Exercise
5.1
Question
1:
In which
of the following situations, does the list of numbers involved make as
arithmetic progression and why?
(i) The
taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for
each additional km.
(ii) The
amount of air present in a cylinder when a vacuum pump removes
of the air remaining in the cylinder at a time.
(iii) The
cost of digging a well after every metre of digging, when it costs Rs 150 for
the first metre and rises by Rs 50 for each subsequent metre.
(iv)The
amount of money in the account every year, when Rs 10000 is deposited at
compound interest at 8% per annum.
Answer:
(i) It
can be observed that
Taxi fare
for 1st km = 15
Taxi fare
for first 2 km = 15 + 8 = 23
Taxi fare
for first 3 km = 23 + 8 = 31
Taxi fare
for first 4 km = 31 + 8 = 39
Clearly
15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding
term.
(ii) Let
the initial volume of air in a cylinder be V lit. In each stroke, the
vacuum pump removes of air remaining in the cylinder at a time.
In other words, after every stroke, only part of air will remain.
Therefore,
volumes will be
Clearly,
it can be observed that the adjacent terms of this series do not have the same
difference between them. Therefore, this is not an A.P.
(iii)
Cost of digging for first metre = 150
Cost of
digging for first 2 metres = 150 + 50 = 200
Cost of
digging for first 3 metres = 200 + 50 = 250
Cost of
digging for first 4 metres = 250 + 50 = 300
Clearly,
150, 200, 250, 300 … forms an A.P. because every term is 50 more than the
preceding term.
(iv) We
know that if Rs P is deposited at r% compound interest per annum for n
years, our money will be after n years.
Therefore,
after every year, our money will be
Clearly, adjacent
terms of this series do not have the same difference between them. Therefore,
this is not an A.P.
Question
2:
Write
first four terms of the A.P. when the first term a and the common
difference d are given as follows
(i) a
= 10, d = 10
(ii) a
= − 2, d = 0
(iii) a
= 4, d = − 3
(iv) a
= − 1 d =
(v) a
= − 1.25, d = − 0.25
Answer:
(i) a
= 10, d = 10
Let the
series be a1, a2, a3, a4,
a5 …
a1 = a = 10
a2 = a1 + d
= 10 + 10 = 20
a3 = a2 + d
= 20 + 10 = 30
a4 = a3 + d
= 30 + 10 = 40
a5 = a4 + d =
40 + 10 = 50
Therefore,
the series will be 10, 20, 30, 40, 50 …
First
four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a
= −2, d = 0
Let the
series be a1, a2, a3, a4
…
a1 = a = −2
a2 = a1 + d
= − 2 + 0 = −2
a3 = a2 + d
= − 2 + 0 = −2
a4 = a3 + d
= − 2 + 0 = −2
Therefore,
the series will be −2, −2, −2, −2 …
First
four terms of this A.P. will be −2, −2, −2 and −2.
(iii) a
= 4, d = −3
Let the
series be a1, a2, a3, a4
…
a1 = a = 4
a2 = a1 + d
= 4 − 3 = 1
a3 = a2 + d
= 1 − 3 = −2
a4 = a3 + d
= − 2 − 3 = −5
Therefore,
the series will be 4, 1, −2 −5 …
First
four terms of this A.P. will be 4, 1, −2 and −5.
(iv) a
= −1, d =
Let the
series be a1, a2, a3, a4
…
Clearly,
the series will be
………….
First
four terms of this A.P. will be .
(v) a
= −1.25, d = −0.25
Let the
series be a1, a2, a3, a4
…
a1 = a = −1.25
a2 = a1 + d
= − 1.25 − 0.25 = −1.50
a3 = a2 + d
= − 1.50 − 0.25 = −1.75
a4 = a3 + d
= − 1.75 − 0.25 = −2.00
Clearly,
the series will be 1.25, −1.50, −1.75, −2.00 ……..
First
four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.
Question
3:
For the
following A.P.s, write the first term and the common difference.
