In this post, you will find the NCERT solutions for class 10 maths ex 5.1. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 5.1
1. In which of the following situations, does
the list of numbers involved make an arithmetic progression, and why?
(i)
The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8
for each additional km.
(ii)
The amount of air present in a cylinder when a vacuum pump removes 1/4 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 ₹ 150
for the first metre and rises by ₹ 50 for each subsequent metre.
(iv)
The amount of money in the account every year, when ₹ 10,000 is deposited at
compound Interest at 8% per annum.
Solution:
(i) Taxi fare for the
first km = ₹ 15
Taxi
fare for 2 km = 15 + 8 = ₹ 23
Taxi
fare for 3 km = 23 + 8 = ₹ 31
Taxi fare for 4
km = 31 + 8 = ₹ 39
Therefore,
the sequence is 15, 23, 31, 39…
The above
sequence is an arithmetic progression because difference between any two
consecutive terms is equal that is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …
)
(ii)
Let the amount of
air initially present in the cylinder = V
Amount
of air left when vacuum pump again pumps out air
= 
Therefore,
the sequence we get is 
Let us find the
difference between consecutive terms.
The
difference between consecutive terms is not equal.
Therefore,
the above sequence is not an arithmetic progression.
(iii)
The cost of digging 1 metre of well
= ₹ 150
The cost of
digging 2 metres of well = 150 + 50 = ₹ 200
The cost of
digging 3 metres of well = 200 + 50 = ₹ 250
The cost of
digging 4 metres of well = 250 + 50 = ₹ 300
Therefore, we get
a sequence of the form 150, 200, 250, 300, …
This sequence is
an arithmetic progression because the difference between any two consecutive
terms is equal that is 50. (200 – 150 = 250 – 200 = 300 – 250 = 50 … )
(iv) Amount in the bank account after one year
= 
… (1)
Amount in the
bank account after two years = 
… (2)
Amount in the
bank account after three years = 
… (3)
Amount in the
bank account after four years = 
… (4)
Here, we can find
that (2) − (1) ≠ (3) − (2)
(The difference
between any two consecutive terms is not equal.)
Therefore, the
sequence is not an arithmetic progression.
2. Write first four terms of the AP, 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 = 1/2
(v)
a = –1.25, d = –0.25
Solution:
(i) a = 10, d = 10
Let
the first four terms of the AP be a, a + d,
a + 2d and a + 3d.
First
term = a = 10
Second term = a + d
= 10 + 10 = 20
Third term = a + 2d
= 10 + 2 × 10 = 30
Fourth term = a + 3d
= 10 + 3 × 10 = 40
Therefore, the
first four terms are: 10, 20, 30, 40
(ii)
a = –2, d = 0
Let
the first four terms of the AP be a, a + d,
a + 2d and a + 3d.
First
term = a = –2
Second term = a + d
= –2 + 0 = –2
Third term = a + 2d
= –2 + 2 × 0 = –2
Fourth term = a + 3d
= –2 + 3 × 0 = –2
Therefore, the
first four terms are: –2, –2, –2, –2
(iii)
a = 4, d = –3
Let
the first four terms of the AP be a, a + d,
a + 2d and a + 3d.
First
term = a = 4
Second term = a + d
= 4 – 3 = 1
Third term = a + 2d
= 4 – 2 × 3 = –2
Fourth term = a + 3d
= 4 – 3 × 3 = –5
Therefore, the
first four terms are: 4, 1, –2, –5
(iv)
a = –1, d = ½
Let
the first four terms of the AP be a, a + d,
a + 2d and a + 3d.
First
term = a = –1
Second term = a + d
= –1 + ½ = −½
Third term = a + 2d
= −1 + 2 × ½ = 0
Fourth term = a + 3d
= −1 + 3 × ½ = ½
Therefore, the
first four terms are: –1, −½, 0, ½
(v)
a = –1.25, d = –0.25
Let
the first four terms of the AP be a, a + d,
a + 2d and a + 3d.
First
term = a = –1.25
Second term = a + d
= –1.25 – 0.25 = –1.50
Third term = a + 2d
= –1.25 – 2 × 0.25 = –1.75
Fourth term = a + 3d
= –1.25 – 3 × 0.25 = –2.00
Therefore, the
first four terms are: –1.25, –1.50, –1.75, –2.00
3. For the following APs, write the first term
and the common difference:
(i) 3, 1, –1, –3 …
(ii) –5, –1, 3, 7 …
(iv) 0.6, 1.7, 2.8, 3.9 …
Solution:
(i) 3, 1, –1, –3 …
First
term = a = 3
Common difference
(d) = Second term – first term =
Third term – second term, and so on
Therefore, the common
difference (d) = 1 – 3 = –2
(ii) –5, –1, 3,
7…
First
term = a = –5
Common difference
(d) = Second term – first term =
Third term – second term, and so on
Therefore, the common
difference (d) = –1 – (–5) = –1 + 5 =
4
(iii)
First
term = a = 1/3
Common difference
(d) = Second term – first term =
Third term – second term, and so on
Therefore,
the common difference (d) = 
(iv) 0.6,
1.7, 2.8, 3.9 …
First
term = a = 0.6
Common difference
(d) = Second term – first term =
Third term – second term, and so on
Therefore,
the common difference (d) = 1.7 − 0.6
= 1.1
4. Which of the following are APs? If they
form an AP, find the common difference d
and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 …
(iii) −1.2, −3.2, −5.2, −7.2 …
(iv) −10, −6, −2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 …
(vi) 0.2, 0.22, 0.222, 0.2222 …
(vii) 0, −4, −8, −12 …
(viii) 
(x) a, 2a, 3a, 4a
…
(xi) a,
a2, a3, a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52,72
…
(xv) 11, 52, 72,
73 …
Solution: (i) 2, 4, 8, 16 …
Here,
4 – 2 ≠ 8 − 4
It is not an AP
because the difference between consecutive terms is not equal.
