Lesson: Arithmetic Progressions
Exercise 5.1 (18)
Question: 1
Choose the correct answer from the given four
options:
In an AP, if , then is
d 4,
n 7,
n
a 4
a
a.
6
b.
7
c.
20
d.
28
Solution:
(d)
Given:
d 4,
n 7,
n
a 4
By the formula of term,
th
n
n
a a n 1 d ......(1)
Now, put the values in equation (1),
4 a 7 1 4
4 a 6 4
4 a 24
a 4 24
a 28
Thus, the value of is .
a
28
Question: 2
In an AP, if , then will be
a 3.5,
d 0,
n 101
n
a
a.
0
b.
3.5
c.
103.5
d.
104.5
Solution:
(b)
Given:
a 3.5,
d 0,
n 101
By the formula of term,
th
n
n
a a n 1 d ......(1)
Now, put the values in equation (1),
n
a 3 5 101. 1 0
n
a 3 5. 100 0
n
a 3.5
Thus, the value of is .
n
a
3.5
Note:
If , then all terms are same.
d 0
Question: 3
The list of numbers is
10,
6,
2,
2,.....
a. an AP with
d 16
b. an AP with
d 4
c. an AP with
d 4
d. not an AP
Solution:
(b)
A sequence of numbers in an AP is
1 2 3
....d d d .....
Where
1 2 1
d a a ,
2 3 2
d a a ,
3 4 3
d a a
1 2 1
d a a
6 10
6 10
4
2 3 2
d a a
2 6
2 6
4
3 4 3
d a a
2 2
2 2
4
So, .
1 2 3
d d d 4
Thus, the given sequence is an AP with .
d 4
Question: 4
The term of the AP: is
th
11
5,
5
,
2
0,
5
,
2
........
a.
20
b.
20
c.
30
d.
30
Solution:
(b)
Given:
5,
5
,
2
0,
5
,
2
........
Compare the given AP with the AP .
1
a ,
2
a ,
3
a ,
n
...... .,a.
1
a 5,
2
5
a ,
2
3
a 0
The first term,
1
a a 5
Common difference,
5
d 5
2
5
5
2
5 10
2
5
2
n 11
By the formula of term,
th
n
n
a a n 1 d ......(1)
Now, put the values in equation (1).
11
5
a 5 11 1
2
11
5
a 5 10
2
11
a 5 25
11
a 20
Thus, the 11
th
term is .
20
Question: 5
The first four terms of an AP, whose first term is
2
and the common difference is , are
2
a.
2,
0,
2,
4
b.
2,
4,
8,
16
c.
2,
4,
6,
8
d.
2,
4,
8,
16
Solution:
(c)
Given: and .
1
a 2
d 2
Now, the sequence will be
2 1
a a d
2 2
4
3 2
a a d
4 2
6
4 3
a a d
6 2
8
Thus, the first four terms are .
2,
4,
6,
8
Question: 6
The 21
st
term of the AP whose first two terms are
and is
3
4
a.
17
b.
137
c.
143
d.
143
Solution:
(b)
Given:
1
a a 3,
2
a 4
2 1
d a a
4 3
4 3
7
By the formula of term,
th
n
n
a a n 1 d ......(1)
Now, put the values in equation (1).
21
a 3 21 1 7
3 20 7
3 140
137
Thus, the 21
st
term is .
137
Question: 7
If the 2
nd
term of an AP is 13 and the 5
th
term is 25,
what is its 7
th
term?
a.
30
b.
33
c.
37
d.
38
Solution:
(b)
Given: and
2
a 13
5
a 25
Now,
2
a a 2 1 d
13 a d
a d 13 ......(1)
5
a a 5 1 d
25 a 4d
a 4d 25 ......(2)
Subtract equation (1) from (2),
a 4d a d 25 13
a 4d a d 25 13
3d 12
d 4
Put the value of in equation (1),
d
a d 13
a 4 13
a 13 4
a 9
Now, find the value of .
7
a
7
a a 6d
9 6 4
9 24
33
Thus, the 7
th
term is .
33
Question: 8
Which term of the AP is ?
21,
42,
63,
....84, .....
210
a.
th
9
b.
th
10
c.
th
11
d.
th
12
Solution:
(b)
Given AP:
21,
42,
63,
....84, .....
Compare the given AP with the AP .
1
a ,
2
a ,
3
a ,
n
...... .,a.
1
a 21,
2
a 42,
3
a 63
The first term,
1
a a 21
Common difference
d 42 21 21
n
a 210
By the formula of term,
th
n
n
a a n 1 d ......(1)
Now, put the values in equation (1).
210 21 n 1 21
210 21 n 1 21
189 n 1 21
189
n 1
21
n 1 9
n 10
Thus, .
n 10
Question: 9
If the common difference of an AP is , then what
5
is ?
18 13
a a
a.
5
b.
20
c.
25
d.
30
Solution:
(c)
Given:
d 5
Now, find the value of .
18 13
a a
18 13
a a a 18 1 d a 13 1 d
a 17d a 12d
a 17d a 12d
5d
5 5
25
Thus, .
18 13
a a 25
Question: 10
What is the common difference of an AP in which
?
18 14
a a 32
a.
8
b.
8
c.
4
d.
4
Solution:
(a)
Given:
18 14
a a 32
Now, find the value of .
d
18 14
a a 32
a 18 1 d a 14 1 d 32
a 17d a 13d 32
4d 32
d 8
Thus, .
d 8
Question: 11
Two APs have the same common difference. The
first term of one of these is and that of the other
1
is . Then the difference between their 4
th
terms is
8
a.
1
b.
8
c.
7
d.
9
Solution:
(c)
Given: and .
1
a 1
1
d 8
Let the same common difference of two APs be .
d
So, .
1
d d,
1
d d
4 4 1 1 1 1
a a a 4 1 d a 4 1 d
  1 3d 8 3d
1 3d 8 3d
7
Thus, the difference between their 4
th
terms is .
7
Question: 2
If 7 times the 7
th
term of an AP is equal to 11 times
its 11
th
term, then its 18
th
term will be
a.
7
b.
11
c.
18
d.
0
Solution:
(d)
Given:
7 11
7a 11a
Now,
7 11
7a 11a
7 a 7 1 d 11 a 11 1 d
7 a 6d 11 a 10d
7a 42d 11a 110d
4a 68d 0
a 17d 0
18
a 0
Know that .
18
a a 17d
Thus, 18
th
term is .
0
Question: 13
The 4
th
term from the end of the AP:
11,
8,
is
....5, ....,
49
a.
37
b.
40
c.
43
d.
58
Solution:
(b)
Given AP:
11,
8,
5,
........,49
Reversing the AP,
....49, ....,
5,
8,
11
Compare the given AP with the AP
1
a ,
2
a ,
3
....a , ....,
n
a
.
1
a 49
The first term,
a 49
n 4
Common difference,
d 8 5
8 5
3
By the formula of term,
th
n
4
a 49 4 1 3
49 3 3
49 9
40
Thus, the value of is .
4
a
40
Question: 14
The famous mathematician associated with finding
the sum of the first natural numbers is
100
a. Pythagoras
b. Newton
c. Gauss
d. Euclid
Solution:
(c)
Gauss is the famous mathematician who is known
for finding the sum of the first natural
100
numbers, i.e., .
........1 2 3 100
Let .
a 1,
d 1,
n 100
Now, find the value of .
n
S
n
n
S 2a n 1 d
2
100
100
S 2 1 100 1 1
2
50 2 99
50 101
5050
Thus, the sum of the first natural numbers is
100
.
5050
Question: 15
If the first term of an AP is and the common
5
difference is 2, then the sum of the first 6 terms is
a.
0
b.
5
c.
6
d.