(i) 3, 1,
− 1, − 3 …
(ii) − 5,
− 1, 3, 7 …
(iii)
(iv) 0.6,
1.7, 2.8, 3.9 …
Answer:
(i) 3, 1,
−1, −3 …
Here,
first term, a = 3
Common
difference, d = Second term − First term
= 1 − 3 =
−2
(ii) −5,
−1, 3, 7 …
Here,
first term, a = −5
Common
difference, d = Second term − First term
= (−1) −
(−5) = − 1 + 5 = 4
(iii)
Here,
first term,
Common
difference, d = Second term − First term
(iv) 0.6,
1.7, 2.8, 3.9 …
Here,
first term, a = 0.6
Common
difference, d = Second term − First term
= 1.7 −
0.6
= 1.1
Question
4:
Which of
the following are APs? If they form an A.P. find the common difference d
and write three more terms.
(i) 2, 4,
8, 16 …
(ii)
(iii) −
1.2, − 3.2, − 5.2, − 7.2 …
(iv) −
10, − 6, − 2, 2 …
(v)
(vi) 0.2,
0.22, 0.222, 0.2222 ….
(vii) 0,
− 4, − 8, − 12 …
(viii)
(ix) 1,
3, 9, 27 …
(x) a,
2a, 3a, 4a …
(xi) a,
a2, a3, a4 …
(xii)
(xii)
(xiii)
(xiv) 12,
32, 52, 72 …
(xv) 12,
52, 72, 73 …
Answer:
(i) 2, 4,
8, 16 …
It can be
observed that
a2 − a1 = 4 − 2 =
2
a3 − a2 =
8 − 4 = 4
a4 − a3 = 16 − 8
= 8
i.e., ak+1−
ak is not the same every time. Therefore, the given numbers
are not forming an A.P.
(ii)
It can be
observed that
i.e., ak+1−
ak is same every time.
Therefore,
and the given numbers are in A.P.
Three
more terms are
(iii)
−1.2, −3.2, −5.2, −7.2 …
It can be
observed that
a2 − a1 = (−3.2)
− (−1.2) = −2
a3 − a2 = (−5.2)
− (−3.2) = −2
a4 − a3 = (−7.2)
− (−5.2) = −2
i.e., ak+1−
ak is same every time. Therefore, d = −2
The given
numbers are in A.P.
Three
more terms are
a5 = − 7.2 − 2 = −9.2
a6 = − 9.2 − 2 = −11.2
a7 = − 11.2 − 2 = −13.2
(iv) −10,
−6, −2, 2 …
It can be
observed that
a2 − a1 = (−6) −
(−10) = 4
a3 − a2 = (−2) −
(−6) = 4
a4 − a3 = (2) −
(−2) = 4
i.e., ak+1
− ak is same every time. Therefore, d = 4
The given
numbers are in A.P.
Three
more terms are
a5 = 2 + 4 = 6
a6 = 6 + 4 = 10
a7 = 10 + 4 = 14
(v)
It can be
observed that
i.e., ak+1
− ak is same every time. Therefore,
The given
numbers are in A.P.
Three
more terms are
(vi) 0.2,
0.22, 0.222, 0.2222 ….
It can be
observed that
a2 − a1 = 0.22 −
0.2 = 0.02
a3 − a2 = 0.222 −
0.22 = 0.002
a4 − a3 = 0.2222
− 0.222 = 0.0002
i.e., ak+1
− ak is not the same every time.
Therefore,
the given numbers are not in A.P.
(vii) 0,
−4, −8, −12 …
It can be
observed that
a2 − a1 = (−4) −
0 = −4
a3 − a2 = (−8) −
(−4) = −4
a4 − a3 = (−12) −
(−8) = −4
i.e., ak+1
− ak is same every time. Therefore, d = −4
The given
numbers are in A.P.
Three
more terms are
a5 = − 12 − 4 = −16
a6 = − 16 − 4 = −20
a7 = − 20 − 4 = −24
(viii)
It can be
observed that
i.e., ak+1
− ak is same every time. Therefore, d = 0
The given
numbers are in A.P.
Three
more terms are
(ix) 1,
3, 9, 27 …
It can be
observed that
a2 − a1 = 3 − 1 =
2
a3 − a2 = 9 − 3 =
6
a4 − a3 = 27 − 9
= 18
i.e., ak+1
− ak is not the same every time.
