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# Ex 5.1 NCERT Solutions- Arithmetic Progressions Class 10 Notes | EduRev

## Class 10 : Ex 5.1 NCERT Solutions- Arithmetic Progressions Class 10 Notes | EduRev

The document Ex 5.1 NCERT Solutions- Arithmetic Progressions Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Ques 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.
Sol: (i) Let us consider,
The first term (T1) = Fare for the first km = ₹ 15 since, the taxi fare beyond the first km is ₹ 8 for each additional km. ⇒ T1 = 15
∴ Fare for 2 km = ₹15 + 1 × ₹ 8 ⇒ T2 = a + 8 [where a = 15]
Fare for 3 km = ₹ 15 + 2 × ₹ 8 ⇒ T3 = a + 16
Fare for 4 km = ₹ 15 + 3 × ₹ 8 ⇒ T4 = a + 24
Fare for 5 km = ₹ 15 + 4 × ₹ 8 ⇒ T5 = a + 32
Fare for n km = ₹ 15 + (n - 1) 8 ⇒ Tn = a + (n - 1) 8
We see that above terms form an A.P.
(ii) Let the amount of air in the cylinder = x
∴ Air removed in 1st stroke = ⇒ Air left after 1st stroke Air left after 2nd stroke Air left after 3rd stroke Air left after 4th stroke Thus, the terms are: Here,  Since The above terms are not in A.P.
(iii) Here, The cost of digging for 1st metre = ₹ 150
The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200
The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250
The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300
∴ The terms are: 150, 200, 250, 300, ...
Since, 200 - 150 = 50
And 250 - 200 = 50
⇒ (200 - 150) = (250 - 200)
∴The above terms form an A.P.
(iv) ∵ The amount at the end of 1st year The amount at the end of 2nd year = The amount at the end of 3rd year = The amount at the end of 4th year = ∴ The terms are
Obviously, ∴ The above terms are not in A.P.

Ques 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 = 0
(v) a = - 1.25, d = - 0.25
Sol: (i) ∵ Tn = a + (n - 1) d
∴ For a = 10 and d = 10, we have:
T1 = 10 + (1 - 1) × 10 = 10 + 0 = 10
T2 = 10 + (2 - 1) × 10 = 10 + 10 = 20
T3 = 10 + (3 - 1) × 10 = 10 + 20 = 30
T4 = 10 + (4 - 1) × 10 = 10 + 30 = 40
Thus, the first four terms of A.P. are:
10, 20, 30, 40.
(ii) ∵ Tn = a + (n - 1) d
∴ For a = - 2 and d = 0, we have:
T1 = -2 + (1 - 1) × 0 = -2 + 0 = -2
T2 = -2 + (2 - 1) × 0 = -2 + 0 = -2
T3 = -2 + (3 - 1) × 0 = -2 + 0 = -2
T4 = -2 + (4 - 1) × 0 = -2 + 0 = -2
∴ The first four terms are:
- 2, - 2, - 2, - 2.
(iii) ∵ Tn = a + (n - 1) d
∴ For a = 4 and d = - 3, we have:
T1 = 4 + (1 - 1) × (- 3) = 4 + 0 = 4
T2 = 4 + (2 - 1) × (- 3) = 4 + (- 3) = 1
T3 = 4 + (3 - 1) × (- 3) = 4 + (- 6) = - 2
T4 = 4 + (4 - 1) × (- 3) = 4 + (- 9) = - 5
Thus, the first four terms are:
4, 1, - 2, - 5.
(iv) ∵ Tn = a + (n - 1) d
For a = - 1 and d = 1/2, we get ∴ The first four terms are: (v) ∵ Tn = a + (n - 1) d
For a = - 1.25 and d = - 0.25, we get
T1 = - 1.25 + (1 - 1) × (- 0.25) = - 1.25 + 0 = - 1.25
T2 = - 1.25 + (2 - 1) × (- 0.25) = - 1.25 + (- 0.25) = - 1.50
T3 = - 1.25 + (3 - 1) × (- 0.25) = - 1.25 + (- 0.50) = - 1.75
T= - 1.25 + (4 - 1) × (- 0.25) = - 1.25 + (- 0.75) = - 2.0
Thus, the four terms are:
- 1.25, - 1.50, - 1.75, - 2.0

