Exercise 5.2
Question 4:
Which term of the A.P. 3, 8, 13, 18,… is 78?
3, 8, 13, 18,…Given
a = 3
d = a2 − a1 = 8 − 3 = 5
Let nth term of this A.P. be last term i;e 78.
an = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16th term of this A.P. is 78.
Question 5:
Find the number of terms in each of the following A.P.
I. 7, 13, 19, …, 205
Given
a = 7
d = a2 − a1 = 13 − 7 = 6
Let there are n terms in this A.P.
an = 205
Since, an = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.
Given
Let there are n terms in this A.P.
Therefore, an = −47 and we know that,
Since, an = a + (n − 1) d
Therefore, this given A.P. has 27 terms in it.
Question 6:
Check whether − 150 is a term of the A.P. 11, 8, 5, 2,…
Given that, a = 11, d = a2 − a1 = 8 − 11 = −3Let −150 be the nth term of this A.P.
Since, an = a + (n − 1) d
Clearly, n is not an integer.
Therefore, −150 is not a term of this A.P.
Question 7:
Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73
Given, a11 = 38, a16 = 73Since, an = a + (n − 1) d
a11 = a + (11 − 1) d
38 = a + 10d .............(1)
Similarly,
a16 = a + (16 − 1) d
73 = a + 15d .............(2)
On subtracting (1) from (2),
35 = 5d
d = 7
From equation (1),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a31 = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31st term is 178.
Question 8:
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term
Given, a3 = 12, a50 = 106
Since, an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a + 2d .............(I)
Similarly, a50 = a + (50 − 1) d
106 = a + 49d ............(II)
On subtracting (I) from (II),
94 = 47d
d = 2
From equation (I),
12 = a + 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64