Therefore,
the given numbers are not in A.P.
(x) a,
2a, 3a, 4a …
It can be
observed that
a2 − a1 = 2a
− a = a
a3 − a2 = 3a
− 2a = a
a4 − a3 = 4a
− 3a = a
i.e., ak+1
− ak is same every time. Therefore, d = a
The given
numbers are in A.P.
Three
more terms are
a5 = 4a + a = 5a
a6 = 5a + a = 6a
a7 = 6a + a = 7a
(xi) a,
a2, a3, a4 …
It can be
observed that
a2 − a1 = a2
− a = a (a − 1)
a3 − a2 = a3
− a2 = a2 (a − 1)
a4 − a3 = a4
− a3 = a3 (a − 1)
i.e., ak+1
− ak is not the same every time.
Therefore,
the given numbers are not in A.P.
(xii)
It can be
observed that
i.e., ak+1
− ak is same every time.
Therefore,
the given numbers are in A.P.
And,
Three
more terms are
(xiii)
It can be
observed that
i.e., ak+1
− ak is not the same every time.
Therefore,
the given numbers are not in A.P.
(xiv) 12,
32, 52, 72 …
Or, 1, 9,
25, 49 …..
It can be
observed that
a2 − a1 = 9 − 1 =
8
a3 − a2 = 25 − 9
= 16
a4 − a3 = 49 − 25
= 24
i.e., ak+1
− ak is not the same every time.
Therefore,
the given numbers are not in A.P.
(xv) 12,
52, 72, 73 …
Or 1, 25,
49, 73 …
It can be
observed that
a2 − a1 = 25 − 1
= 24
a3 − a2 = 49 − 25
= 24
a4 − a3 = 73 − 49
= 24
i.e., ak+1
− ak is same every time.
Therefore,
the given numbers are in A.P.
And, d
= 24
Three
more terms are
a5 = 73+ 24 = 97
a6 = 97 + 24 = 121
a7 = 121 + 24 = 145
Exercise
5.2
Question
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
|
…..
|
Answer:
I. a = 7, d = 3, n
= 8, an = ?
We know
that,
For an
A.P. an = a + (n − 1) d
= 7 + (8
− 1) 3
= 7 + (7)
3
= 7 + 21
= 28
Hence,
an = 28
II. Given that
a = −18, n = 10, an
= 0, d = ?
We know
that,
an = a + (n − 1) d
0 = − 18
+ (10 − 1) d
18 = 9d
Hence,
common difference, d = 2
III. Given that
d = −3, n = 18, an
= −5
We know
that,
an = a + (n − 1) d
−5 = a
+ (18 − 1) (−3)
−5 = a
+ (17) (−3)
−5 = a
− 51
a = 51 − 5 = 46
Hence, a
= 46
IV. a = −18.9, d =
2.5, an = 3.6, n = ?
We know
that,
an = a + (n − 1) d
3.6 = −
18.9 + (n − 1) 2.5
3.6 +
18.9 = (n − 1) 2.5
22.5 = (n
− 1) 2.5
Hence, n
= 10
V. a = 3.5, d = 0, n
= 105, an = ?
We know
that,
an = a + (n − 1) d
an = 3.5 + (105 − 1) 0
an = 3.5 + 104 × 0
an = 3.5
Hence, an
= 3.5
Question
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
II 11th term of the A.P. is
A. 28 B. 22 C. − 38 D.
Answer:
I. Given that
A.P. 10,
7, 4, …
First
term, a = 10
Common
difference, d = a2 − a1 = 7 − 10
= −3
We know
that, an = a + (n − 1) d
a30 = 10 + (30 − 1) (−3)
a30 = 10 + (29) (−3)
a30 = 10 − 87 = −77
Hence,
the correct answer is C.
II. Given that, A.P.
First
term a = −3
Common
difference, d = a2 − a1
We know
that,
Hence,
the answer is B.
Question
3:
In the
following APs find the missing term in the boxes
I.
II.
III.
IV.
V.
Answer:
|
|
nth
Term Of An Arithmetic Progression
|
I.
