Selina Concise Class 10 Math Chapter 10 Arithmetic Progression Solutions
Exercise 10F
(Q1) The 6th term of an arithmetic progression is 16 and the 14th term is 32. Determine the 36th term.
= Solution:
Given that, The 6th term of an arithmetic progression is 16.
∴ t6 = 16
And the 14th term is 32
∴ t14 = 32
Let us consider the first term ‘a’ and the common difference ‘d’.
We know that,
tn = a + (n – 1) d
We have to find t6 –
t6 = a + (6 – 1) d
t6 = a + 5d
16 = a + 5d
And we have find t14 –
t14 = a + (14 – 1) d
t14 = a + 13d
32 = a + 13d —– (2)
From equation (1) – (2)
We get,
a + 5d = 16
a + 13d = 32
(-) (-) (-)
____________
– 8d = -16
d = 16/8
d = 2
Put d = 2 in equation (i), we get,
16 = a + 5(2)
16 = a + 10
16 – 10 = a
6 = a
∴ a = 6
Now, We have to find 36th term –
(Q3) An arithmetic progression of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the arithmetic progression.
Solution:
Given that, an arithmetic progression of 50 terms –
∴ n = 50
And also given that, the third (3rd) term is 12 –
t3 = 12
We know that,
tn = a + (n – 1) d
t3 = a + (3 – 1) d
t3 = a + 2d
12 = a + 2d —- (1)
Also, given that, last term (l) = 106
t50 = 106
We know that,
tn = a + (n – 1) d
t50 = a + (50 – 1) d
106 = a + 49d —– (2)
Subtract equation (1) from (2) we get,
12 = a + 2d
106 = a + 49d
(-) (-) (-)
____________
-94 = -47d
94/47 = d
∴ d = 2
Put d = 2 in equation (1), we get,
12 = a + 2×2
12 = a + 4
12 – 4 = a
8 = a
∴ a = 8
Now,
∴ We have to find the 29th term –
t29 = a + 28d
= 8 + 28 × d
= 8 + 28 × 2
= 8 + 56
t29 = 64
(Q5) Find the sum of first 10 terms of the arithmetic progression –
4 + 6 + 8 + —–
Solution:
Given arithmetic progression is 4 + 6 + 8 + —-
∴ a = 4
t1 = 4, t2 = 6, t3 = 8 —-
t2 – t1 = 6 – 4 = 2 = d1
t3 – t2 = 8 – 6 = 2 = d2
d1 = d2 = d
d = 2
Now, we have to find the sum of first 10 terms –
Sn = n/2 [2a + (n – 1) d]
S10 = 10/2 [2×4 + (10 – 1) 2] (∵ n = 10)
= 5 [8 + 9 (2)]
= 5 [8 + 18]
= 5 [26]
S10 = 130
(Q6) Find the sum of first 20 terms of an arithmetic progression whose first term is 3 and the last term is 57.
Solution:
Given that, the sum of first 20 terms of an arithmetic progression
∴ n = 20
And also given that,
First term, (t1) = a = 3
And the last term l = 57.
We know that,
S = n/2 (a + l)
= 20/2 (3 + 57)
= 20/2 (60)
= 20×30
S = 600
∴ The sum o first 20 terms of an arithmetic progression is 500.
(Q9) Find the general term (nth term) and 23rd term of the sequence 3, 1, -1, -3, —
Solution:
Given sequence is 3, 1, -1, -3, —-
Here, t1 = 3, t2 = 1, t3 =-1, t4 = -3, —-
t2 – t1 = 1 – 3= -2 = d1
t3 – t2 = -1 -1 = -2 = d2
d1 = d2 = d
∴ d = -2
t1 = a = 3
∴ The given sequence is in arithmetic progression.
∴ We know that the general form of arithmetic progression –
tn = a + (n – 1) d.
tn = 3 + (n – 1) (-2)
= 3 + (-2n + 2)
= 3 – 2n + 2
tn = 5 – 2n
Now, We have to find 23rd term –
t23 = 5 – 2n
= 5 – 2 × (23)
= 5 – 46
t23 = – 41
(Q10) Which term of the sequence 3, 8, 13, —- is 78?
Solution:
The given sequence is 3, 8, 13, —-
First term (t1) = a = 3
t1 = 3, t2 = 8, t3 = 13, ——
t2 – t1 = 8 – 3 = 5 = d1
t3 – t2 = 13 – 8 = 5 = d2
d1 = d2 = d
∴ d = 5
Given that the last term is –
Last term (l) = 78
∴ tn = 78
We know that,
tn = a + (n – 1) d
78 = 3 + (n – 1) (5)
= 3 + 5n – 5
78 = 5n – 2
78 + 2 = 5n
80 = 5n
80/5 = n
16 = n
∴ n = 16
∴ The 16th term of the given sequence is 78.
(Q12) How many two digit numbers are divisible by 3?
Solution:
The two – digit numbers are divisible by 3 are –
12, 15, 18, 21, 24, 27, —–, 99.
Here, t1 = 12, t2 = 15, t3 = 18, —
t2 – t1 = 15 – 12 = 3 = d1
t3 – t2 = 18 – 15 = 3 = d2
d1 = d2 = d
∴ d = 3 and t1 = a = 12
∴ The last term is 99.
tn = 99
We know that,
tn = a + (n – 1) d
99 = 12 + (n – 1) (3)
99 = 12 + 3n – 3
99 = 9 + 3n
99 – 9 = 3n
90 = 3n
90/3 = n
30 = n
∴ n = 30
∴ The 30 two-digit number are divisible by 3.