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 an the nth term of the A.P:
|
|
a |
d |
n |
an |
|
(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 an = a
+ (n – 1)d,
an = 7 + (8 – 1)3
an = 7 + 21 = 28
(ii) Here,
a = -18, n = 10 and an = 0 Therefore, substituting the value in an = a + (n – 1)d,
0 = -18 + (10 – 1)d
18 = 9d
d = 2
(iii) Here,
d = -3, n = 18 and an = -5 Therefore, substituting the value in an = 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 an = 3.6 Therefore, substituting the value in an = 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 an = a + (n – 1)d,
an = 3.5 + (105 – 1)0
an = 3.5 + 0 = 3.5
2. Choose
the correct choice in the following and justify:
(i) 30th 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 an = a + (n – 1)d,
a30 = 10 + (30 – 1)×-3
a30 = 10 – 87 = -77
Hence, option (C) is correct.
(ii) 11th 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 an = a + (n – 1)d,
a11 = -3 + (11 – 1)×5/2
a11 = -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 a3 = 26,
To find a2
a3
= a + (3 – 1)d = 26 [given that]
2 + 2d =
26
d = 12
Therefore,
a2 = a + (2 – 1)d = 2 + 12 = 14
(ii) _,
13, _, 3
Given, a2
= 13 and
a4 = 3,
To find a1
and
a3
a2
= a + (2 – 1)d = 13
[given that]
a + d = 13
a = 13 –
d …………(1)
and
a4 = 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
a1 = 18
and
a3 = a + (3 –
1)d = 18 + 2(-5) = 8
(iii) 5,
_, _, 9 ½
Given, a =
5 and
a4 = 9 ½,
To find a2
and
a3
a4
= a + (4 – 1)d = 9 ½ [given that]
5 + 3d =
19/2
3d = 19/2
– 5
3d = 9/2
d = 3/2
Therefore
a2 = a + d = 5
+ 3/2 = 6 ½ and a3 = a + 2d =
5 + 2(3/2) = 8
(iv) -4,_,
_, _, _, 6
Given, a =
- 4 and a6
= 6,
To find a2,
a3, a4 and
a5
a6
= a + (6 – 1)d = 6
[given that]
-4 + 5d =
6
5d = 10
d = 2
Therefore
a2 = a + d =
-4 + 2 = -2
a3
= a + 2d = -4 + 2×2 = 0
a4
= a + 3d = -4 + 3×2 = 2
a5
= a + 4d = -4 + 4×2 = 4
(v) _, 38,
_, _, _, 22
Given, a2
= 38 and a6
= -22,
To find a1,
a3, a4
and
a5
a2
= a + (2 – 1)d = 38
[given that]
a + d = 38
a = 38 –
d …………(1)
and
a6 = -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,
a1 = a = 53
a3
= a + 2d = 53 + 2×-15 = 23
a4
= a + 3d = 53 + 2×-15 = 8
a5
= 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, an = 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,
an = 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,
an = -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
an = -150
a + (n –
1)d = -150
11 + (n –
1)×-3 = -150
11 – 3n +
3 = -150
-3n = -164
n = 164/3
=
Here n is
not an integer. Hence, -150 is not the term of A.P.: 11, 8, 5, 2, ….
7. Find the 31st term of an AP
whose 11th term is 38 and the 16th term is 73.
Here, a11
= 38 and
a16 = 73,
To find a31
a11
= a + (11 – 1)d = 38
[given that]
a + 10d =
38
a = 38 –
10d ……………(1)
and
a16 = 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
a31 = a + 30d = -32 + 30×7 = 178
Hence, the
31st
term of A.P is 178.
8. An AP consists of 50 terms of which 3rd
term is 12 and the last term is 106. Find the 29th term.
Given, a3
= 12 and a50
= 106,
To find a29
a3
= a + (3 – 1)d = 12
[given that]
a + 2d =
12
a = 12 –
2d …………(1)
and
a50 = 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
a29 = a + 28d = 8 + 28×2 = 64
Hence, 29th term of the A.P. 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?
Given, a3
= 4 and a9
= -8,
To find n:
where an = 0
a3
= a + (3 – 1)d = 4
[given that]
a + 2d = 4
a = 4 –
2d …………(1)
and
a9 = -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 an = 0,
an
= 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 17th term of an AP
exceeds its 10th term by 7. Find the common difference.
Let the
first term = a
and common difference
= d
According
to question,
a17
= a10 + 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 54th 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,
an
= a54 + 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, 65th
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 100th terms is 100, what is
the difference between their 1000th 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 100th term = A100 – a100 =
100
(A + 99d)
– (a + 99d) = 100
A – a = 100
Difference
between their 1000th term = A1000 – a1000
(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 three-digit 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 an
= 994
an
= 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 an
= 248
an
= 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
An = A + (n –
1)D
An
= 63 + (n – 1)2
and the
first term of the second AP = a = 3
and the common difference
= d = 10 – 3 = 7
Therefore, an = a + (n – 1)d
an
= 3 + (n – 1)7
According
to question,
An
= an
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 7th term exceeds the 5th term by 12.
Let the
first term of AP = a and
the common difference = d
Third term
= 16
a3
= 16
a + 2d =
16 ………(1)
7th
term exceeds the 5th term by 12,
a7
= a5 + 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 20th term from the
last term of the A.P.: 3, 8, 13, … , 253.
The 20th
term from the last term of the AP: 3, 8, 13, … , 253 =
the 20th 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,
a20 = a + 19d
a20
= 253 + 19×-5 = 253 – 95 = 158
18. The sum of the 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.
Let the
first term of the AP = a and
the common difference = d
According
to the first condition,
a4
+ a8 = 24
a + 3d + a
+ 7d = 24
2a + 10d =
24
a + 5d =
12
a = 12 –
5d …………(1)
According
to the second condition,
a6
+ a10 = 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, an = 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, an = 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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