For this
A.P.,
a = 2
a3 = 26
We know
that, an = a + (n − 1) d
a3 = 2 + (3 − 1) d
26 = 2 +
2d
24 = 2d
d = 12
a2 = 2 + (2 − 1) 12
= 14
Therefore,
14 is the missing term.
II.
For this
A.P.,
a2 = 13 and
a4 = 3
We know
that, an = a + (n − 1) d
a2 = a + (2 − 1) d
13 = a
+ d (I)
a4 = a + (4 − 1) d
3 = a
+ 3d (II)
On
subtracting (I) from (II), we obtain
−10 = 2d
d = −5
From
equation (I), we obtain
13 = a
+ (−5)
a = 18
a3 = 18 + (3 − 1) (−5)
= 18 + 2
(−5) = 18 − 10 = 8
Therefore,
the missing terms are 18 and 8 respectively.
III.
For this
A.P.,
We know
that,
Therefore,
the missing terms are and 8 respectively.
IV.
For this
A.P.,
a = −4 and
a6 = 6
We know
that,
an = a + (n − 1) d
a6
= a + (6 − 1) d
6 = − 4 +
5d
10 = 5d
d = 2
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
Therefore,
the missing terms are −2, 0, 2, and 4 respectively.
V.
For this
A.P.,
a2 = 38
a6 = −22
We know
that
an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a
+ d (1)
a6 = a + (6 − 1) d
−22 = a
+ 5d (2)
On
subtracting equation (1) from (2), we obtain
− 22 − 38
= 4d
−60 = 4d
d = −15
a = a2 − d
= 38 − (−15) = 53
a3 = a + 2d = 53 + 2
(−15) = 23
a4 = a + 3d = 53 + 3
(−15) = 8
a5 = a + 4d = 53 + 4
(−15) = −7
Therefore,
the missing terms are 53, 23, 8, and −7 respectively.
Question
4:
Which
term of the A.P. 3, 8, 13, 18, … is 78?
Answer:
3, 8, 13,
18, …
For this
A.P.,
a = 3
d = a2 − a1
= 8 − 3 = 5
Let nth
term of this A.P. be 78.
an = a + (n − 1) d
78 = 3 +
(n − 1) 5
75 = (n
− 1) 5
(n
− 1) = 15
n = 16
Hence, 16th
term of this A.P. is 78.
Question
5:
Find the
number of terms in each of the following A.P.
I. 7, 13,
19, …, 205
II.
Answer:
I. 7, 13,
19, …, 205
For this
A.P.,
a = 7
d = a2 − a1
= 13 − 7 = 6
Let there
are n terms in this A.P.
an = 205
We know
that
an = a + (n − 1) d
Therefore,
205 = 7 + (n − 1) 6
198 = (n
− 1) 6
33 = (n
− 1)
n = 34
Therefore,
this given series has 34 terms in it.
II.
For this
A.P.,
Let there
are n terms in this A.P.
Therefore,
an = −47 and we know that,
Therefore,
this given A.P. has 27 terms in it.
Question
6:
Check
whether − 150 is a term of the A.P. 11, 8, 5, 2, …
Answer:
For this
A.P.,
a = 11
d = a2 − a1
= 8 − 11 = −3
Let −150
be the nth term of this A.P.
We know
that,
Clearly, n
is not an integer.
Therefore,
−150 is not a term of this A.P.
Question
7:
Find the
31st term of an A.P. whose 11th term is 38 and the 16th
term is 73
Answer:
Given
that,
a11 = 38
a16 = 73
We know
that,
an = a + (n − 1) d
a11 = a + (11 − 1) d
38 = a
+ 10d (1)
Similarly,
a16 = a + (16 − 1) d
73 = a
+ 15d (2)
On
subtracting (1) from (2), we obtain
35 = 5d
d = 7
From
equation (1),
38 = a
+ 10 × (7)
38 − 70 =
a
a = −32
a31 = a + (31 − 1) d
= − 32 +
30 (7)
= − 32 +
210
= 178
Hence, 31st
term is 178.