(ii) 2, 5/2, 3, 7/2 …
Here, 5/2 – 2 = 3 – 5/2 = 1/2
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = 1/2
Fifth
term = 7/2 + ½ = 4
Sixth
term = 4 + 1/2 = 9/2
Seventh
term = 9/2 + ½ = 5
Therefore,
the next three terms are 4, 9/2 and 5.
(iii) −1.2, −3.2, −5.2, −7.2 …
Here, −3.2 − (−1.2) = −5.2 − (−3.2) = −7.2 −
(−5.2) = −2
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = −2
Fifth
term = −7.2 – 2 = −9.2
Sixth
term = −9.2 – 2 = −11.2
Seventh
term = −11.2 – 2 = −13.2
Therefore, the
next three terms are −9.2, −11.2 and −13.2.
(iv) −10, −6, −2, 2 …
Here, −6 − (−10) = −2 − (−6) = 2 − (−2) = 4
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = 4
Fifth
term = 2 + 4 = 6
Sixth
term = 6 + 4 = 10
Seventh
term = 10 + 4 = 14
Therefore, the
next three terms are 6, 10 and 14.
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 …
Here, 3 + √2 − 3 = √2, 3 + 2√2 – (3 + √2) = √2 and 3 + 3√2
– (3 + 2√2) = √2
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = √2
Fifth
term = 3 + 3√2 + √2 = 3 + 4√2
Sixth
term = 3 + 4√2 + √2 = 3 + 5√2
Seventh
term = 3 + 5√2 + √2 = 3 + 6√2
Therefore,
the next three terms are (3 + 4√2), (3 + 5√2) and (3 + 6√2).
(vi) 0.2, 0.22, 0.222, 0.2222 …
Here, 0.22 − 0.2 ≠ 0.222 − 0.22
It is not an AP
because the difference between consecutive terms is not equal.
(vii) 0, −4, −8, −12 …
Here, −4 – 0 = −8 − (−4) = −12 − (−8) = −4
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = −4
Fifth term = −12
– 4 = −16
Sixth term = −16
– 4 = −20
Seventh
term = −20 – 4 = −24
Therefore, the next
three terms are −16, −20 and −24.
(viii)
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = 0
Fifth
term = 
Sixth
term =
Seventh
term =
Therefore,
the next three terms are 
(ix) 1, 3, 9, 27 …
Here, 3 – 1 ≠ 9 – 3
It
is not an AP because the difference between consecutive terms is not equal.
(x) a, 2a, 3a, 4a …
Here,
2a – a = 3a − 2a = 4a − 3a = a
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = a
Fifth
term = 4a + a = 5a
Sixth
term = 5a + a = 6a
Seventh
term = 6a + a = 7a
Therefore,
the next three terms are 5a, 6a and 7a.
(xi) a, a2, a3, a4 …
Here, a2 – a ≠ a3 − a2
It is not an AP
because the difference between consecutive terms is not equal.
(xii) √2, √8, √18, √32 …
⇒ √2, 2√2,
3√2, 4√2
Here, 2√2 – √2 =
3√2 – 2√2 = 4√2 – 3√2 =√2
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = √2
Fifth
term = 4√2+ √2 = 5√2 = √50
Sixth
term = 5√2+ √2 = 6√2 = √72
Seventh
term = 6√2+ √2= 7√2 = √98
Therefore,
next three terms are √50, √72, √98.
(xiii) √3, √6, √9, √12 …
Here, √6 – √3 ≠ √9 – √6
It is not an AP
because the difference between consecutive terms is not equal.
(xiv) 12, 32, 52,
72 …
Here, 32 – 12 ≠ 52
– 32
It is not an AP
because the difference between consecutive terms is not equal.
(xv) 12, 52, 72,
73 …
⇒ 1, 25, 49, 73 …
Here, 52 – 12 = 72
– 52 = 73 – 72 = 24
It is an AP
because the difference between consecutive terms is equal.
Common
difference (d) = 24
Fifth
term = 73 + 24 = 97
Sixth
term = 97 + 24 = 121
Seventh
term = 121 + 24 = 145
Therefore, the next three terms are 97, 121 and 145.