Ques 3: For the following APs, 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, ...
Sol: (i) We have: 3, 1, - 1, - 3, .....
⇒ T1 = 3 ⇒ a = 3
T2 = 1
T3 = - 1
T4 = - 3
∴ T2 - T1 = 1 - 3 = - 2
T4 - T3 = - 3 - (- 1) = - 3 + 2 = - 2 ⇒ d = - 2
Thus, a = 3 and d = - 2
(ii) We have: - 5, - 1, 3, 7, ..... Thus, a = -5 and d = 4
(iii) We have: .....  Thus, a =1/3 and d = 4/3
(iv) We have: 0.6, 1.7, 2.8, 3.9, .....
⇒ T1 = 0.6⇒ a = 0.6
T2 = 1.7 ⇒ d = T2 - T1 = 1.7 - 0.6 = 1.1
T3 = 2.8
T4 = 3.9 ⇒ d = T4 - T3 = 3.9 - 2.8 = 1.1
Thus, a = 0.6 and d = 1.1

Ques 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) ...
(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) √2,√8 √18 √ 32...
(xiii) √3, √6 √9, √12, ...

(xiv) 12, 32, 52, 72, ...
(xv) 12, 52, 72, 73, ...
Sol: (i) We have: 2, 4, 8, 16, ..... ∴ T2 - T1 ≠ T4 - T3
∴ The given numbers do not form an A.P.
(ii) We have: , .....
∴T1 = 2, T2 = 5/2, T3 = 3, T4 = 7/2
T2 - T1 = T3 - T2 = T4 - T3 = ∵ T2 - T1 = T3 - T2 = T4 - T3 = 1/2 ⇒ d = 1/2
∴The given numbers form an A.P.
∴T5 = T6 = T7 = Thus, d = 1/2 and T5 = 4, T6 = 9/2 and T7 = 5
(iii) We have: - 1.2, - 3.2, - 5.2, - 7.2, .....
∴T1 = -1.2, T2 = -3.2, T3 = -5.2, T4 = -7.2
T2 - T1 = -3.2 + 1.2 = -2
T3 - T2 = -5.2 + 3.2 = -2
T4 - T3 = -7.2 + 5.2 = -2
∵ T2 - T1 = T3 - T2 = T4 - T3 = -2 ⇒ d = -2
∴ The given numbers form an A.P.
Such that d = - 2.
Now, T5 = T4 + (- 2) = - 7.2 + (- 2) = - 9.2
T6 = T5 + (- 2) = - 9.2 + (- 2) = - 11.2
T7 = T6 + (- 2) = - 11.2 + (- 2) = - 13.2
Thus, d = - 2 and T5 = - 9.2, T6 = - 11.2 and T7 = - 13.2
(iv) We have: - 10, - 6, - 2, 2, .....
∴ T1 = -10, T2 = -6, T3 = -2, T4 = 2
T2 - T1 = -6 + 10 = 4
T3 - T2 = -2 + 6 = 4
T4 - T3 = 2 + 2 = 4
∵ T2 - T1 = T3 - T2 = T4 - T3 = 4 ⇒ d = 4
∴The given numbers form an A.P.
Now, T5 = T4 + 4 = 2 + 4 = 6
T6 = T5 + 4 = 6 + 4 = 10
T7 = T6 + 4 = 10 + 4 = 14
Thus, d = 4 and T5 = 6, T6 = 10, T7 = 14
(v) We have:   ⇒ The given numbers form an A.P.
Now, T5 = T4+√2
= 3+3√2+√2 = 3+4√2
T6 = T5+ √2
= 3+4√2+√2 = 3+5√2
T7 = T6+√2
= 3+5√2+√2 = 3+6√2
Thus, d = √2 and T5 = 3+4√2, T6 = 3+5√2, T7 = 3.+6√2.
(vi) We have: 0.2, 0.22, 0.222, 0.2222, ..... Since,
