**(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 = – 16B. is an A.P., Reason d = 4C. is an A.P., Reason d = – 4D. is not an A.P.**

**Solution:**

B. is an A.P., Reason d = 4**Explanation:**

First term a = – 10

Second term t_{1} = – 6

Third term t_{2} = – 2

Fourth term t_{3} = 2

Common difference d = t_{1} – a = – 6 – (– 10) = – 6 + 10 = 4

Common difference d = t_{2} – t_{1} = – 2 – (– 6) = – 2 + 6 = 4

Common difference d = t_{3} – t_{2} = 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, 4B. – 2, 4, – 8, 16C. – 2, – 4, – 6, – 8D. – 2, – 4, – 8, – 16**

**Solution:**

C. – 2, – 4, – 6, – 8**Explanation:**

Given first term t_{1} = – 2

Common difference d = – 2

By using formula t_{n + 1} = t_{n} + d

t_{2} = t_{1} + d = – 2 + (– 2) = – 2 – 2 = – 4

t_{3} = t_{2} + d = – 4 + (– 2) = – 4 – 2 = – 6

t_{4} = t_{3} + 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. 464B. 465C. 462D. 461**

**Solution:**

B. 465**Explanation:**

List of first 30 natural number is

1, 2, 3, ……..,30

First term a = 1

Second term t_{1} = 2

Third term t_{2} = 3

Common difference d = t_{3} – t_{2} = 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 S_{30}

⇒ S_{30} = 15 [ 2 + 29]

⇒ S_{30} = 15 × 31

⇒ S_{30} = 465

Hence, Correct answer is (B)

**(4) For a given A.P. t _{7} = 4, d = – 4 then a = . . .A. 6B. 7C. 20D. 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

⇒ t_{7} = 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. 0B. 3.5C. 103.5D. 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. – 143B. 143C. 137D. 17**

**Solution:**

C. 137**Explanation:**

Given first term a = – 3

Second term t_{1} = 4

Common difference d = t_{1} – a = 4 – (– 3) = 4 + 3 = 7

We need to find t_{21} 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,

⇒ t_{21} = – 3 + (21 – 1) × 7

⇒ t_{21} = – 3 + 20 × 7

⇒ t_{21} = – 3 + 140 = 137

Hence, correct answer is (C)

**(7) If for any A.P. d = 5 then t _{18} – t_{13} = …**

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, t_{18} – t_{13} = [a + (18 – 1) × 5] – [ a + (13 – 1) × 5]

⇒ t_{18} – t_{13} = [ 17 × 5] – [ 12 × 5]

⇒ t_{18} – t_{13} = 85 – 60 = 25

Thus, correct answer is (C)

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

**Solution:**

A. 45**Explanation:**

First five multiples of 3 are

3, 6, 9, 12, 15

First term a = 3

Second term t_{1} = 6

Third term t_{2} = 9

Common difference d = t_{2} – t_{1} = 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 S_{5}

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

**Solution:**

A. – 75**Explanation:**

First term a = 15

Second term t_{1} = 10

Third term t_{2} = 5

Common difference d = t_{2} – t_{1} = 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 S_{10}

⇒S_{10} = 5 [ 30 + 9 × (– 5)]

⇒S_{10} = 5 [ 30 – 45]

⇒S_{10} = 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. 42B. 38C. 21D. 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