Practice Set 3.2
1, 8, 15, 22, . . . Here a = square , t_1 = square , t_2 = square , t_3 = square…
3, 6, 9, 12, . . . Here t_1 = square , t_2 = square , t_3 = square , t_4 = square…
- 3, - 8, - 13, - 18, . . . Here t_3 = square , t_2 = square , t_4 = square , t_1 =…
70, 60, 50, 40, . . . Here t_1 = square , t_2 = square , t_3 = square therefore a =…
Decide whether following sequence is an A.P., if so find the 20th term of the…
Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this…
Find the 19th term of the following A.P. 7, 13, 19, 25, . . .
Find the 27th term of the following A.P. 9, 4, - 1, - 6, - 11, . . .…
Find how many three digit natural numbers are divisible by 5.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the…
11, 8, 5, 2, . . . In this A.P. which term is number - 151?
In the natural numbers from 10 to 250, how many are divisible by 4?…
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.…
Practice Set 3.2
Question 1.
Write the correct number in the given boxes from the following A. P.
1, 8, 15, 22, . . .
Here 
Answer:
1, 8, 15, 22, . . .
First term a = 1
Second term t1 = 8
Third term t2 = 15
Fourth term t3 = 22
We know that d = tn + 1 – tn
Thus, t2 – t1 = 15 – 8 = 7
t3 – t2 = 22 – 15 = 7
Thus, d = 7
Question 2.
Write the correct number in the given boxes from the following A. P.
3, 6, 9, 12, . . .
Here 
Answer:
3,6,9,12, . . .
First term a = 3
Second term t1 = 6
Third term t2 = 9
Fourth term t3 = 12
We know that d = tn + 1 – tn
Thus, t2 – t1 = 9 – 6 = 3
t3 – t2 = 12 – 9 = 3
Thus, d = 3
Question 3.
Write the correct number in the given boxes from the following A. P.
– 3, – 8, – 13, – 18, . . .
Here 
Answer:
– 3, – 8, – 13, – 18, . . .
First term a = – 3
Second term t1 = – 8
Third term t2 = – 13
Fourth term t3 = – 18
We know that d = tn + 1 – tn
Thus, t2 – t1 = – 13 – ( – 8) = – 13 + 8 = – 5
t3 – t2 = – 18 – ( – 13) = – 18 + 13 = – 5
Thus, d = – 5
Question 4.
Write the correct number in the given boxes from the following A. P.
70, 60, 50, 40, . . .
Here 
Answer:
70, 60, 50, 40, . . .
First term a = 70
Second term t1 = 60
Third term t2 = 50
Fourth term t3 = 40
We know that d = tn + 1 – tn
Thus, t2 – t1 = 50 – 60 = – 10
t3 – t2 = 40 – 50 = – 10
Thus, d = – 10
Question 5.
Decide whether following sequence is an A.P., if so find the 20th term of the progression.
– 12, – 5, 2, 9, 16, 23, 30, . . .
Answer:
Given A.P. is – 12, – 5, 2, 9, 16, 23, 30, . . .
Where first term a = – 12
Second term t1 = – 5
Third term t2 = 2
Common Difference d = t2 – t1 = 2 – ( – 5) = 2 + 5 = 7
We know that, nth term of an A.P. is
tn = a + (n – 1)d
We need to find the 20th term,
Here n = 20
Thus, t20 = – 12 + (20 – 1)× 7
t20 = – 12 + (19)× 7 = – 12 + 133 = 121
Thus, t20 = 121
Question 6.
Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.
Answer:
Given A.P. is 12, 16, 20, 24, . . .
Where first term a = 12
Second term t1 = 16
Third term t2 = 20
Common Difference d = t2 – t1 = 20 – 16 = 4
We know that, nth term of an A.P. is
tn = a + (n – 1)d
We need to find the 24th term,
Here n = 24
Thus, t24 = 12 + (24 – 1)× 4
t24 = 12 + (23)× 4 = 12 + 92 = 104
Thus, t24 = 104
Question 7.
Find the 19th term of the following A.P.
7, 13, 19, 25, . . .
Answer:
Given A.P. is 7, 13, 19, 25, . . .
