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Practice each question, then open the answer or worked solution.
In an AP, if d = –4, n = 7, an = 4, then a is
(A) 6 (B) 7 (C) 20 (D) 28
(D) 28
Explanation:
We know that nth term of an AP is
an = a + (n – 1)d
where,
a = first term
an is nth term
d is the common difference
According to the question,
4 = a + (7 – 1)(- 4)
4 = a – 24
a = 24 + 4 = 28
In an AP, if a = 3.5, d = 0, n = 101, then an will be
(A) 0 (B) 3.5 (C) 103.5 (D) 104.5
(B) 3.5
Explanation:
We know that nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
an = 3.5 + (101 – 1)0
= 3.5
(Since, d = 0, it’s a constant A.P)
The list of numbers – 10, – 6, – 2, 2,… is
(A) an AP with d = – 16
(B) an AP with d = 4
(C) an AP with d = – 4
(D) not an AP
(B) an AP with d = 4
Explanation:
According to the question,
a1 = – 10
a2 = – 6
a3 = – 2
a4 = 2
a2 – a1 = 4
a3 – a2 = 4
a4 – a3 = 4
a2 – a1 = a3 – a2 = a4 – a3 = 4
Therefore, it’s an A.P with d = 4
The 11th term of the AP: –5, (–5/2), 0, 5/2, …is
(A) –20 (B) 20 (C) –30 (D) 30
(B) 20
Explanation:
First term, a = – 5
Common difference,
d = 5 – (-5/2) = 5/2
n = 11
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a11 = – 5 + (11 – 1)(5/2)
a11 = – 5 + 25 = 20
The first four terms of an AP, whose first term is –2 and the common difference is –2, are
(A) – 2, 0, 2, 4
(B) – 2, 4, – 8, 16
(C) – 2, – 4, – 6, – 8
(D) – 2, – 4, – 8, –16
(C) – 2, – 4, – 6, – 8
Explanation:
First term, a = – 2
Second Term, d = – 2
a1 = a = – 2
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
Hence, we have,
a2 = a + d = – 2 + (- 2) = – 4
Similarly,
a3 = – 6
a4 = – 8
So the A.P is
– 2, – 4, – 6, – 8
The 21st term of the AP whose first two terms are –3 and 4 is
(A) 17 (B) 137 (C) 143 (D) –143
(B) 137
Explanation:
First two terms of an AP are a = – 3 and a2 = 4.
We know, nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a2 = a + d
4 = – 3 + d
d = 7
Common difference, d = 7
a21 = a + 20d
= – 3 + (20)(7)
= 137
If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(A) 30 (B) 33 (C) 37 (D) 38
(B) 33
Explanation:
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a2 = a + d = 13 …..(1)
a5 = a + 4d = 25 …… (2)
From equation (1) we have,
a = 13 – d
Using this in equation (2), we have
13 – d + 4d = 25
13 + 3d = 25
3d = 12
d = 4
a = 13 – 4 = 9
a7 = a + 6d
= 9 + 6(4)
= 9 + 24 = 33
Which term of the AP: 21, 42, 63, 84… is 210?
(A) 9th (B) 10th (C) 11th (D) 12th
(B) 10th
Explanation:
Let nth term of the given AP be 210.
According to question,
first term, a = 21
common difference, d = 42 – 21 = 21 and an = 210
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
210 = 21 + (n – 1)21
189 = (n – 1)21
n – 1 = 9
n = 10
So, 10th term of an AP is 210.
If the common difference of an AP is 5, then what is a18 – a13?
(A) 5 (B) 20 (C) 25 (D) 30
(C) 25
Explanation:
Given, the common difference of AP i.e., d = 5
Now,
As we know, nth term of an AP is
an = a + (n – 1)d
where a = first term
an is nth term
d is the common difference
a18 -a13 = a + 17d – (a + 12d)
= 5d
= 5(5)
= 25