15
Solution:
(a)
Given:
a 5,
d 2,
n 6
By the formula of ,
n
S
n
n
S 2a n 1 d
2
6
6
S 2 5 6 1 2
2
3 10 5 2
3 10 10
0
Thus, .
6
S 0
Question: 16
The sum of the first 16 terms of the AP:
.1 .0, .6,2, ....
is
a.
320
b.
320
c.
352
d.
400
Solution:
(a)
Given AP:
10,
6,
...2, ....
Compare the given AP with the AP
1
a ,
2
a ,
3
a ,.
n
...... ,a.
1
a 10,
2
a 6,
3
a 2
The first term,
a 10
Common difference,
d 6 10
4
n 16
By the formula of ,
n
S
n
n
S 2a n 1 d
2
16
16
S 2 10 16 1 4
2
8 20 15 4
8 20 60
8 40
320
Thus, .
16
S 320
Question: 17
In an AP is and , then is
a 1,
n
a 20
n
S 399
n
a.
19
b.
21
c.
38
d.
42
Solution:
(c)
Given: and
a 1,
n
a 20
n
S 399
By the formula of ,
n
S
n
n
S 2a n 1 d
2
n
n
S a a n 1 d
2
n
n
399 a a
2
n
399 1 20
2
399 2
n
21
n 38
Thus, .
n 38
Question: 18
The sum of the first five multiples of 3 is
a.
45
b.
55
c.
65
d.
75
Solution:
(a)
1
st
five multiples of are
3
3,
6,
9,
12,
....15, .....
Compare the given AP with the AP =.
1
a ,
2
a ,
n
........, ,a
1
a 3,
2
a 6,
3
a 9
The first term,
a 3
Common difference,
d 6 3
3
n 5
By the formula of ,
n
S
n
n
S 2a n 1 d
2
5
5
S 2 3 5 1 3
2
5
6 4 3
2
5
18
2
45
Thus, the sum of the first five multiples of is .
3
45
Exercise 5.2 (8)
Question: 1
Which of the following form is an AP? Justify your
(i)
1,
1,
1,
1,
....
(ii)
0,
2,
0,
....
2,
(iii)
1,
1,
2,
2,
3,
3,....
(iv)
11,
22,
33,....
(v)
1
,
2
1
,
3
1
,
4
....
(vi)
2,
2
2 ,
3
2 ,
4
2 ,....
(vii)
3,
12,
27,
48,....
Solution:
(i) Given AP:
1,
1,
1,
1,....
Compare the sequence with AP
1
a ,
2
a ,
3
a ,
n
......,a
1
a 1,
2
a 1,
3
a 1,
4
a 1
Now, find the common difference.
2 1
a a 1 1 1 1 0
3 2
a a 1 1 1 1 0
4 3
a a 1 1 1 1 0
Clearly, the common difference is same for
each pair of terms.
Thus, the given sequence is form an AP.
(ii) Given AP:
0,
2,
0,
2,....
Compare the sequence with AP
1
a ,
2
a ,
3
...a , ....,
.
n
a
1
a 0,
2
a 2,
3
a 0,
4
a 2
Now, find the common difference.
2 1
a a 2 0 2
3 2
a a 0 2 2
4 3
a a 2 0 2
Clearly, the common difference is not same for
each pair of terms.
Thus, the given sequence is not form an AP.
(iii) Given AP:
1,
1,
2,
2,
3,
3,....
Compare the sequence with AP
1
a ,
2
a ,
3
...a , ....,
.
n
a
1
a 1,
2
a 1,
3
a 2,
4
a 2
Now, find the common difference.
2 1
a a 1 1 0
3 2
a a 2 1 1
4 3
a a 2 2 0
Clearly, the common difference is not same for
each pair of terms.
Thus, the given sequence does not form an AP.
(iv) Given AP:
11,
22,
33,....
Compare the sequence with AP
1
a ,
2
a ,
3
...a , ....,
.
n
a
1
a 11,
2
a 22,
3
a 33
Now, find the common difference.
2 1
a a 22 11 11
3 2
a a 33 22 11
Clearly, the common difference is same for
each pair of terms.
Thus, the given sequence is form an AP.
(v) Given AP:
1
,
2
1
,
3
1
,
4
....
Compare the sequence with AP
1
a ,
2
a ,
3
...a , ....,
.
n
a
1
1
a ,
2
2
1
a ,
3
3
1
a
4
Now, find the common difference.
2 1
1 1 2 3 1
a a
3 2 6 6
3 2
1 1 3 4 1
a a
4 3 12 12
Clearly, the common difference is not same for
each pair of terms.
Thus, the given sequence does not form an AP.
(vi) Given AP:
2,
2
2
3
2 ,
4
2 ,....
Compare the sequence with AP
1
a ,
2
a ,
3
...a , ....,
.
n
a
1
a 2,
2
2
a 2
3
3
a 2 ,
4
4
a 2
Now, find the common difference.
2
2 1
a a 2 2 4 2 2
3 2
3 2
a a 2 2 8 4 4
4 3
4 3
a a 2 2 16 8 8
Clearly, the common difference is not same for
each pair of terms.
Thus, the given sequence does not form an AP.
(vii) Given AP:
3,
12,
27,
48,....
Compare the sequence with AP
1
a ,
2
a ,
3
a ,.
.
n
......,a
1
a 3,
2
a 12,
3
a 27,
4
a 48
Now, find the common difference.
2 1
a a 12 3 2 3 3 3 2 1 3
3 2
a a 27 12 3 3 2 3 3 3 2 3
4 3
a a 48 27 4 3 3 3 3 4 3 3
Clearly, the common difference is same for
each pair of terms.
Thus, the given sequence forms an AP.
Question: 2
Justify whether it is true to say that
1,
3
,
2
2,
5
,
2
....
forms an AP as .
2 1 3 2
a a a a
Solution:
Given AP:
1,
3
,
2
2,
5
,
2
....
Compare the sequence with AP .
1
a ,
2
a ,
3
a ,
n
.......,a
1
a 1,
2
3
a ,
2
3
a 2,
4
5
a
2
Now, find the common difference.
2 1
3 3 2 1
a a 1
2 2 2
3 2
3 4 3 1
a a 2
2 2 2
4 3
5 5 4 9
a a 2
2 2 2
Thus, but .
2 1 3 2
a a a a
2 1 3 2 4 3
a a a a a a
Hence, the given sequence does not form an AP
and the given statement is false.
Question: 3
For the AP: , can we find directly
3,
7,
.11, ....
without actually finding and ? Give
30 20
a a
30
a
20
a
Solution:
Given AP:
3, 7, 11,.....
Compare the given AP with the AP .
1
a ,
2
a ,
3
a ,
n
...... ,a.
1
a 3,
2
a 7,
3
a 11
The first term,
a 3
Common difference,
d 7 3
7 3
4
30
a a 30 1 d
a 29d
And
20
a a 20 1 d
a 19d
So,
30 20
a a a 29d a 19d
a 29d a 19d
10d
10 4
40
Thus, we can find directly without actually
30 20
a a
finding and .
30
a
20
a
Hence, .
30 20
a a 40
Question: 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 10
th
terms is the same
as the difference between their 21
st
terms, which is
the same as the difference between any two
corresponding terms. Why?
Solution:
Given: and
1
a 2
1
a 7
Let be the same common difference of two APs.
d
So, and .
1
d d
1
d d
Now, find the difference between tenth terms,
10 10 1 1 1 1
a a a 10 1 d a 10 1 d
2 9d 7 9d
2 9d 7 9d
5
Now, find the difference between twenty first
terms,
21 21 1 1 1 1
a a a 21 1 d a 21 1 d
2 20d 7 20d
2 20d 7 20d
5
Thus,
10 10 21 21
a a a a 5
Now,
n n 1 1 1 1
a a a n 1 d a n 1 d
2 n 1 d 7 n 1 d
2 nd d 7 nd d
2 nd d 7 nd d
5
Thus, .
n n
a a 5
Hence, the difference between any two
corresponding terms is same as .
