Maharashtra Board Class 10 Math Part 1 Solution Chapter 3 Practice Set 3.2 – Arithmetic Progression
Balbharati Maharashtra Board Class 10 Math Part 1 Solution Chapter 3: Arithmetic Progression. Marathi or English Medium Students of Class 10 get here Arithmetic Progression full Practice set Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 1 Solution |
Chapter |
Arithmetic Progression |
Practice set |
3.2 |
Practice Set 3.2
(1) Write the correct number in the given boxes from the following A.P.
(i) 1, 8, 15, 22, …
Here a = , t1 = , t2 = , t3 = ,
t2 – t1 = – =
t3 – t2 = – =
∴ d =
Solution:
1, 8, 15, 22, ….
Here a = 1, t1 = 1, t2 = 8, t3 = 15, t4 = 22
t2 – t1 = 8 – 1 = 7
t3 – t2 = 15 – 8 = 7
∴ d = 7
(ii) 3, 6, 9, 12, …..
Here, t1 = , t2 = , t3 = , t4 = ,
t2 – t1 = , t3 – t2 =
∴ d =
Solution:
3, 6, 9, 12, ……
Here, t1 = 3, t2 = 6, t3 = 9, t4 = 12,
t2 – t1 = 6-3, t3 – t2 = 9-6 = 3
∴ d = 3
(iii) -3, -8, -13, -18, …..
Here, t3 = , t2 = , t4 = , t1 = ,
t2 – t1 = , t3 – t2 =
∴ a = , d =
Solution:
3, -8, -13, -18, …..
t3 = -13, t2 = -8, t4 = -18, t1 = -5,
t2 – t1 = -8, -(-3) = -5
t3 – t2 = -13 – (-8) = -5
∴ a = -3, d = -5
(iv) 70, 60, 50, 40, …….
Here, t1 = , t2 = , t3 = , …….
∴ a = , d =
Solution:
70, 60, 50, 40, …….
t1 = 70, t2 = 60, t3 = 50
∴ t2 – t1 = 60 – 70 = -10
t3 – t2 = 50 – 60 = -10
∴ 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, …..
Solution:
-12, -5, 2, 9, 16, 23, 30, …..
a = 12
t2 – t1 = -5 – (-12)
= 7
t3 – t2 = 2 – (-5)
t4 – t3 = 9 – 2 = 7
t5 – t4 = 16 – 9 = 7
∴ d = 7
∴ The above series is in A.P.
Now, tn = a + (n – 1) d. for n = 20
tn = -12 + (20 – 1) = 7
= – 12 + 140 – 7
= 140 – 19
= 121
∴ The 30th term is 121.
(3) Given Arithmetic Progression 12, 16, 20, 24, …… Find the 24th term of this progression.
Solution:
12, 16, 20, 24, ……
∴ a = 12; t2 – t1 = 16 – 12 = 4
t3 – t2 = 20 – 16 = 4
∴ d = 4
for n = 24
tn = a + (n – 1) d
= 12 + (24 – 1) 4
= 12 + 96 – 4
= 104
∴ 24th term of the AP is 104
(4) Find the 19th term of the following A.P. 7, 13, 19, 25, ……
Solution:
7, 13, 19, 25, ……
a = 7; t2 – t1 = 13 – 7 = 6
t3 – t2 = 19 – 13 = 6
∴ d = 6
for n = 19
tn = a + (n – 1) d
= 7 + (19 – 1) × 6
= 7 + 18 × 6
= 7 + 108
= 115
∴ The 19th term of the AP is 115
(5) Find the 27th term of the following A.P. 9, 4, -1, -6, -11, ……
Solution:
9, 4, -1, -6, -11, ……
a = 9; t2 – t1 = 4 – 9 = -5
t3 – t2 = -1-4 = -5
∴ d = -5
for n = 27
tn = a + (n – 1) d
= 9 + (27 – 1) × -5
= 9 + 26 × -5
= 9 – 130
= -121
∴ The 27th term of the AP is -121
(6) Find how many three digit natural numbers are divisible by 5.
Solution:
3 digit natural number divisible by 5 are –
100, 105, 110, ….. 995
∴ a = 100; t2 – t1 = 105 – 100 = 5
t3 – t2 = 110 – 105 = 5
∴ d = 5
∴ for tn = 995
tn = a + (n – 1) d
Or, 995 = 100 + (n – 1) 5
Or, 895 = 5n – 5
Or, 5n = 900
Or, n = 900/5 = 180
∴ There are 180 three digit numbers that are divisible by 5.
(7) The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P.
Solution:
Given, 11th term = 16
21st term = 29
t21 = 29
t11 = 16
∴ t11 = a + (n – 1) d: where n = 11
Or, 16 = a + (11 – 1) d
Or, 16 = a + 10d – (i)
∴ t21 = a + (n – 1) d; where n = 21
Or, 29 = a + (21 – 1) d
Or, 29 = a + 20d — (ii)
Now, from equation (i) substituting the value 7 d we get.
a + 10d = 16
Or, a + 10 × 13/10 = 16
Or, a = 16 – 13 = 3
∴ t41 = a + (n – 1) d
= 3 + (41 – 1) × 13/10
= 3 + 40 × 13/10
= 3 + 52
= 55
∴ The 41st term 7 the AP is 55.
(8) 11, 8, 5, 2, ….. In this A.P. which term is number -151?
Solution:
11, 8, 5, 2, …..
Given, tn = -151 a = 11
t2 – t1 = 8 – 11 = -3
t3 – t2 = 5 – 8 = -3
∴ d = -3
∴ tn = a + (n – 1) d
Or, -151 = 11 + (n – 1) × -3
Or, – 151 = 11 – 3n + 3
Or, 3n = 14 + 151
Or, n = 165/3 = 55
∴ -151 is the 55th term 7 the AP.
(9) In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:
Natural numbers from 10 to 25 divisible by 4 are –
12, 16, 20, ….. 248
Now, a = 12; t2 – t1 = 16 – 12 = 4
t3 – t2 = 20 – 16 = 4
∴ d = 4
For, tn = 248.
(Since 10 and 250 are not divisible by 4 we take natural numbers from within then divisibility by 4.)
tn = a + (n – 1) d
Or, 248 = 12 + (n – 1) × 4
Or, 248 = 12 + 4n – 4
Or, 4n = 248 – 8
Or, n = 240/4 = 60
∴ There are 60 natural numbers between 10 and 250 divisible by 4.
(10) In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
Solution:
t17 – t10 = 7
∴ t17 = a + (n – 1) d
= a + (17 – 1) d
= a + (17 – 1) d
= a + 16d
t10 = a + (n – 1) d
= a + (10 – 1) d
= a + 9d
Now, t17 – t10 = 7
Or, a + 16d – a – 9d = 7
Or, 7d = 7
Or, d = 7/7 = 1
∴ The common difference is 1.