Where first term a = 7
Second term t1 = 13
Third term t2 = 19
Common Difference d = t2 – t1 = 19 – 13 = 6
We know that, nth term of an A.P. is
tn = a + (n – 1)d
We need to find the 19th term,
Here n = 19
Thus, t19 = 7 + (19 – 1)× 6
t19 = 7 + (18)× 6 = 7 + 108 = 115
Thus, t19 = 115
Question 8.
Find the 27th term of the following A.P.
9, 4, – 1, – 6, – 11, . . .
Answer:
Given A.P. is 9, 4, – 1, – 6, – 11, . . .
Where first term a = 9
Second term t1 = 4
Third term t2 = – 1
Common Difference d = t2 – t1 = – 1 – 4 = – 5
We know that, nth term of an A.P. is
tn = a + (n – 1)d
We need to find the 27th term,
Here n = 27
Thus, t27 = 9 + (27 – 1)× ( – 5)
t27 = 9 + (26)× ( – 5) = 9 – 130 = – 121
Thus, t27 = – 121
Question 9.
Find how many three digit natural numbers are divisible by 5.
Answer:
List of three digit number divisible by 5 are
100, 105,110,115,……….. 995
Let us find how many such number are there?
From the above sequence, we know that
tn = 995, a = 100
t1 = 105, t2 = 110
Thus, d = t2 – t1 = 110 – 105 = 5
Now, By using nth term of an A.P. formula
tn = a + (n – 1)d
we can find value of “n”
Thus, on substituting all the value in formula we get,
995 = 100 + (n – 1)× 5
⇒ 995 – 100 = (n – 1)× 5
⇒ 895 = (n – 1) × 5
⇒ n = 179 + 1 = 180
Question 10.
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.
Answer:
Given: t11 = 16 and t21 = 29
To find: t41
Using nth term of an A.P. formula
tn = a + (n – 1)d
we will find value of “a” and “d”
Let, t11 = a + (11 – 1) d
⇒ 16 = a + 10 d …..(1)
t21 = a + (21 – 1) d
⇒ 29 = a + 20 d …..(2)
Subtracting eq. (1) from eq. (2), we get,
⇒ 29 – 16 = (a – a) + (20 d – 10 d)
⇒ 13 = 10 d
Substitute value of “d” in eq. (1) to get value of “a”
⇒ 16 = a + 13
⇒ a = 16 – 13 = 3
Now, we will find the value of t41 using nth term of an A.P. formula
⇒ t41 = 3 + 4 × 13 = 3 + 52 = 55
Thus, t41 = 55
Question 11.
11, 8, 5, 2, . . . In this A.P. which term is number – 151?
Answer:
By, given A.P. 11, 8, 5, 2, . . .
we know that
a = 11, t1 = 8, t2 = 5
Thus, d = t2 – t1 = 5 – 8 = – 3
Given: tn = – 151
Now, By using nth term of an A.P. formula
tn = a + (n – 1)d
we can find value of “n”
Thus, on substituting all the value in formula we get,
– 151 = 11 + (n – 1)× ( – 3)
⇒ – 151 – 11 = (n – 1)× ( – 3)
⇒ – 162 = (n – 1) × ( – 3)
⇒ n = 54 + 1 = 55
Question 12.
In the natural numbers from 10 to 250, how many are divisible by 4?
Answer:
List of number divisible by 4 in between 10 to 250 are
12, 16,20,24,……….. 248
Let us find how many such number are there?
From the above sequence, we know that
tn = 248, a = 12
t1 = 16, t2 = 20
Thus, d = t2 – t1 = 20 – 16 = 4
Now, By using nth term of an A.P. formula
tn = a + (n – 1)d
we can find value of “n”
Thus, on substituting all the value in formula we get,
248 = 12 + (n – 1)× 4
⇒ 248 – 12 = (n – 1)× 4
⇒ 236 = (n – 1) × 4
⇒ n = 59 + 1 = 60
Question 13.
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
Answer:
Given: t17 = 7 + t10 ……(1)
In t17, n = 17
In t10, n = 10
By using nth term of an A.P. formula,
tn = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
tn = nth term
Thus, on using formula in eq. (1) we get,
⇒ a + (17 – 1)d = 7 + (a + (10 – 1)d)
⇒ a + 16 d = 7 + (a + 9 d)
⇒ a + 16 d – a – 9 d = 7
⇒ 7 d = 7
Thus, common difference “d” = 1