5
Question: 5
Is a term of the AP: ? Justify your
0
31,
28,
25,.....
Solution:
Given AP:
31,
28,
25,.....
Compare the given AP with the AP
1
a ,
2
a ,
3
....a , ....,
n
a
1
a 31,
2
a 28,
3
a 25
The first term,
a 31
Common difference,
d 28 31
3
0 31 n 1 3
0 31 3 n 1
31 3 n 1
31
n 1
3
31
n 1
3
Take LCM on right hand side.
31 3
n
3
34
n
3
1
n 11
3
This is not a natural number.
Thus, is not a natural number so is not a term
n
n
a
of the given AP.
Question: 6
The taxi fare after each km, when the fare is Rs 15
for the first km and Rs 8 for each additional km,
does not form an AP as the total fare (in Rs) after
each km is
15,
8,
8,
8,......
Is the statement true? Give reasons.
Solution:
Given:
are not the total fare for km,
15,
8,
8,
8,......
1,
2,
3,
4,
respectively.
Total fare for 1 km
Rs.15
Total fare for 2 km
Rs Rs R.15 .8 s.23
Total fare for 3 km
Rs Rs R.23 .8 s.31
Total fare for 4 km
Rs Rs R.31 .8 s.39
Total fare for are
1km,
2km,
3km,
..4km, ....
15,
23,
respectively.
31,
39,.....
Now, find the common difference.
2 1
a a 23 15 8
3 2
a a 31 23 8
4 3
a a 39 31 8
Thus, the total fare from an AP as .
15,
23,
31,
39,....
Question: 7
In which of the following situations, does the lists
of numbers involved form an AP? Give reasons for
(i) The fee charged from a student every month
by a school for the whole session, when the
monthly fee is Rs 400.
(ii) The fee charged every month by a school from
Classes I to XII, when the monthly fee for
Class I is Rs 250, and it increases by Rs 50 for
the next higher class.
(iii) The amount of money in the account of Varun
at the end of every year when Rs 1000 is
deposited at simple interest of 10% per annum.
(iv) The number of bacteria in a certain food item
after each second, when they double in every
second.
Solution:
(i) The monthly fee of a student is Rs. 400.
For the whole session, the fee charged from a
student is 400, 400, 400, 400, ….
So,
2 1
a a 400 400 0
3 2
a a 400 400 0
4 3
a a 400 400 0
Thus, the sequence of numbers is an AP.
(ii) Given:
Fee for class 1
st
Rs.250
Fee for class 2
nd
Rs 250 50. Rs.300
Fee for class 3
rd
Rs 300 50. Rs.350
Fee for class 4
th
Rs 350 50. Rs.400
So, sequence consist
250,
300,
350,
.400, ....
12
terms.
Compare the given AP with the AP
1
a ,
2
a ,
3
a ,.
.
n
..... ,a.
1
a 250,
2
a 300,
3
a 350
2 1
a a 300 250 50
3 2
a a 350 300 50
4 3
a a 400 350 50
Thus, the common difference is .
50
Thus, the numbers are in
250,
300,
350,
.400, ....
AP.
(iii) Given: Simple Interest
Rs.1000
Rate per annum
10%
Now, calculate the simple interest.
PRT
SI
100
1000 10 1
100
100
Thus, at the end of year, in the account of
Varun Rs. 100 is credited.
Money at the beginning of 1
st
year
Rs.1000
Money at the end of 1
st
year with interest
Rs 1000 10. .0 Rs 1100
Money at the end of 2
nd
year
Rs 1100 10. .0 Rs 1200
Money at the end of 3
rd
year
Rs 1200 10. .0 Rs 1300
Money at the end of 4
th
year
Rs 1300 10. .0 Rs 1400
Amount of money at the end of each year
1000,
1100,
1200,
1300,
.1400, ....
Compare the given AP with the AP
1
a ,
2
a ,
3
a ,.
.
n
..... ,a.
1
a 1000,
2
a 1100,
3
a 1200,
4
a 1300
2 1
a a 1100 1000 100
3 2
a a 1200 1100 100
4 3
a a 1300 1200 100
Thus, the difference is 100.
Hence, the list is an AP.
(iv) Let the number of bacteria present initially
x
Number of bacteria present after sec
1
2x
Number of bacteria present after sec
2
2 2x 4x
Number of bacteria present after sec
3
2 4x 8x
Number of bacteria present after sec
4
2 8x 16x
So, the number of bacteria are given by
x,
2x,
4x,
8x,
.16x, ....
Compare the given AP with the AP
1
a ,
2
a ,
3
a ,
.
n
...... ,a.
1
a x,
2
a 2x,
3
a 4x,
4
a 8x
2 1
a a 2x x x
3 2
a a 4x 2x 2x
4 3
a a 8x 4x 4x
Thus, the differences are not equal.
Hence, the list of numbers does not form an
AP.
Question: 8
Justify whether it is true to say that the following
are the terms of an AP.
th
n
(i)
2n 3
(ii)
2
3n 5
(iii)
2
1 n n
Solution:
(i) Let .
n
a 2n 3
Now, take ,
n 1
1
a 2 1 3 2 3 1
Take ,
n 2
2
a 2 2 3 4 3 1
Take ,
n 3
3
a 2 3 3 6 3 3
Take ,
n 4
4
a 2 4 3 8 3 5
Now, find the differences,
1 2 1
d a a
1 1
1 1
2
2 3 2
d a a
3 1
2
3 4 3
d a a
5 3
2
Thus, .
1 2 3
d d d 2
Hence, forms term of an AP.
n
a 2n 3
th
n
(ii) Let .
2
3n 5
Now, take ,
n 1
2
1
a 3 1 5 3 5 8
Take ,
n 2
2
2
a 3 2 5 12 5 17
Take ,
n 3
2
3
a 3 3 5 27 5 32
Take ,
n 4
2
4
a 3 4 5 48 5 53
Now, find the differences,
1 2 1
d a a
17 8
9
2 3 2
d a a
32 17
15
3 4 3
d a a
53 32
21
Thus, .
1 2 3
d d d
Hence, the numbers do not form an
8,
17,
32,.....
AP.
(iii) Let .
2
1 n n
Now, take ,
n 1
2
1
a 1 1 1 1 1 1 3
Take ,
n 2
2
2
a 1 2 2 1 2 4 7
Take ,
n 3
2
3
a 1 3 3 1 3 9 13
Take ,
n 4
2
4
a 1 4 4 1 4 16 21
Now, find the differences,
1 2 1
d a a
7 3
4
2 3 2
d a a
13 7
6
3 4 3
d a a
21 13
8
Thus, .
1 2 3
d d d
Hence, the numbers do not form
3,
7,
13,
...21, ....
an AP.
Exercise 5.3 (35)
Question: 1
Match the APs given in column A with suitable
common differences given in column B.
Column A
Column B
1
A
2,
2,
6,
..10, ....
1
B
2
3
2
A
a 18,
n 10,
n
a 0
2
B
5
3
A
a 0,
10
a 6
3
B
4
4
A
2
a 13,
4
a 3
4
B
4
5
B
2
6
B
1
2
7
B
5
Solution:
. Given:
1
A
2,
2,
6,
..10, ....
1
a 2
Now, find the common difference.
1
d 2 2
4
2
d 6 2
6 2
4
Hence, matches with .
