(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
(ii) 11th term of the AP: – 3, -1/2, 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48 1/2
Solution:
(i) From the given,
A.P. = 10, 7, 4, …
Therefore, we can find,
First term, a = 10
Common difference, d = a2 − a1 = 7 − 10 = −3
As we know, for an A.P.,
an = a + (n − 1) d
Putting the values;
a30 = 10 + (30 − 1)(−3)
a30 = 10 + (29)(−3)
a30 = 10 − 87 = −77
Hence, the correct answer is option C.
(ii) From the given,
A.P. = -3, -1/2, ,2 …
Therefore, we can find,
First term a = – 3
Common difference, d = a2 − a1 = (-1/2) – (-3)
⇒ (-1/2) + 3 = 5/2
As we know, for an A.P.,
an = a + (n − 1) d
Putting the values;
a11 = -3 + (11 – 1)(5/2)
a11 = -3 + (10)(5/2)
a11 = -3 + 25
a11 = 22
Hence, the correct answer is option B.