1 =
236/4
n – 1 = 59
n = 60
Hence,
there are 60 multiples
of 4 between
10 and
250.
15. For what value of n, are the nth terms
of two APs: 63, 65, 67, …
and
3, 10, 17, … equal?
Let the
first term of the first AP = A = 63
and the common difference
= D = 65 – 63 = 2
Therefore
A_{n} = A + (n –
1)D
A_{n}
= 63 + (n – 1)2
and the
first term of the second AP = a = 3
and the common difference
= d = 10 – 3 = 7
Therefore, a_{n} = a + (n – 1)d
a_{n}
= 3 + (n – 1)7
According
to question,
A_{n}
= a_{n}
63 + (n –
1)2 = 3 + (n – 1)7
63 + 2n –
2 = 3 + 7n – 7
65 = 5n
n = 13
Hence, the
13th
term of both the APs are equal.
16. Determine the AP Whose third term is
16 and the 7^{th} term exceeds the 5^{th} term by 12.
Let the
first term of AP = a and
the common difference = d
Third term
= 16
a_{3}
= 16
a + 2d =
16 ………(1)
7^{th}
term exceeds the 5^{th} term by 12,
a_{7}
= a_{5} + 12
a + 6d = a
+ 4d + 12
2d = 12
d = 6
Substituting
the value of d in equation (1),
a + 2×6 =
16
a = 4
Hence, A.P. = a, a + d, a + 2d, …….. = 4,
10, 16, ……
17. Find the 20^{th} term from the
last term of the A.P.: 3, 8, 13, … , 253.
The 20^{th}
term from the last term of the AP: 3, 8, 13, … , 253 =
the 20^{th} term
from the beginning of the AP: 253, … , 13, 8, 3
In A.P.:
253, … , 13, 8, 3, the first term = 253
and the common difference
= 3 – 8 = 5
Therefore,
a_{20} = a + 19d
a_{20}
= 253 + 19×5 = 253 – 95 = 158
18. The sum of the 4^{th} and 8^{th}
terms of an AP is 24 and the sum of the 6^{th} and 10^{th}
terms is 44. Find the first three terms of the AP.
Let the
first term of the AP = a and
the common difference = d
According
to the first condition,
a_{4}
+ a_{8} = 24
a + 3d + a
+ 7d = 24
2a + 10d =
24
a + 5d =
12
a = 12 –
5d …………(1)
According
to the second condition,
a_{6}
+ a_{10} = 44
a + 5d + a
+ 9d = 44
2a + 14d =
44
a + 7d =
22
Substituting
the value of a in equation (1),
(12 – 5d)
+ 7d = 22
2d = 10
d = 5
Substituting
the value of d in equation (1),
a = 12 –
5×5
a = 13
इस A.P.
के
प्रथम
तीन
पद: a, a + d, a + 2d = 13, 8, 3
हैं।
19. Subba Rao started work in 1995 at an
annual salary of ₹5000
and received and
increment of ₹200 each year. In which year did his
income reach ₹7000?
Starting
monthly salary = a = ₹5000 And increment every year (common
difference) = d = ₹200
Let his
salary become ₹7000 in n years.
Therefore, a_{n} = 7000
a + (n –
1)d = 7000
5000 + (n
– 1)(200) = 7000
(n –
1)(200) = 2000
n – 1 = 10
n = 11
Hence, in
the 11th
year his salary became ₹7000.
20. Ramkali save
₹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.
First week
savings = a
= ₹5 and
increased weekly savings (common difference)
= d = ₹1.75
The weekly
savings in n years become ₹20.75.
Therefore, a_{n} = 20.75
a + (n – 1)d
= 20.75
5 + (n –
1)(1.75) = 20.75
(n –
1)(1.75) = 15.75
n – 1 =
15.75/1.75 = 9
n = 10
Hence, the
value of n is 10.
Download Class10 NCRT Arithmetic Progressions Exercise – 5.2 pdf
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