1
A
4
B
. Given:
2
A
a 18,
n 10,
n
a 0
By the formula of term,
th
n
0 18 10 1 d
9d 18
d 2
Hence, matches with .
2
A
5
B
. Given:
3
A
a 0,
10
a 6
By the formula of term,
th
n
10
a 0 10 1 d
6 9d
6
d
9
2
d
3
Hence, matches with .
3
A
1
B
. Given:
4
A
2
a 13,
4
a 3
a 2 1 d 13
a d 13
a 13 d ......(1)
a 4 1 d 3
a 3d 3
13 d 3d 3
2d 3 13
2d 10
d 5
Hence, matches with .
4
A
2
B
Question: 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
....
(ii)
5,
14
,
3
13
,
3
4,......
(iii)
3,
2 3,
..3 3, ....
(iv)
a b,
a 1 b,
a 1 b 1 ,......
(v)
a,2a 1,
3a 2,
..4a 3, ....
Solution:
(i) Given:
0,
1
,
4
1
,
2
..
3
,
4
....
1 2 1
1 1
d a a 0
4 4
2 3 2
1 1 2 1 1
d a a
2 4 4 4
3 4 3
3 1 3 2 1
d a a
4 2 4 4
So, .
1 2 3
1
d d d
4
Thus, the given numbers form an AP.
5 4
a a d
3 1
4 4
3 1 4
1
4 4
6 5
a a d
4 1
4 4
4 1 5
4 4
7 6
a a d
5 1
4 4
5 1 6 3
4 4 2
Thus, the next three terms are and .
1,
5
4
3
2
(ii) Given:
5,
14
,
3
13
,
3
4,......
1 2 1
14 14 15 1
d a a 5
3 3 3
2 3 2
13 14 13 14 1
d a a
3 3 3 3
3 4 3
13 12 13 1
d a a 4
3 3 3
So, .
1 2 3
1
d d d
3
Thus, the given numbers form an AP.
5 4
a a d
1
4
3
12 1 11
3 3
6 5
a a d
11 1
3 3
11 1 10
3 3
7 6
a a d
10 1
3 3
10 1 9
3
3 3
Thus, the next three terms are and .
11
,
3
10
3
3
(iii) Given:
3,
2 3,
..3 3, ....
1 2 1
d a a 2 3 3 3
2 3 2
d a a 3 3 2 3 3
So, .
1 2
d d 3
Thus, the given numbers form an AP.
4 3
a a d
3 3 3
4 3
5 4
a a d
4 3 3
5 3
6 5
a a d
5 3 3
6 3
Thus, the next three terms are and
4 3,
5 3
.
6 3
(iv) Given:
a b,
a 1 b
a 1 b 1 ,......
1 2 1
d a a a 1 b a b a 1 b a b 1
2 3 2
d a a
a 1 b 1 a 1 b
a 1 b 1 a 1 b
1
So, .
1 2
d d 1
Thus, the given numbers form an AP.
4 3
a a d
a 1 b 1 1
a 2 b 1
5 4
a a d
a 2 b 1 1
a 2 b 2
6 5
a a d
a 2 b 2 1
a 3 b 2
Thus, the next three terms are
a 2 b 1 ,
and
a 2 b 2
a 3 b 2
(v) Given:
a,
2a 1,
3a 2,
..4a 3, ....
1 2 1
d a a 2a 1 a a 1
2 3 2
d a a 3a 2 2a 1
3a 2 2a 1
a 1
3 4 3
d a a 4a 3 3a 2
4a 3 3a 2
a 1
So, .
1 2 3
d d d a 1
Thus, the given numbers form an AP.
5 4
a a d
4a 3 a 1
5a 4
6 5
a a d
5a 4 a 1
6a 5
7 6
a a d
6a 5 a 1
7a 6
Thus, the next three terms are and
5a 4 ,
6a 5
.
7a 6
Question: 3
Write the first three terms of the APs when and
a
d
are as given below:
(i)
1
a ,
2
1
d
6
(ii)
a 5,
d 3
(iii)
a 2,
1
d
2
Solution:
(i) Given:
1
a ,
2
1
d
6
By the formula of term,
th
n
1 1
n 1
2 6
1 n 1
2 6 6
3 1 n
6
4 n
6
Now, .
n
4 n
a
6
Put .
n 1
1
4 1 3 1
a
6 6 2
2
4 2 2 1
a
6 6 3
3
4 3 1 1
a
6 6 6
Thus, the first three terms are and .
1
,
2
1
3
1
6
(ii) Given:
a 5,
d 3
By the formula of term,
th
n
5 n 1 3
5 3n 3
2 3n
Now, .
n
a 2 3n
Put .
n 1
1
a 2 3 1 2 3 5
2
a 2 3 2 2 6 8
3
a 2 3 3 2 9 11
Thus, the first three terms are and .
5,
8
11
(iii) Given:
a 2,
1
d
2
By the formula of term,
th
n
1
2 n 1
2
n 1
2
2 2
n 1
2
2 2
Now, .
n
1 n
a
2
Put .
n 1
1
1 1 2 2 2 2 2
a 2
2
2 2 2 2
2
1 2 3 3 2 3 2
a
2
2 2 2 2
3
1 3 4 4 2 4 2
a 2 2
2
2 2 2 2
Thus, the first three terms are and
2,
3 2
2
2 2
Question: 4
Find and such that the following numbers are
a,b
c
in AP: .
a,
7,
b,
23,
c
Solution:
Given AP: .
a,
7,
b,
23,
c
Now, find the difference.
1 2 1
d a a 7 a
2 3 2
d a a b 7
3 4 3
d a a 23 b
4 5 4
d a a c 23
The numbers are in AP.
So .
1 2 3 4
d d d d
Now,
2 3
d d
b 7 23 b
b b 30
2b 30
b 15
Now,
2 1
d d
b 7 7 a
15 7 7 a
8 7 a
a 7 8
a 1
Now,
4 2
d d
c 23 b 7
c 23 15 7
38 7
31
Hence, and .
a 1,
b 15
c 31
Question: 5
Determine the AP whose fifth term is 19 and the
difference of the eighth term from the thirteenth
term is 20.
Solution:
Given: .
5
a 19,
13 8
a a 20
Now,
5
a 19
a 5 1 d 19
a 4d 19 ......(1)
Also, .
13 8
a a 20
13 8
a a 20
a 13 1 d a 8 1 d 20
a 12d a 7d 20
12d 7d 20
5d 20
d 4
Now,
a 4d 19
a 4 4 19
a 19 16
a 3
Now, the AP is given by
3,
3 4 ,
3 2 4 ,
.
..3 3 4 , ....
Hence, the required AP is .
3,
7,
11,
15,....
Question: 6
The 26
th
, 11
th
and the last term of an AP are and
0,
3
, respectively. Find the common difference and
1
5
the number of terms.
Solution:
Given: and .
26
a 0,
11
a 3
n
1
a
5
Now,
26
a 0
a 26 1 d 0
a 25d 0 ......(1)
11
a 3
a 11 1 d 3
a 10d 3 ......(2)
Thus, .
n
1
a
5
n
1
a
5
1
a n 1 d ......(3)
5
Subtract equation (2) from (1),
a 25d a 10d 0 3
a 25d a 10d 3
15d 3
3 1
d
15 5
From (2),
a 10d 3
1
a 10 3
5
a 2 3
From (3),
1 1
5 n 1
5 5
25 n 1 1
25 1 n 1
n 1 26
n 27
Thus, and .
1
d
5
n 27
Question: 7
The sum of the 5
th
and the 7
th
terms of an AP is 52
and the 10
th
term is 46. Find the AP.
