Practice Set 3.2 Arithmetic Progression – Maharashtra Board Class 10 Maths 1 Solutions
Practice Set 3.2 – Arithmetic Progression
Formula:
Where, a = first term, d = common difference, n = number of terms and tn = nth term.
1. Write the correct number in the given boxes from the following A.P.
(i) 1, 8, 15, 22, ...
Here 1 a = 1, t1 = 8, t2 = 15, t3 = 15,
t2 − t1 = 8 − 1 = 7
t3 − t2 = 15 − 8 = 7 ∴ d = 7
Correct filling according to the question:
Here a = 1, t1 = 1, t2 = 8, t3 = 15
t2 − t1 = 8 − 1 = 7
t3 − t2 = 15 − 8 ∴ d = 7
(ii) 3, 6, 9, 12, ...
Here t1 = 3, t2 = 6, t3 = 9, t4 = 12,
t2 − t1 = 3, t3 − t2 = 3 ∴ d = 3
Therefore, d = 3.
(iii) −3, −8, −13, −18, ...
Here t3 = −13, t2 = −8, t4 = −18, t1 = −3,
t2 − t1 = −8 − (−3) = −5
t3 − t2 = −13 − (−8) = −5
∴ a = −3, d = −5
(iv) 70, 60, 50, 40, ...
Here t1 = 70, t2 = 60, t3 = 50, ...
∴ a = 70, d = −10
2. Decide whether following sequence is an A.P., if so find the 20th term of the progression.
−12, −5, 2, 9, 16, 23, 30, ...
Common difference:
−5 − (−12) = 7
2 − (−5) = 7
9 − 2 = 7
Since the common difference is constant, the given sequence is an A.P.
Therefore, a = −12, d = 7 and n = 20.
Using the formula:
t20 = −12 + (20 − 1) × 7
= −12 + 19 × 7
= −12 + 133
∴ t20 = 121
Answer: The sequence is an A.P. and its 20th term is 121.
3. Given Arithmetic Progression 12, 16, 20, 24, ... Find the 24th term of this progression.
Here, a = 12
d = 16 − 12 = 4
n = 24
Using,
t24 = 12 + (24 − 1) × 4
= 12 + 23 × 4
= 12 + 92
∴ t24 = 104
Answer: 104
4. Find the 19th term of the following A.P.
7, 13, 19, 25, ...
a = 7
d = 13 − 7 = 6
n = 19
t19 = a + (n − 1)d
= 7 + (19 − 1) × 6
= 7 + 18 × 6
= 7 + 108
∴ t19 = 115
Answer: 115
5. Find the 27th term of the following A.P.
9, 4, −1, −6, −11, ...
a = 9
d = 4 − 9 = −5
n = 27
t27 = 9 + (27 − 1)(−5)
= 9 + 26(−5)
= 9 − 130
∴ t27 = −121
Answer: −121
6. Find how many three digit natural numbers are divisible by 5.
The smallest three-digit number divisible by 5 is 100.
The largest three-digit number divisible by 5 is 995.
Therefore, the A.P. is:
100, 105, 110, ..., 995
Here, a = 100, d = 5 and tn = 995.
Using,
995 = 100 + (n − 1)5
995 − 100 = 5(n − 1)
895 = 5(n − 1)
179 = n − 1
n = 180
Answer: There are 180 three-digit natural numbers divisible by 5.
7. The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Given:
t11 = 16
t21 = 29
Using tn = a + (n − 1)d:
t11 = a + 10d = 16 ...(1)
t21 = a + 20d = 29 ...(2)
Subtracting (1) from (2):
10d = 13
d = 13/10
From equation (1):
a + 10(13/10) = 16
a + 13 = 16
a = 3
Now,
t41 = a + (41 − 1)d
= 3 + 40(13/10)
= 3 + 52
∴ t41 = 55
Answer: 55
8. 11, 8, 5, 2, ... In this A.P. which term is number −151?
Given:
a = 11
d = 8 − 11 = −3
Let −151 be the nth term.
Therefore,
−151 = 11 + (n − 1)(−3)
−151 = 11 − 3(n − 1)
−162 = −3(n − 1)
54 = n − 1
n = 55
Answer: −151 is the 55th term.
9. In the natural numbers from 10 to 250, how many are divisible by 4?
The first number greater than 10 divisible by 4 is 12.
The last number less than 250 divisible by 4 is 248.
The A.P. is:
12, 16, 20, 24, ..., 248
Here, a = 12, d = 4 and tn = 248.
Using,
248 = 12 + (n − 1)4
236 = 4(n − 1)
59 = n − 1
n = 60
Answer: 60 natural numbers are divisible by 4.
10. In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
Given:
t17 = t10 + 7
Using the formula:
t17 = a + 16d
t10 = a + 9d
Therefore,
a + 16d = a + 9d + 7
16d = 9d + 7
16d − 9d = 7
7d = 7
∴ d = 1
Answer: The common difference is 1.
Final Answers
| Question | Answer |
|---|---|
| 1(i) | a = 1, d = 7 |
| 1(ii) | d = 3 |
| 1(iii) | a = −3, d = −5 |
| 1(iv) | a = 70, d = −10 |
| 2 | 121 |
| 3 | 104 |
| 4 | 115 |
| 5 | −121 |
| 6 | 180 |
| 7 | 55 |
| 8 | 55th term |
| 9 | 60 |
| 10 | d = 1 |
Comments
Post a Comment