Question
8:
An A.P.
consists of 50 terms of which 3rd term is 12 and the last term is
106. Find the 29th term
Answer:
Given
that,
a3 = 12
a50 = 106
We know
that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a
+ 2d (I)
Similarly,
a50 = a + (50 − 1) d
106 = a
+ 49d (II)
On
subtracting (I) from (II), we obtain
94 = 47d
d = 2
From
equation (I), we obtain
12 = a
+ 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore,
29th term is 64.
Question
9:
If the 3rd
and the 9th terms of an A.P. are 4 and − 8 respectively. Which term
of this A.P. is zero.
Answer:
Given
that,
a3 = 4
a9 = −8
We know
that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
4 = a
+ 2d (I)
a9 = a + (9 − 1) d
−8 = a
+ 8d (II)
On
subtracting equation (I) from (II), we obtain
−12 = 6d
d = −2
From
equation (I), we obtain
4 = a +
2 (−2)
4 = a
− 4
a = 8
Let nth
term of this A.P. be zero.
an = a + (n − 1) d
0 = 8 + (n
− 1) (−2)
0 = 8 − 2n
+ 2
2n =
10
n = 5
Hence, 5th
term of this A.P. is 0.
Question
10:
If 17th
term of an A.P. exceeds its 10th term by 7. Find the common
difference.
Answer:
We know
that,
For an
A.P., an = a + (n − 1) d
a17 = a + (17 − 1) d
a17 = a + 16d
Similarly,
a10 = a + 9d
It is
given that
a17 − a10 = 7
(a
+ 16d) − (a + 9d) = 7
7d
= 7
d = 1
Therefore,
the common difference is 1.
Question
11:
Which
term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th
term?
Answer:
Given
A.P. is 3, 15, 27, 39, …
a = 3
d = a2 − a1
= 15 − 3 = 12
a54 = a + (54 − 1) d
= 3 +
(53) (12)
= 3 + 636
= 639
132 + 639
= 771
We have
to find the term of this A.P. which is 771.
Let nth
term be 771.
an = a + (n − 1) d
771 = 3 +
(n − 1) 12
768 = (n
− 1) 12
(n
− 1) = 64
n = 65
Therefore,
65th term was 132 more than 54th term.
Alternatively,
Let nth
term be 132 more than 54th term.
Question
12:
Two APs
have the same common difference. The difference between their 100th
term is 100, what is the difference between their 1000th terms?
Answer:
Let the
first term of these A.P.s be a1 and a2
respectively and the common difference of these A.P.s be d.
For first
A.P.,
a100 = a1 + (100 −
1) d
= a1
+ 99d
a1000 = a1 + (1000 −
1) d
a1000 = a1 + 999d
For
second A.P.,
a100 = a2 + (100 −
1) d
= a2
+ 99d
a1000 = a2 + (1000 −
1) d
= a2
+ 999d
Given
that, difference between
100th
term of these A.P.s = 100
Therefore,
(a1 + 99d) − (a2 + 99d) = 100
a1 − a2 = 100 (1)
Difference
between 1000th terms of these A.P.s
(a1
+ 999d) − (a2 + 999d) = a1 − a2
From
equation (1),
This
difference, a1 − a2 = 100
Hence,
the difference between 1000th terms of these A.P. will be 100.
Question
13:
How many
three digit numbers are divisible by 7?
Answer:
First
three-digit number that is divisible by 7 = 105
Next
number = 105 + 7 = 112
Therefore,
105, 112, 119, …
All are
three digit numbers which are divisible by 7 and thus, all these are terms of
an A.P. having first term as 105 and common difference as 7.
The
maximum possible three-digit number is 999. When we divide it by 7, the
remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit
number that is divisible by 7.
The
series is as follows.
105, 112,
119, …, 994
Let 994
be the nth term of this A.P.
a = 105
d = 7
an = 994
n = ?
an = a + (n − 1) d
994 = 105
+ (n − 1) 7
889 = (n
− 1) 7
(n −
1) = 127
n = 128
Therefore,
128 three-digit numbers are divisible by 7.
Question
14:
How many
multiples of 4 lie between 10 and 250?