Solution:
Given:
5 7
a a 52
5 7
a a 52
a 5 1 d a 7 1 d 52
2a 4d 6d 52
2a 10d 52 ......(1)
Also,
10
a 46
10
a 46
a 10 1 d 46
a 9d 46 ......(2)
Subtract (2) from (1),
a 5d a 9d 26 46
a 5d a 9d 20
4d 20
d 5
Put the value of in equation (1),
d
a 5d 26
a 5 5 26
a 26 25
a 1
Thus, the AP is .
a,
a d,
.a 2d, ....
Hence, the AP is .
1,
6,
11,
16,....
Question: 8
Find the 20
th
term of the AP whose 7
th
term is 24
less than the 11
th
term, the first term being 12.
Solution:
According to the question,
7 11
a a 24
a 7 1 d 24 a 11 1 d
6d 10d 24
4d 24
d 6
Now, find the value of .
20
a
20
a a n 1 d
12 20 1 6
12 19 6
12 114
126
Thus, the term of the AP is .
th
20
126
Question: 9
If the 9
th
term of an AP is zero, prove that its 29
th
term is twice its 19
th
term.
Solution:
Let be the first term and be the common
a
d
difference of an AP.
9
a 0
a 9 1 d 0
a 8d 0
a 8d ......(1)
To prove that .
29 19
a 2a
29
a a 29 1 d
a 28d
8d 28d
20d ......(2)
Now,
19
a a 19 1 d
8d 18d
10d
But
29
a 20d
2 10d
19
2a
Thus, .
29 19
a 2a
Hence, the term is twice the 19
th
term in the
th
29
AP.
Question: 10
Find whether 55 is a term of the AP: or
7,
10,
..13, ....
not. If yes, find which term it is.
Solution:
Given: .
n
a 55
Let .
a 7,
d 10 7 3
n
a 55
a n 1 3 55
7 n 1 3 55
n 1 3 55 7
n 1 3 48
n 1 16
n 17
Which is a natural number.
Hence, 55 is the 17
th
term of the AP.
Question: 11
Determine so that
k
2
k 4k 8,
2
2k 3k 6,
are three consecutive terms of an AP.
2
3k 4k 4
Solution:
Given: The numbers are in AP.
1 2
d d d
1 2 1
d a a
2 2
2k 3k 6 k 4k 8
2 2
2k 3k 6 k 4k 8
2
k k 2
2 3 2
d a a
2 2
3k 4k 4 2k 3k 6
2 2
3k 4k 4 2k 3k 6
2
k k 2
The given terms are in AP.
2 1
d d
2 2
k k 2 k k 2
2k 2 2
2k 0
k 0
Hence, the given sequence of numbers are in
k 0
AP.
Question: 12
Split 207 into three parts such that these are in AP
and the product of the two smaller parts is 4623.
Solution:
Let the three terms consecutive terms be
a d ,
a,
.
a d
According to the question,
a d a a d 207
3a 207
a 69
Also,
a d a 4623
69 d 69 4623
4623
69 d
69
69 d 67
d 2
So,
AP a d ,
a, a d
69 2 ,
69,
69 2
67,
69,
71
Thus, can be divided into .
207
67,
69,
71
These three terms form an AP.
Question: 13
The angles of a triangle are in AP. The greatest
angle is twice the least. Find all the angles of the
triangle.
Solution:
The three terms are in AP be .
a d ,
a,
a d
By the angle sum property,
a d a a d 180
3a 180
a 60
The greatest angle is twice of the smallest.
a d 2 a d
a d 2a 2d
a d 2a 2d 0
a 3d 0
3d a ......(1)
Put the value of in equation (1),
d
3d a
3d 60
60
d
3
d 20
The terms are,
a d,
a,
a d
60 20 ,
60 ,
60 20
40 ,
60 ,
80
Thus, the angles of triangle are and
40 ,
60 ,
.
80
Question: 14
If the terms of the two APs: and
th
n
9,
7,
5,...
24,
21,
are the same, find the value of . Also, find
18,....
n
that term.
Solution:
The first AP is: .
9,
7,
Let
a 9,
d 7 9 2
By the formula of term,
th
n
9 n 1 2
9 2 n 1
9 2n 2
11 2n
Second AP is:
24,
21,
18,....
Let
1
a 24,
1
d 21 24 3
By the formula of term,
th
n
n 1 1
a a n 1 d
24 n 1 3
24 3n 3
27 3n
According to the question,
n n
a a
n 16
Now, find the 16
th
term of the 1
st
AP.
16 1
a a n 1 d
9 16 1 2
9 2 15
9 30
21
Now, find the 16
th
term of the 2
nd
AP.
16
a 24 16 1 3
24 15 3
24 45
21
Thus, the 16
th
term of both APs is equal to .
21
Question: 15
If sum of the 3
rd
and the 8
th
terms of an AP is 7 and
the sum of the 7
th
and the 14
th
terms is , find the
3
10
th
term.
Solution:
Let be the first term and be the common
a
d
difference of an AP.
3 8
a a 7
a 3 1 d a 8 1 d 7
a 2d a 7d 7
2a 9d 7 ......(1)
Also,
7 14
a a 3
7 14
a a 3
a 7 1 d a 14 1 d 3
a 6d a 13d 3
2a 19d 3 ......(2)
Subtract equation (2) from (1),
2a 9d 2a 19d 7 3
10d 10
d 1
Put the value of in equation (1),
d
2a 9d 7
2a 9 1 7
2a 9 7
2a 7 9
a 8
Now, find the value of term.
th
10
10
a a 10 1 d
8 9 1
8 9
1
Thus, the 10
th
term of AP is .
1
Question: 16
Find the 12
th
term from the end of the AP:
2,
4,
.
6,
,..... 100
Solution:
Given AP:
2,
4,
6,
......, 100
Consider the AP in reverse order,
100,.
.....,
6,
4,
2
The first term
a 100
The common difference
n 1 n
d a a
4 6
4 6
2
n 12
Now, find the term.
th
12
12
a a n 1 d
100 12 1 2
100 11 2
100 22
78
Thus, the term is .
th
12
78
Question: 17
Which term of the AP: is the first negative
53,
48,
term?
Solution:
Given AP:
53,
48,
43,.....
Compare the given AP with the AP .
1
a ,
1
a ,
3
a ,
n
...... .,a.
1
a 53,
2
a 48,
3
a 43
The first term,
1
a a 53
Common difference,
d 48 53
5
The term of the AP is the first negative term.
th
n
n
a 0
a n 1 d 0
53 n 1 5 0
5 n 1 53
5n 53 5
5n 58
58
n
5
n 11.6
So, .
n 12
Hence, the first negative term of the AP is the 12
th
term,
12
a a n 1 d
53 12 1 5
53 5 11
53 55
2
Thus, .
12
a 2
Question: 18
How many numbers lie between 10 and 300, which
when divided by 4 leave a remainder 3?
Solution:
The least number between and which when
10
300
divided by leaves remainder is .
4
3
11
The largest number between and which
10
300
when divided by leaves remainder is
4
3
.
296 3 299
1
st
term or number
11
2
nd
term or number
15
3
rd
term or number
19
Last term or number
299
Now, the AP becomes.
11,
15,
19,
....,299
Here, .
n
a 299,
a 11,
d 15 11 4
By the formula of term,
th
n
299 11 n 1 4
n 1 4 299 11
n 1 4 288
n 1 72
n 72 1
n 73
Thus, the required number is .
73
Question: 19
Find the sum of the two middle most terms of the
AP:
4
,
3
1,
2
,
3
1
...,4
3
Solution:
Given AP:
4
,
3
1,
2
,
3
.
1
...,4
3
Here, .
