RD Sharma Solutions: Arithmetic Progressions (Exercise 5D)

# Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

``` Page 1

1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv)
1 1 1
, , ,.....
15 12 10
to 11 terms.
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
Page 2

1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv)
1 1 1
, , ,.....
15 12 10
to 11 terms.
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30

180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here,
1
15
a and
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i)
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is
1
7 10 14 ..... 84.
2
Here,
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
Page 3

1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv)
1 1 1
, , ,.....
15 12 10
to 11 terms.
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30

180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here,
1
15
a and
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i)
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is
1
7 10 14 ..... 84.
2
Here,
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,

84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
Page 4

1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv)
1 1 1
, , ,.....
15 12 10
to 11 terms.
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30

180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here,
1
15
a and
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i)
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is
1
7 10 14 ..... 84.
2
Here,
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,

84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,

230
5 1 3 230 1
3 2 230
3 230 2 228
76
n
n
a
n a a n d
n
n
n
Required sum
76
5 230
2 2
n
n
S a l
76
235
2
8930
3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its
first 20 terms.
Sol:
Let
n
a be the nth term of the AP.
5 6
n
a n
Putting 1, n we get
First term,
1
5 6 1 1 a a
Putting 2, n we get
2
5 6 2 7 a
Let d be the common difference of the AP.
2 1
7 ( 1) 7 1 6 d a a
Sum of first n tern of the AP,
n
S
2
2 1 1 6 2 1
2 2
2 6 6
2
2 3
2 3
n
n n
n S a n d
n
n
n n
n n
Putting 20, n we get
2
20
2 20 3 20 40 1200 1160 S
4. The sum of the first n terms of an AP is
2
3 6 . n n Find the nth term and the 15
th
term of this
AP.
Sol:
Page 5

1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv)
1 1 1
, , ,.....
15 12 10
to 11 terms.
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30

180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here,
1
15
a and
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i)
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is
1
7 10 14 ..... 84.
2
Here,
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,

84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,

230
5 1 3 230 1
3 2 230
3 230 2 228
76
n
n
a
n a a n d
n
n
n
Required sum
76
5 230
2 2
n
n
S a l
76
235
2
8930
3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its
first 20 terms.
Sol:
Let
n
a be the nth term of the AP.
5 6
n
a n
Putting 1, n we get
First term,
1
5 6 1 1 a a
Putting 2, n we get
2
5 6 2 7 a
Let d be the common difference of the AP.
2 1
7 ( 1) 7 1 6 d a a
Sum of first n tern of the AP,
n
S
2
2 1 1 6 2 1
2 2
2 6 6
2
2 3
2 3
n
n n
n S a n d
n
n
n n
n n
Putting 20, n we get
2
20
2 20 3 20 40 1200 1160 S
4. The sum of the first n terms of an AP is
2
3 6 . n n Find the nth term and the 15
th
term of this
AP.
Sol:

Let
n
S denotes the sum of first n terms of the AP.
2
2
1
2
2
3 6
3 1 6 1
3 2 1 6 1
3 3
n
n
S n n
S n n
n n n
n
t h
n term of the AP,
n
a
1
2 2
3 6 3 3
6 3
n n
S S
n n n
n
Putting 15, n we get
15
6 15 3 90 3 93 a
Hence, the
t h
n term is 6 3 n and 15
th
term is 93.
5. The sum of the  first n terms of an AP is given by
2
3
n
S n n . Find its
(i) nth term,
(ii) first term and
(iii) common difference.
Sol:
Given:
2
3 ......
n
S n n i
Replacing n by 1 n in (i), we get:
2
1
2
2
3 1 1
3 2 1 1
3 7 4
n
S n n
n n n
n n
(i) Now,
1 n n n
T S S
2 2
3 3 7 4 6 4 n n n n n
t h
n term, 6 4 .......
n
T n i i
(ii) Putting 1 n in (ii), we get:
1
6 1 4 2 T
(iii) Putting 2 n in (ii), we get:
2
6 2 4 8 T
Common difference,
2 1
8 2 6 d T T
```

## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests

## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests

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