Answer:
First
multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore,
12, 16, 20, 24, …
All these
are divisible by 4 and thus, all these are terms of an A.P. with first term as
12 and common difference as 4.
When we
divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible
by 4.
The
series is as follows.
12, 16,
20, 24, …, 248
Let 248
be the nth term of this A.P.
Therefore,
there are 60 multiples of 4 between 10 and 250.
Question
15:
For what
value of n, are the nth terms of two APs 63, 65, 67,
and 3, 10, 17, … equal
Answer:
63, 65,
67, …
a = 63
d = a2 − a1
= 65 − 63 = 2
nth term of this A.P. = an
= a + (n − 1) d
an= 63 + (n − 1) 2 = 63 + 2n
− 2
an = 61 + 2n (1)
3, 10,
17, …
a = 3
d = a2 − a1
= 10 − 3 = 7
nth term of this A.P. = 3 + (n
− 1) 7
an = 3 + 7n − 7
an = 7n − 4 (2)
It is
given that, nth term of these A.P.s are equal to each other.
Equating
both these equations, we obtain
61 + 2n
= 7n − 4
61 + 4 =
5n
5n
= 65
n = 13
Therefore,
13th terms of both these A.P.s are equal to each other.
Question
16:
Determine
the A.P. whose third term is 16 and the 7th term exceeds the 5th
term by 12.
Answer:
=a3
= 16
a + (3 − 1) d = 16
a + 2d = 16 (1)
a7 − a5 = 12
[a+
(7 − 1) d] − [a + (5 − 1) d]= 12
(a
+ 6d) − (a + 4d) = 12
2d
= 12
d = 6
From
equation (1), we obtain
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore,
A.P. will be
4, 10,
16, 22, …
Question
17:
Find the
20th term from the last term of the A.P. 3, 8, 13, …, 253
Answer:
Given
A.P. is
3, 8, 13,
…, 253
Common
difference for this A.P. is 5.
Therefore,
this A.P. can be written in reverse order as
253, 248,
243, …, 13, 8, 5
For this
A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
a20 = 253 + (19) (−5)
a20 = 253 − 95
a = 158
Therefore,
20th term from the last term is 158.
Question
18:
The sum
of 4th and 8th terms of an A.P. is 24 and the sum of the
6th and 10th terms is 44. Find the first three terms of
the A.P.
Answer:
We know
that,
an = a + (n − 1) d
a4 = a + (4 − 1) d
a4 = a + 3d
Similarly,
a8 = a + 7d
a6 = a + 5d
a10 = a + 9d
Given
that, a4 + a8 = 24
a + 3d + a + 7d
= 24
2a
+ 10d = 24
a + 5d = 12 (1)
a6 + a10 = 44
a + 5d + a + 9d
= 44
2a
+ 14d = 44
a + 7d = 22 (2)
On
subtracting equation (1) from (2), we obtain
2d
= 22 − 12
2d
= 10
d = 5
From
equation (1), we obtain
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a2 = a + d = − 13 + 5
= −8
a3 = a2 + d
= − 8 + 5 = −3
Therefore,
the first three terms of this A.P. are −13, −8, and −3.
Question
19:
Subba Rao
started work in 1995 at an annual salary of Rs 5000 and received an increment
of Rs 200 each year. In which year did his income reach Rs 7000?
Answer:
It can be
observed that the incomes that Subba Rao obtained in various years are in A.P.
as every year, his salary is increased by Rs 200.
Therefore,
the salaries of each year after 1995 are
5000,
5200, 5400, …
Here, a
= 5000
d = 200
Let after
nth year, his salary be Rs 7000.
Therefore,
an = a + (n − 1) d
7000 =
5000 + (n − 1) 200
200(n
− 1) = 2000
(n
− 1) = 10
n = 11
Therefore,
in 11th year, his salary will be Rs 7000.
Question
20:
Ramkali
saved Rs 5 in the first week of a year and then increased her weekly saving by
Rs 1.75. If in the nth week, her week, her weekly savings
become Rs 20.75, find n.
Answer:
Given
that,
a = 5
d = 1.75
an = 20.75
n = ?
an = a + (n − 1) d
n − 1 = 9
n = 10
Hence, n
is 10.