4
a ,
3
1 4 1 4 3 4 1
d
1 3 1 3 3 3
n
13
a
3
By the formula of term,
th
n
13 4 1
n 1
3 3 3
13 4 n 1
n 1 13 4
n 17 1
n 18
Thus,
The middle mo
18 18
th a
st
term in1 8terms
nd th 1
2 2
and
9th
10th term
The required sum is:
9 10
a a a 9 1 d a 10 1 d
2a 8d 9d
2a 17d
4 1
2 17
3 3
8 17 9
3
3 3
Thus, the sum of two middle terms is .
3
Question: 20
The first term of an AP is and the last term is
5
45
. If the sum of the terms of the AP is , then find
120
the number of terms and the common difference.
Solution:
Let be the first term and be the common
a
d
difference.
Given:
a 5,
n
a 45,
n
S 120
Now,
n
n
S 2a n 1 d
2
n
n
S a a n 1 d
2
n n n
n
S a a a last term
2
n
n
120 a a
2
n
120 5 45
2
n
120 40
2
120 2
n
40
n 6
Thus, the number of terms
6
Now, find the value of .
d
45 5 6 1 d
45 5 5d
5d 50
d 10
Thus, the common difference is .
10
Question: 21
Find the sum:
(i)
...1 5 .2 8 236
(ii)
1
....upt
2 3
4 4 4
n n n
on terms
(iii)
a b 3a 2b 5a 3
....to1
b
a
1t
b a b a b
erms
Solution:
(i) Given series:
...1 5 .2 8 236
a 1,
n
a 236
, ,
1
d 2 1 3
2
d 5 2 5 2 3
3
d 8 5 8 5 3
Now,
1 2 3
d d d d 3
By the formula of term,
th
n
236 1 n 1 3
3 n 1 236 1
3 n 1 237
n 1 79
n 80
Now, the sum of series.
n
n
S 2a n 1 d
2
80
2 1 80 1 3
2
40 2 79 3
40 2 237
40 235
9400
Thus, the sum of all terms is .
9400
(ii) Given series:
1
....upt
2 3
4 4 4
n n n
on terms
and
1
a 4
n
n n
1
2 1
d 4 4
n n
2 1
4 4
n n
1
n
2
3 2
d 4 4
n n
3 2
4 4
n n
1
n
Now, find the sum.
n
n
S 2a n 1 d
2
n 1 1
2 4 n 1
2 n n
n 1
n 2
8
2 n n
n 1
7
2 n
n 7n 1 7n 1
2 n 2
Thus, the sum is .
7n 1
2
(iii) Given series:
a b 3a 2b 5a 3
....to1
b
a
1t
b a b a b
erms
a b
a first term ,
a b
n 11
3a 2b a b
a b
3a 2b a b
a b
2a b
a b
Now, find the sum.
n
n
S 2a n 1 d
2
11
2 a b
11 2a b
S 11 1
2 a b a b
2a 2b 10 2a b
11
2 a b
11
2a 2b 20a 10b
2 a b
11
22a 12b
2 a b
11 2
11a 6b
2 a b
11
11a 6b
a b
Thus, .
11
11
S 11a 6b
a b
Question: 22
Which term of the AP: will be ?
2,
7,
12,....
77
Find the sum of this AP up to the term .
77
Solution:
Given AP:
2,
7,
12,....
Here,
a 2,
n
a 77
1
d 7 2 7 2 5
2
d 12 7 12 7 5
Now,
n
a 77
By the formula of term,
th
n
a n 1 d 77
2 n 1 5 77
2 5n 5 77
5n 3 77
5n 3 77
5n 77 3
5n 80
n 16
Now, calculate the sum of terms.
16
16
16
S 2 2 16 1 5
2
8 4 15 5
8 4 75
8 79 632
Thus, the sum is .
632
Question: 23
If , show that form an AP. Also,
n
a 3 4n
1
a ,
2
a ,...
find .
20
S
Solution:
Given:
n
a 3 4n
Now,
1
a 3 4 1 3 4 1
2
a 3 4 2 3 8 5
3
a 3 4 3 3 12 9
4
a 3 4 4 3 16 13
Now,
1 2 1
d a a 5 1 5 1 4
2 3 2
d a a 9 5 9 5 4
3 4 3
d a a 13 9 13 9 4
Now, .
1 2 3
d d d 4
Now, .
1
a ,
2
a ,
3
a ,...
Here, .
a 1,
d 4,
n 20
Now, find the sum of terms.
20
20
20
S 2 1 20 1 4
2
10 78
780
Thus, .
20
S 780
Question: 24
In an AP, if , find the AP.
n
S n 4n 1
Solution:
Given:
n
S n 4n 1
1 1
a S ,
2 2 1
a S S,
.
3 3 2
a S S
Now,
n n n 1
a S S
2
2
4n n 4 n 1 n 1
2 2
4n n 4 n 1 2n n 1
2 2
4n n 4n 4 8n n 1
2 2
4n n 4n 7n 3
Now, .
n
a 8n 3
1
a 8 1 3 8 3 5
2
a 8 2 3 16 3 13
3
a 8 3 3 24 3 21
4
a 8 4 3 32 3 29
Thus, the required AP is .
5,
13,
21,
29,...
Question: 25
In an AP, if and , find the value
2
n
S 3n 5n
k
a 164
of .
k
Solution:
Given: and
2
n
S 3n 5n
k
a 164
By the formula,
n n n 1
a S S
2
2
3n 5n 3 n 1 5 n 1
2 2
3n 5n 3 n 1 2n 5 n 1
2 2
3n 5n 3n 3 6n 5n 5
6n 2
Now, .
k
a 6k 2
Also,
164 6k 2
6k 164 2
6k 162
k 27
Thus, the value of is .
k
27
Question: 26
If denotes the sum of first terms of an AP,
n
S
n
prove that .
12 8 4
S 3 S S
Solution:
Let be the first term and be the common
a
d
difference of an AP.
n
n
S 2a n 1 d
2
12
12
S 2a 12 1 d
2
12
S 6 2a 11d ......(1)
Now,
8
8
S 2a 8 1 d
2
8
S 4 2a 7d ......(2)
And
4
4
S 2a 4 1 d
2
4
S 2 2a 3d ......(3)
Now, prove the condition.
Take RHS,
4
S 2 2a 3d
3 8a 28d 4a 6d
6 2a 11d
12
S
LHS RHS
Hence proved.
Question: 27
Find the sum of the first terms of an AP whose
17
4
th
and 9
th
terms are and , respectively.
15
30
Solution:
Given: .
4
a 15,
9
a 30
Now, the 4
th
term is
4
a 15
a 4 1 d 15
a 3d 15 ......(1)
The 9
th
term is
9
a 30
a 9 1 d 30
a 8d 30 ......(2)
Subtract equation (1) from (2),
a 8d a 3d 30 15
a 8d a 3d 30 15
5d 15
d 3
Now, put the value of in equation (1),
d
a 3d 15
a 3 3 15
a 15 9
a 6
Now, find the sum of terms.
17
n
n
S 2a n 1 d
2
17
17
S 2 6 17 1 3
2
17
12 16 3
2
17
60
2
510
Thus, the sum is .
510
Question: 28
If sum of the first 6 terms of an AP is 36 and that of
the first 16 terms is 256, find the sum of first 10
terms.
Solution:
Let be the first term and be the common
a
d
difference of an AP.
Sum of first 6 terms.
6
S 36
6
2a 6 1 d 36
2
3 2a 6 1 d 36
2a 5d 12 ......(1)
Sum of first 16 terms.
16
S 256
16
2a 16 1 d 256
2
8 2a 16 1 d 256
2a 15d 32 ......(2)
Subtract equation (1) from (2),
2a 15d 2a 5d 32 12
2a 15d 2a 5d 32 12
10d 20
d 2
Now, put the value of in equation (1),
d
2a 5d 12
2a 5 2 12
2a 12 10
a 1
Now, find the sum of terms.