T2 - T1 ≠ T4 - T3
∴ The given numbers do not form an A.P.
(vii) We have: 0, - 4, - 8, - 12, .....
∴ T1 = 0, T2 = -4, T3 = -8, T4 = -12
T2 - T1 = -4 - 0 = -4
T3 - T2 = -8 + 4 = -4
T4 - T3 = -12 + 8 = -4
∵ T2 - T1 = T3 - T2 = T4 - T3 = -4 ⇒ d = -4
∴The given numbers form an A.P.
Now, T5 = T4 + (- 4) = - 12 + (- 4) = - 16
T6 = T5 + (- 4) = - 16 + (- 4) = - 20
T7 = T6 + (- 4) = - 20 + (- 4) = - 24
Thus, d = - 4 and T5 = - 16, T6 = - 20, T7 = - 24
(viii) We have:  T2 - T1 = 0
T3 - T2 = 0
T4 - T3 = 0
∵ T2 - T1 = T3 - T2 = T4 - T3 = 0 ⇒ d = 0
∴ The given number form an A.P.
Now,
T5 = T6 = T7 = Thus, (ix) We have: 1, 3, 9, 27, .....
Here, ∵ T2 - T1 ≠ T4 - T3
∴The given numbers do not form an A.P.
(x) We have: a, 2a, 3a, 4a, .....
∴ T1 = a, T2 = 2a, T3 = 3a, T4 = 4a
T2 - T1 = 2a - a = a
T3 - T2 = 3a - 2a = a
T4 - T3 = 4a - 3a = a
∵  T2 - T1 = T3 - T2 = T4 - T3 = a ⇒ d = a
∴The numbers form an A.P.
Now, T5 = T4 + a = 4a + a = 5a
T6 = T5 + a = 5a + a = 6a
T7 = T6 + a = 6a + a = 7a
Thus, d = a and T5 = 5a, T6 = 6a, T7 = 7a
(xi) We have: a, a2, a3, a4, ..... Since,
T2 - T1 ≠ T4 - T3
∴ The given terms are not in A.P.
(xii) We have:   ∴ The given numbers form an A.P.
Now, Thus, (xiii) We have:  ∴ T2 - T1 ≠ T4 - T3
⇒ The given terms do not form an A.P.
(xiv) We have: 12, 32, 52, 72, ..... ∵ T2 − T1 ≠ T4 − T3
∴ The given terms do not form an A.P.
(xv) We have: 12, 52, 72, 73, .....
∴ T1 = 12 = 1, T2 = 52 = 25, T3 = 72 = 49, T4 = 73
T2 - T1 = 25 - 1 = 24
T3 - T2 = 49 - 25 = 24
T4 - T3 = 73 - 49 = 24
∵ T2 - T1 = T3 - T2 = T4 - T3 = 24 ⇒ d = 24
∴ The numbers form an A.P.
Now, T5 = T4 + 24 = 73 + 24 = 97
T6 = T5 + 24 = 97 + 24 = 121
T7 = T6 + 24 = 121 + 24 = 145
Thus, d = 24 and T= 97, T6 = 121, T7 = 145
nth Term of an A.P.

The nth term Tn of the A.P. with first term ‘a’ and common difference ‘d’ is given by
Tn = a + (n - 1) d
‘Tn’ is also called the general term of the A.P. If there are ‘m’ terms in the A.P., then ‘Tm’ represents the last term which is generally denoted by ‘l’.

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## Mathematics (Maths) Class 10

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