Exercise
5.3
Question
1:
Find the sum of the
following APs.
(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv)
,………, to 11 terms
Answer:
(i)2, 7, 12 ,…,
to 10 terms
For this A.P.,
a = 2
d = a2 − a1 = 7 − 2 = 5
n = 10
We know that,
(ii)−37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a2 − a1 = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that,
(iv)
.
…….. , to 11 terms
For this A.P.,
n = 11
We know that,
Question 2:
(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv)
,………, to 11 termsAnswer:
 | Use Of Formula For Sum Of n Terms Of An Arithmetic Progression | Read now to understand this topic better » |
For this A.P.,
a = 2
d = a2 − a1 = 7 − 2 = 5
n = 10
We know that,
(ii)−37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a2 − a1 = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that,
(iv)
.
…….. , to 11 termsFor this A.P.,
n = 11
We know that,
Question 2:
Find the sums given
below
(i) 7 +
+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer:
(i)7 +
+ 14 + …………+ 84
For this A.P.,
a = 7
l = 84
Let 84 be the nth term of this A.P.
l = a + (n − 1)d
22 = n − 1
n = 23
We know that,
(ii)34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a2 − a1 = 32 − 34 = −2
l = 10
Let 10 be the nth term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
(iii)(−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
a = −5
l = −230
d = a2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
Question 3:
(i) 7 +
+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer:
(i)7 +
+ 14 + …………+ 84For this A.P.,
a = 7
l = 84
Let 84 be the nth term of this A.P.
l = a + (n − 1)d
22 = n − 1
n = 23
We know that,
(ii)34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a2 − a1 = 32 − 34 = −2
l = 10
Let 10 be the nth term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
(iii)(−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
a = −5
l = −230
d = a2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
Question 3:
In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x)Given l = 28, S = 144 and there are total 9 terms. Find a.
Answer:
(i) Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
∴ 50 = 5 + (n − 1)3
45 = (n − 1)3
15 = n − 1
n = 16
(ii) Given that, a = 7, a13 = 35
As an = a + (n − 1) d,
∴ a13 = a + (13 − 1) d
35 = 7 + 12 d
35 − 7 = 12d
28 = 12d
(iii)Given that, a12 = 37, d = 3
As an = a + (n − 1)d,
a12 = a + (12 − 1)3
37 = a + 33
a = 4
(iv) Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d (i)
On multiplying equation (1) by 2, we obtain
30 = 2a + 4d (iii)
On subtracting equation (iii) from (ii), we obtain
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9 = 8
(v)Given that, d = 5, S9 = 75
As
,
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
an = a + (n − 1)d
a9 = a + (9 − 1) (5)
(vi) Given that, a = 2, d = 8, Sn = 90
As,
90 = n [2 + (n − 1)4]
90 = n [2 + 4n − 4]
90 = n (4n − 2) = 4n2 − 2n
4n2 − 2n − 90 = 0
4n2 − 20n + 18n − 90 = 0
4n (n − 5) + 18 (n − 5) = 0
(n − 5) (4n + 18) = 0
Either n − 5 = 0 or 4n + 18 = 0
n = 5 or
However, n can neither be negative nor fractional.
Therefore, n = 5
an = a + (n − 1)d
a5 = 2 + (5 − 1)8
= 2 + (4) (8)
= 2 + 32 = 34
(vii) Given that, a = 8, an = 62, Sn = 210
n = 6
an = a + (n − 1)d
62 = 8 + (6 − 1)d
62 − 8 = 5d
54 = 5d
(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n (i)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n2 + 10n
2n2 − 10n − 28 = 0
n2 − 5n −14 = 0
n2 − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we obtain
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8
(ix)Given that, a = 3, n = 8, S = 192
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6
(x)Given that, l = 28, S = 144 and there are total of 9 terms.
(16) × (2) = a + 28
32 = a + 28
a = 4
Question 4:
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Answer:
Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 − a1 = 17 − 9 = 8
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
or n = 12
n cannot be
.
As the number of terms can neither be negative nor fractional,
therefore, n = 12 only.