10
10
10
S 2 1 10 1 2
2
5 2 9 2
5 20
100
Thus, the sum of first terms is .
10
100
Question: 29
Find the sum of all the terms of an AP whose
11
middle most term is .
30
Solution:
Given: .
n 11
Middle term
th
11 1
6 term
2
Middle term
30
6
a 30
a 6 1 d 30
a 5d 30 ......(1)
Now, find the sum of terms.
11
11
11
S 2a 11 1 d
2
11
2a 10d
2
11 a 5d
11 30
330
Thus, the sum of all terms is .
11
330
Question: 30
Find the sum of last ten terms of the AP:
8,
10,
12,
.
.....,126
Solution:
Given AP:
8,
10,
12,
.....,126
Reverse the order of given AP:
126,
.....,
12,
10,
8
a 126,
d 10 12 2,
n 10
Now, find the sum of last terms.
10
10
10
S 2 126 10 1 2
2
5 252 9 2
5 252 18
5 234
1170
Thus, the sum of last terms of AP is .
10
1170
Question: 31
Find the sum of the first seven numbers which are
multiples of 2 as well as of 9.
Solution:
The numbers that are multiples of as well as
2
9
are
18,
36,
54,
....,7 terms
So,
n 7,
a 18,
d 36 18 18
Now, find the sum of terms.
7
7
7
S 2 18 7 1 18
2
7 18
2 6
2
7 18
8
2
504
Thus, the sum is .
504
Question: 32
How many terms of the AP: are
15,
13,
11,....
needed to make the sum ? Explain the reason
55
Solution:
Given AP:
15,
13,
11,....
n
S 55,
a 15
d 13 15
13 15
2
By the formula of sum,
n
S 55
n
2a n 1 d 55
2
n
2a n 1 d 55
2
n 2 15 n 1 2 55 2
n 30 2n 2 110
n 30 2n 2 110 0
2
30n 2n 2n 110 0
2
2n 32n 110 0
Now, solve the quadratic equation.
2
n 11n 5n 55 0
n n 11 5 n 11 0
n 11 n 5 0
That is,
or
5 0n
n 11 0
11n
or
5n
Thus,
5
or
11
terms of AP make the sum
55
.
Question: 33
The sum of the first
n
terms of an AP whose first
term is 8 and the common difference is 20 is equal
to the sum of first
2
n
terms of another AP whose
first term is
30
and the common difference is 8.
Find
n
.
Solution:
Given:
For 1
st
AP:
8,a
20d
For 2
nd
AP:
30,a
8d
According to the question,
2n n
S S
2
2 1 2 2 1
2 2
n n
a n d a n d
2 8 1 20 2 2 30 2 1 8n n
16 20 20 2 60 16 8n n
20 4 2 68 16n n
32 20 136 4n n
12 132n
11n
Thus,
11
n
.
Question: 34
Kanika was given her pocket money on Jan 1
st
, 2008.
She puts Re 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day
3, and continued doing so till the end of the month,
from this money into her piggy bank. She also spent
Rs 204 of her pocket money, and found that at the
end of the month she still had Rs 100 with her. How
much was her pocket money for the month?
Solution:
Let Rs
x
be the pocket money for the month.
The money she deposited in piggy bank and spent
Rs204
Money put in piggy bank from
Jan to Jan1 31 1 2 3 4. . .... 31
Now,
1,a
1,d
31n
Now, the sum of
31
terms.
31
31
2 1 31 2 1
2
S
31
2 31 2
2
31
2 29
2
31 32
2
31 16
496
Thus, the money deposited in piggy bank
Rs 496
Money spent
Rs204
Let the money she still have
Rs100
496 204 100x
100 496 204x
800x
Thus, Rs 800 is her monthly pocket money.
Question: 35
Yasmeen saves Rs 32 during the first month, Rs 36
in the second month and Rs 40 in the third month.
If she continues to save in this manner, in how
many months will she save Rs 2000?
Solution:
1
st
month saving of Yasmeen
Rs32
2
nd
month saving of Yasmeen
Rs36
3
rd
month saving of Yasmeen
Rs 40
4
th
month saving of Yasmeen
Rs 44
So,
32 36 40 44 .... 2000
Also,
32,a
36 32 4d
Now, by the formula of
n
S
.
2000
n
S
2 1 2000
2
n
a n d
2 32 1 2000 2n n d
64 4 4 4000n n
60 4 4000n n
15 1000n n
2
15 1000 0n n
2
40 25 1000 0n n n
40 25 40 0n n n
40 or 25n n
40
is not a natural number.
Thus, she saves Rs
2000
in
25
months.
Exercise 5.4 (10)
Question: 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:
Let
a
be the first term and
d
be the common
difference of an AP.
According to the question,
5 7
167S S
5 7
2 5 1 2 7 1 167
2 2
a d a d
5 2 4 7 2 6 167 2a d a d
10 20 14 42 167 2a d a d
24 62 167 2a d
2 12 31 167 2a d
12a 31d 167 ......(1)
Given,
10
235
S
10
235S
10
2 10 1 235
2
a d
10
2 10 1 235
2
a d
2a 9d 47 ......(2)
Multiply equation (2) by 6,
12a 54d 282 ......(3)
Subtract equation (1) from (3),
12 54 12 31 282 167a d a d
23 115d
5d
Put the value of
d
in equation (2),
2 9 47a d
2 9 5 47a
2 47 45a
2 2a
1a
Now, find the sum of
20
terms.
20
20
2 20 1
2
S a d
10 2 1 19 5
10 2 95
10 97
970
Thus, the sum is
970
.
Question: 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.
Solution:
(i) Integers between 1 and 500 which are
multiples of 2 as well as of 5 are
10,
20,
30,
....,490
.
10, 10, 490
n
a d a
By the formula of
n
a
,
490
n
a
1 490a n d
10 1 10 490n
1 1 49n
49n
By the formula of sum of
49
terms,
49
49
2 10 49 1 10
2
S
49
20 48 10
2
49
500
2
12250
Thus, the sum is
12250
.
(ii) Integers from 1 and 500 which are multiples of
2 as well as of 5 are
10,
20,
30,
....,500
.
10,a
10,d
500
n
a
500
n
a
1 500a n d
10 1 10 500n
1 1 50n
50n
By the formula of sum of
50
terms,
50
50
2 10 50 1 10
2
S
50
20 49 10
2
25 20 490
25 510
12750
Thus, the sum is
12750
.
(iii) Sum of integers from 1 and 500 which are
multiples of 2 or 5 (not of 10)
Sum of
integers which are multiple of
2
Sum of
integers which are multiple of
5
Sum of
integers which are multiple of
10
... 500 ... 5002 4 6 5 10 15 10 20 ... 503 00
1 2 3
S S S
For
1
2 4 6 .... 500
S
2,a
2,d
500
n
a
By the formula of
n
a
,
500
n
a
1 500a n d
2 1 2 500n
2 500n
250n
So,
1 250
S S
.
1 250
S S
250
2 2 250 1 2
2
125 4 249 2
125 4 498
62750
For
2
5 10 15 .... 500
S
5,a
5,d
500
n
a
By the formula of
n
a
,
500
n
a
1 500a n d
5 1 5 500n
5 500n
100n
So,
2 100
S S
.
2 100
S S
100
2 5 100 1 5
2
50 10 494
50 505
25250
For
3
10 20 .... 500
S
10,a
10,d
500
n
a
By the formula of
n
a
,
500
n
a
1 500a n d
10 1 10 500n
10 500n
50n
So,
3 50
S S
.
