Choose the correct alternative answer for each of the following sub questions.

(1) Choose the correct alternative answer for each of the following sub questions.
The sequence – 10, – 6, – 2, 2, . . .
A. is an A.P., Reason d = – 16
B. is an A.P., Reason d = 4
C. is an A.P., Reason d = – 4
D. is not an A.P.

Solution:
B. is an A.P., Reason d = 4
Explanation:
First term a = – 10
Second term t₁ = – 6
Third term t₂ = – 2
Fourth term t₃ = 2
Common difference d = t₁ – a = – 6 – (– 10) = – 6 + 10 = 4
Common difference d = t₂ – t₁ = – 2 – (– 6) = – 2 + 6 = 4
Common difference d = t₃ – t₂ = 2 – (– 2) = 2 + 2 = 4
Since, the common difference is same
∴ The given sequence is A.P. with common difference d = 4
Hence, correct answer is (B)

(2) First four terms of an A.P. are ….., whose first term is – 2 and common difference is – 2.
A. – 2, 0, 2, 4
B. – 2, 4, – 8, 16
C. – 2, – 4, – 6, – 8
D. – 2, – 4, – 8, – 16

Solution:
C. – 2, – 4, – 6, – 8
Explanation:
Given first term t₁ = – 2
Common difference d = – 2
By using formula t_{n + 1} = t_n + d
t₂ = t₁ + d = – 2 + (– 2) = – 2 – 2 = – 4
t₃ = t₂ + d = – 4 + (– 2) = – 4 – 2 = – 6
t₄ = t₃ + d = – 6 + (– 2) = – 6 – 2 = – 8
Hence, the A.P. is – 2, – 4, – 6, – 8
∴ correct answer is (C)

(3) What is the sum of the first 30 natural numbers?
A. 464
B. 465
C. 462
D. 461

Solution:
B. 465
Explanation:
List of first 30 natural number is
1, 2, 3, ……..,30
First term a = 1
Second term t₁ = 2
Third term t₂ = 3
Common difference d = t₃ – t₂ = 3 – 2 = 1
number of terms n = 30
Thus, by using sum of n^th term of an A.P. we will find its sum

Where, n = number of terms
a = first term
d = common difference
S_n = sum of n terms
We need to find S₃₀

⇒ S₃₀ = 15 [ 2 + 29]
⇒ S₃₀ = 15 × 31
⇒ S₃₀ = 465
Hence, Correct answer is (B)

(4) For a given A.P. t₇ = 4, d = – 4 then a = . . .
A. 6
B. 7
C. 20
D. 28

Solution:
D. 28
Explanation:
By using n^th term of an A.P. formula
t_n = a + (n – 1) d
where n = number of terms
a = first term
d = common difference
t_n = n^th terms
⇒ t₇ = a + (7 – 1) × (– 4)
⇒ 4 = a + 6 × (– 4)
⇒ 4 = a – 24
⇒ a = 24 + 4 = 28
Thus, the correct answer is (D)

(5) For a given A.P. a = 3.5, d = 0, n = 101, then t_n = . . .
A. 0
B. 3.5
C. 103.5
D. 104.5

Solution:
B. 3.5
Explanation:
Given: a = 3.5, d = 0, n = 101
By using n^th term of an A.P. formula
t_n = a + (n – 1) d
where n = number of terms
a = first term
d = common difference
t_n = n^th terms
Substituting all given value in the formulae we get,
⇒ t_n = 3.5 + (101 – 1) × 0
⇒ t_n = 3.5
Thus, correct answer is (B)

(6) In an A.P. first two terms are – 3, 4 then 21^st term is . . .
A. – 143
B. 143
C. 137
D. 17

Solution:
C. 137
Explanation:
Given first term a = – 3
Second term t₁ = 4
Common difference d = t₁ – a = 4 – (– 3) = 4 + 3 = 7
We need to find t₂₁ where n = 21
Now, by using n^th term of an A.P. formula
t_n = a + (n – 1) d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Substituting all given value in the formulae we get,
⇒ t₂₁ = – 3 + (21 – 1) × 7
⇒ t₂₁ = – 3 + 20 × 7
⇒ t₂₁ = – 3 + 140 = 137
Hence, correct answer is (C)

(7) If for any A.P. d = 5 then t₁₈ – t₁₃ = …
A. 5
B. 20
C. 25
D. 30

Solution:
C. 25
Explanation:
Given d = 5
By using n^th term of an A.P. formula
t_n = a + (n – 1) d
where n = number of terms
a = first term
d = common difference
t_n = n^th terms
Thus, t₁₈ – t₁₃ = [a + (18 – 1) × 5] – [ a + (13 – 1) × 5]
⇒ t₁₈ – t₁₃ = [ 17 × 5] – [ 12 × 5]
⇒ t₁₈ – t₁₃ = 85 – 60 = 25
Thus, correct answer is (C)

(8) Sum of first five multiples of 3 is. . .
A. 45
B. 55
C. 15
D. 75

Solution:
A. 45
Explanation:
First five multiples of 3 are
3, 6, 9, 12, 15
First term a = 3
Second term t₁ = 6
Third term t₂ = 9
Common difference d = t₂ – t₁ = 9 – 6 = 3
Thus, by using sum of n^th term of an A.P. we will find its sum

Where, n = number of terms
a = first term
d = common difference
S_n = sum of n terms
We need to find S₅

(9) 15, 10, 5, . . . In this A.P. sum of first 10 terms is . . .
A. – 75
B. – 125
C. 75
D. 125

Solution:
A. – 75
Explanation:
First term a = 15
Second term t₁ = 10
Third term t₂ = 5
Common difference d = t₂ – t₁ = 5 – 10 = – 5
Number of terms n = 10
Thus, by using sum of n^th term of an A.P. we will find its sum

Where, n = number of terms
a = first term
d = common difference
S_n = sum of n terms
We need to find S₁₀

⇒S₁₀ = 5 [ 30 + 9 × (– 5)]
⇒S₁₀ = 5 [ 30 – 45]
⇒S₁₀ = 5 × (– 15) = – 75

Hence, correct answer is (A)

(10) In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = . . .
A. 42
B. 38
C. 21
D. 19

Solution:
B. 38
Explanation:
Given, first term = 1
Last term = 20
Sum of n terms, S_n = 399
We need to find number of terms n
Using Sum of n terms of an A.P. formula