Lesson 3 Arithmetic Progression | Practice Set 3.1 Solutions | Maharashtra Board Class 10 Maths 1
Lesson number 3. Arithmetic progression
Practice Set 3.1 — Solutions
Q.1: Compare consecutive differences.
Q.2: Start with a = t₁, then add d to get the next terms.
Q.3: Identify a = t₁ and calculate d = t₂ − t₁.
Q.1 Which of the following sequences are A.P.? If they are A.P., find the common difference.
(1) 2, 4, 6, 8, …
Given:
t₁ = 2, t₂ = 4, t₃ = 6, t₄ = 8
Now,
t₂ − t₁ = 4 − 2 = 2
t₃ − t₂ = 6 − 4 = 2
t₄ − t₃ = 8 − 6 = 2
All the differences are equal.
The given sequence is an A.P.
d = 2
(2) 2,
| 5 |
| 2 |
| 7 |
| 3 |
t₁ = 2, t₂ =
| 5 |
| 2 |
| 7 |
| 3 |
t₂ − t₁ =
| 5 |
| 2 |
| 1 |
| 2 |
t₃ − t₂ = 3 −
| 5 |
| 2 |
| 1 |
| 2 |
But,
t₄ − t₃ =
| 7 |
| 3 |
| 7 |
| 3 |
| 9 |
| 3 |
| 2 |
| 3 |
The differences are not equal.
The given sequence is not an A.P.
(3) −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
All differences are equal.
The given sequence is an A.P.
d = 4
(4) 0.3, 0.33, 0.333, …
t₁ = 0.3, t₂ = 0.33, t₃ = 0.333
t₂ − t₁ = 0.33 − 0.3 = 0.03
t₃ − t₂ = 0.333 − 0.33 = 0.003
Since,
0.03 ≠ 0.003
The given sequence is not an A.P.
(5) 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
All differences are equal.
The given sequence is an A.P.
d = −4
(6)
| −1 |
| 5 |
| −1 |
| 5 |
| −1 |
| 5 |
t₁ =
| −1 |
| 5 |
| −1 |
| 5 |
| −1 |
| 5 |
t₂ − t₁ =
| −1 |
| 5 |
| −1 |
| 5 |
t₃ − t₂ =
| −1 |
| 5 |
| −1 |
| 5 |
Therefore,
The given sequence is an A.P.
d = 0
(7) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …
t₁ = 3
t₂ = 3 + √2
t₃ = 3 + 2√2
t₄ = 3 + 3√2
Now,
t₂ − t₁ = (3 + √2) − 3 = √2
t₃ − t₂ = (3 + 2√2) − (3 + √2) = √2
t₄ − t₃ = (3 + 3√2) − (3 + 2√2) = √2
Hence,
The given sequence is an A.P.
d = √2
(8) 127, 132, 137, …
t₁ = 127, t₂ = 132, t₃ = 137
t₂ − t₁ = 132 − 127 = 5
t₃ − t₂ = 137 − 132 = 5
Therefore,
The given sequence is an A.P.
d = 5
Final answers of Q.1
| No. | A.P.? | Common difference |
|---|---|---|
| (1) | Yes | 2 |
| (2) | No | — |
| (3) | Yes | 4 |
| (4) | No | — |
| (5) | Yes | −4 |
| (6) | Yes | 0 |
| (7) | Yes | √2 |
| (8) | Yes | 5 |
Q.2 Write an A.P. whose first term is a and common difference is d.
(1) a = 10, d = 5
Given:
a = 10, d = 5
Therefore,
t₁ = a = 10
t₂ = t₁ + d = 10 + 5 = 15
t₃ = t₂ + d = 15 + 5 = 20
t₄ = t₃ + d = 20 + 5 = 25
Therefore,
10, 15, 20, 25, …
(2) a = −3, d = 0
t₁ = a = −3
t₂ = t₁ + d = −3 + 0 = −3
t₃ = t₂ + d = −3 + 0 = −3
t₄ = t₃ + d = −3 + 0 = −3
Therefore,
−3, −3, −3, −3, …
(3) a = −7, d =
| 1 |
| 2 |
t₁ = a = −7
t₂ = t₁ + d = −7 +
| 1 |
| 2 |
| 13 |
| 2 |
t₃ = t₂ + d = −
| 13 |
| 2 |
| 1 |
| 2 |
| 12 |
| 2 |
t₄ = t₃ + d = −6 +
| 1 |
| 2 |
| 11 |
| 2 |
Therefore,
−7, −
| 13 |
| 2 |
| 11 |
| 2 |
(4) a = −1.25, d = 3
t₁ = a = −1.25
t₂ = t₁ + d = −1.25 + 3 = 1.75
t₃ = t₂ + d = 1.75 + 3 = 4.75
t₄ = t₃ + d = 4.75 + 3 = 7.75
Therefore,
−1.25, 1.75, 4.75, 7.75, …
(5) a = 6, d = −3
t₁ = a = 6
t₂ = t₁ + d = 6 − 3 = 3
t₃ = t₂ + d = 3 − 3 = 0
t₄ = t₃ + d = 0 − 3 = −3
Therefore,
6, 3, 0, −3, …
(6) a = −19, d = −4
t₁ = a = −19
t₂ = t₁ + d = −19 − 4 = −23
t₃ = t₂ + d = −23 − 4 = −27
t₄ = t₃ + d = −27 − 4 = −31
Therefore,
−19, −23, −27, −31, …
Q.3 Find the first term and common difference for each A.P.
(1) 5, 1, −3, −7, …
Given:
t₁ = 5, t₂ = 1, t₃ = −3, t₄ = −7
The first term is:
a = t₁ = 5
Now,
d = t₂ − t₁
= 1 − 5
d = −4
Check:
−3 − 1 = −4
−7 − (−3) = −4
Therefore,
a = 5, d = −4
(2) 0.6, 0.9, 1.2, 1.5, …
t₁ = 0.6, t₂ = 0.9, t₃ = 1.2, t₄ = 1.5
First term:
a = t₁ = 0.6
Common difference:
d = t₂ − t₁
= 0.9 − 0.6
d = 0.3
Therefore,
a = 0.6, d = 0.3
(3) 127, 135, 143, 151, …
t₁ = 127, t₂ = 135, t₃ = 143, t₄ = 151
First term:
a = t₁ = 127
Common difference:
d = t₂ − t₁
= 135 − 127
d = 8
Therefore,
a = 127, d = 8
(4)
| 1 |
| 4 |
| 3 |
| 4 |
| 5 |
| 4 |
| 7 |
| 4 |
t₁ =
| 1 |
| 4 |
| 3 |
| 4 |
| 5 |
| 4 |
| 7 |
| 4 |
First term:
a = t₁ =
| 1 |
| 4 |
Common difference:
d = t₂ − t₁
=
| 3 |
| 4 |
| 1 |
| 4 |
=
| 3 − 1 |
| 4 |
d =
| 1 |
| 2 |
Therefore,
a =
| 1 |
| 4 |
| 1 |
| 2 |
Final Answers — Q.3
(1) a = 5, d = −4
(2) a = 0.6, d = 0.3
(3) a = 127, d = 8
(4)
a =
| 1 |
| 4 |
| 1 |
| 2 |
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