3 50
S S
50
2 10 50 1 10
2
25 20 490
25 20 490
12750
Sum
1 2 3
S S S
62750 25250 12750
75250
Thus, the sum of required integers are
75250
.
Question: 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 15
th
term.
Solution:
Let
a
be the first term and
d
be the common
difference of an AP.
8 2
1
2
a a
1
8 1 2 1
2
a d a d
2 7a d a d
2 14 0a d a d
13 0a d
… (1)
Now,
11 4
1
1
3
a a
11 4
1
1
3
a a
1
11 1 4 1 1
3
a d a d
3 10 3 3a d a d
3 30 3 3 0a d a d
2 27 3a d
… (2)
Multiply equation (1) by 2,
2 26 0
a d
… (3)
Subtract equation (3) from (2),
2 27 2 26 3 0a d a d
2 27 2 26 3a d a d
3d
Put the value of
d
in equation (1),
13 0a d
13 3 0a
39a
By the formula of
n
a
.
15
39 15 1 3a
39 14 3
39 42
3
Thus,
15
3
a
.
Question: 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:
Let
a
be the first term and
d
be the common
difference of an AP.
Total terms
37
The middle most term
th term
37 1 38
19
2 2
By the sum of three middle most terms,
18 19 20
Sum a a a
18 1
225
19 1 20 1
a d a
d a d
225 3 17 18 19a d d d
225 3 54a d
18 75a d
… (1)
Sum of last three terms
37 36 35
Sum a a a
37 1
429
36 1 35 1
a d a
d a d
429 3 36 35 34a d d d
3 105 429a d
3 105 429a d
35 143a d
… (2)
Subtract equation (1) from (2),
35 18 143 75a d a d
35 18 143 75a d a d
17 68d
4d
Put the value of
d
in equation (1),
18 75a d
18 4 75a
75 72a
3a
Thus, the AP is
3,
7,
11,
15,...
.
Question: 5
Find the sum of the integers between 100 and 200
that are
(i) divisible by
9
(ii) not divisible by
9
Solution:
(i) Numbers between
100 200
divisible by
9
are
108,
117,
126,
...,198
.
108,a
117 108 9d
and
198
n
a
By the formula of
th
n
term,
1 198a n d
108 1 9 198n
198
12 1
9
n
22 11n
11n
Now, find the sum of
11
terms.
11
11
2 108 11 1 9
2
S
11
216 90
2
11
306
2
1683
Thus, the sum of
11
terms are
1683
.
(ii) Numbers between
100 200
are
101,
102,
103,
...,199
.
101,a
1,d
199
n
a
By the formula of
th
n
term,
1 199a n d
101 1 1 199n
1 199 101n
1 98n
99n
Now, find the sum of
99
terms.
99
99
2 101 99 1 1
2
S
99
202 98
2
99
202 98
2
99
300
2
14850
The sum of numbers between
100
to
200
that
are not divisible by
9
14850 1683
13167
Thus, the sum of numbers between
100
to
200
that are not divisible by
9
is
13167
.
Question: 6
The ratio of the 11
th
term to the 18
th
term of an AP
is
2: 3
. Find the ratio of the 5
th
term to the 21
st
term,
and also the ratio of the sum of the first five terms
to the sum of the first 21 terms.
Solution:
Let
a
be the first term and
d
be the common
difference of an AP.
11 18
: 2: 3
a a
10 2
17 3
a d
a d
3 30 2 34a d a d
3 2 34 30a a d d
4a d
To find:
5
21
4 4 4 8 1
20 4 20 24 3
a a d d d d
a a d d d d
Now, the ratio is
5 21
: 1:3
a a
.
Now, find the ratio of sum.
5
21
5
2 5 1
2
21
2 21 1
2
a d
S
S
a d
5 2 4
21 2 20
a d
a d
5 2 4 4
21 2 4 20
d d
d d
5 8 4
21 8 20
d d
d d
5 12
21 28
d
d
5 5
5 : 49
7 7 49
Thus,
5 21
: 5: 49
S S
.
Question: 7
Show that the sum of an AP whose first term is
a
,
the second term
b
and the last term
c
, is equal to
2
2
a c b c a
b a
Solution:
Given:
first term ,a a
,d b a
n
a c
By the formula of
th
n
term,
1a n d c
1n b a c a
1
c a
n
b a
1
c a
n
b a
2b c a
n
b a
… (1)
Now, find the sum of
n
terms.
2 1
2
n
n
S a n d
2
2
2 1
2
b c a
b c a
a b a
b a b a
2
2
2
2
b c a
b c a b a
a b a
b a b a
2
2
2
b c a
a c a
b a
2
2
b c a
a c
b a
Hence proved.
Question: 8
Solve the equation
..4 . 4 72 31
x
Solution:
Given series:
4 1 2 ...
x
1
1 4d
1 4
3
2
2 1d
2 1
3
1 2
3
d d
Thus, the list of numbers are in AP.
4a
and
n
a x
By the formula of
n
a
,
n
a x
1a n d x
4 1 3n x
1 3 4n x
4
1
3
x
n
4
1
3
x
n
7
3
x
n
… (1)
By the formula of
n
S
,
2 1
2
n
n
S a n d
7 4 3
2 4
2 3 3
x x
7
8 4
6
x
x
7 4
6
x x
Now,
437
n
S
437
n
S
7 4
437
6
x x
2
3 28 437 6x x
2
3 28 2622 0x x
2
3 2650 0x x
Solve the quadratic equation,
2
3 2650 0x x
2
53 50 2650 0x x x
53 50 53 0x x x
53 50 0x x
53x
or
50x
53
is rejected.
Thus,
50
x
.
Question: 9
Jaspal Singh repays his total loan of Rs 118000 by
paying every month starting with the first
instalment of Rs 1000. If he increases the instalment
by Rs 100 every month, what amount will be paid
by him in the 30th instalment? What amount of
loan does he still have to pay after the 30
th
instalment?
Solution:
The monthly instalments are
1000,
1100,
1200,
...,30 terms
1000,a
100,d
?,
n
a
30n
By the formula of
n
a
,
1
n
a a n d
1000 30 1 100
1000 2900
3900
Thus, the 30
th
instalment is
Rs3900
.
The amount for all 30-instalment paid by Jaspal
Singh
1000 1100 12 ....0 39000
1000,a
100,d
30n
Now, find the sum of
30
instalments.
30
30
2 1000 30 1 100
2
S
15 2000 2900
15 4900
73500
Thus, the loan amount left after
th
30
instalment
118000 73500
44500
.
Question: 10
The students of a school decided to beautify the
school on the Annual Day by fixing colourful flags
on the straight passage of the school. They have 27
flags to be fixed at intervals of every 2 m. The flags
are stored at the position of the middle most flag.
Ruchi was given the responsibility of placing the
flags. Ruchi kept her books where the flags were
stored. She could carry only one flag at a time. How
much distance did she cover in completing this job
and returning back to collect her books? What is
the maximum distance travelled by her carrying a
flag?
Solution:
Total flags
27
Position of the middle most flag
th flag th flag th flag
27 1 28
14
2 2
Distance between the flags
m2
Distance covered by placing the first flag
m2 2 4
Distance covered by placing the second flag
m4 4 8
Distance covered by placing the third flag
m6 6 12
The total distance covered to place
13
flags on
either side is
terms
13
....14 8 12 3
S
Here,
4,a
4,d
13n
Now, find the sum of
13
terms.
13
13
2 13 1
2
S a d
13
2 4 12 4
2
13
8 48
2
13
56
2
13 28
364
The distance covered by Ruchi to place
13
flags on
other side is
364
m.
Total distance to place
27
flags
364 2 728
m
The maximum distance she travelled carrying a
flag
Distance covered in fixing 1
st
or 27
th
flag
m13 2 26