Question 5:
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer:
Given that,
a = 5
l = 45
Sn = 400
n = 16
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
Question 6:
The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer:
Given that,
a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
Question 7:
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer:
d = 7
a22 = 149
S22 = ?
an = a + (n − 1)d
a22 = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Question 8:
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
Given that,
a2 = 14
a3 = 18
d = a3 − a2 = 18 − 14 = 4
a2 = a + d
14 = a + 4
a = 10
= 5610
Question 9:
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer:
Given that,
S7 = 49
S17 = 289
7 = (a + 3d)
a + 3d = 7 (i)
Similarly,
17 = (a + 8d)
a + 8d = 17 (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
= n2
Question 10:
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x)Given l = 28, S = 144 and there are total 9 terms. Find a.
Answer:
(i) Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
∴ 50 = 5 + (n − 1)3
45 = (n − 1)3
15 = n − 1
n = 16
(ii) Given that, a = 7, a13 = 35
As an = a + (n − 1) d,
∴ a13 = a + (13 − 1) d
35 = 7 + 12 d
35 − 7 = 12d
28 = 12d
(iii)Given that, a12 = 37, d = 3
As an = a + (n − 1)d,
a12 = a + (12 − 1)3
37 = a + 33
a = 4
(iv) Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d (i)
On multiplying equation (1) by 2, we obtain
30 = 2a + 4d (iii)
On subtracting equation (iii) from (ii), we obtain
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9 = 8
(v)Given that, d = 5, S9 = 75
As
,25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
an = a + (n − 1)d
a9 = a + (9 − 1) (5)
(vi) Given that, a = 2, d = 8, Sn = 90
As
,90 = n [2 + (n − 1)4]
90 = n [2 + 4n − 4]
90 = n (4n − 2) = 4n2 − 2n
4n2 − 2n − 90 = 0
4n2 − 20n + 18n − 90 = 0
4n (n − 5) + 18 (n − 5) = 0
(n − 5) (4n + 18) = 0
Either n − 5 = 0 or 4n + 18 = 0
n = 5 or
However, n can neither be negative nor fractional.
Therefore, n = 5
an = a + (n − 1)d
a5 = 2 + (5 − 1)8
= 2 + (4) (8)
= 2 + 32 = 34
(vii) Given that, a = 8, an = 62, Sn = 210
n = 6
an = a + (n − 1)d
62 = 8 + (6 − 1)d
62 − 8 = 5d
54 = 5d
(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n (i)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n2 + 10n
2n2 − 10n − 28 = 0
n2 − 5n −14 = 0
n2 − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we obtain
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8
(ix)Given that, a = 3, n = 8, S = 192
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6
(x)Given that, l = 28, S = 144 and there are total of 9 terms.
(16) × (2) = a + 28
32 = a + 28
a = 4
Question 4:
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Answer:
Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 − a1 = 17 − 9 = 8
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
or n = 12n cannot be
.
As the number of terms can neither be negative nor fractional,
therefore, n = 12 only.Question 5:
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer:
Given that,
a = 5
l = 45
Sn = 400
n = 16
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
Question 6:
The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer:
Given that,
a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
Question 7:
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer:
d = 7
a22 = 149
S22 = ?
an = a + (n − 1)d
a22 = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Question 8:
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
Given that,
a2 = 14
a3 = 18
d = a3 − a2 = 18 − 14 = 4
a2 = a + d
14 = a + 4
a = 10
= 5610
Question 9:
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer:
Given that,
S7 = 49
S17 = 289
7 = (a + 3d)
a + 3d = 7 (i)
Similarly,
17 = (a + 8d)
a + 8d = 17 (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
= n2
Question 10:
Show that a1, a2 … , an
, … form an AP where an is defined as
below
(i) an = 3 + 4n
(ii) an = 9 − 5n
Also find the sum of the first 15 terms in each case.
Answer:
(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
= 15 × 35
= 525
(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
= −465
(i) an = 3 + 4n
(ii) an = 9 − 5n
Also find the sum of the first 15 terms in each case.
Answer:
(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
= 15 × 35
= 525
(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
= −465
No comments:
Post a Comment