**1. Which term of the A.P. 5, 14, 23, 32, ……. will be 198 more than its 32 ^{nd} term?**

a) 52^{nd}

b) 55^{th}

c) 54^{th}

d) 56^{th}

**Answer:** c) 54^{th}

**Explanation: **Given A.P.: 5, 14, 23, 32, …….

Thus,

a = 5

d = 14 − 5 = 9

We know that a_{n} = a + (n − 1) d

a_{32} = 5 + (32 − 1) 9

= 5 + (31) 9

= 5 + 279

= 284

Let a_{k} be 198 more than the 32^{nd} term of the A.P.

a_{k} = a_{32} + 198

= 284 + 198

= 482

Now we will find the value of k using an = a + (n − 1) d

a_{k} = a + (k − 1) d

482 = 5 + (k − 1) 9

482 − 5 = (k − 1) 9

477 = (k − 1) 9

477 / 9 = k − 1

53 = k − 1

53 + 1 = k

k = 54

Thus, 54^{th} term of the A.P. is 198 more than its 32^{nd} term.

**2. What is the sum of the first 55 terms of the A.P. 13, 24, 35, ………..?**

a) 17150

b) 17050

c) 17500

d) 17250

**Answer: **b) 17050

**Explanation: **Given A.P.: 13, 24, 35, ……….

Thus,

a = 13

d = 24 − 13 = 11

n = 55

We know that S_{n} = n ÷ 2 [2a + (n - 1) d]

S_{55} = 55 ÷ 2 [2(13) + (55 - 1) 11]

= 55 ÷ 2 [26 + (54) 11]

= 55 ÷ 2 [26 + 594]

= 55 ÷ 2 (620)

= 55 × 310

= 17050

**3. If the fourth term of an A.P. is 28 and the ninth term is 53, then what is the sum of the first 19 terms of the A.P.?**

a) 1102

b) 1112

c) 1108

d) 1122

**Answer:** a) 1102

**Explanation: **Given: a_{4} = 28 and a_{9} = 53

We know that a_{n} = a + (n − 1) d

a_{4} = a + 3d

28 = a + 3d … (1)

a_{9} = a + 8d

53 = a + 8d … (2)

Now, subtracting (1) from (2)

53 − 28 = (a + 8d) − (a + 3d)

25 = 8d − 3d

25 = 5d

d = 25 / 5

d = 5

Putting the value of d = 5 in equation (1) to find the value of a.

28 = a + 3(5)

28 = a + 15

28 − 15 = a

a = 13

Now, we know that the sum of n terms of an A.P. is given by:

S_{n} = n ÷ 2[2a + (n - 1) d]

Here, n = 19

S_{19} = 19 ÷ 2 [2(13) + (19 - 1)5]

= 19 ÷ 2 [26 + (18)5]

= 19 ÷ 2 [26 +90]

= (19 × 116) ÷ 2

= 19 × 58

= 1102

**4. If the sum of three numbers in A.P. is 78 and their product is 16302, then what is the value of the common difference?**

a) −7

b) −17

c) 7

d) 17

**Answer: **c) 7

**Explanation: **Let a − d, a, a + d be the three numbers in A.P.

We are given that their sum is 78.

(a − d) + a + (a + d) = 78

3a = 78

a = 78 / 3

a = 26

We are given that the product of the numbers is 16302.

(a − d) × a × (a + d) = 16302

(26 − d) × 26 × (26 + d) = 16302

(26 − d) × (26 + d) = 16302 / 26

262 − d^{2} = 627

676 − d^{2} = 627

676 − 627 = d^{2}

49 = d^{2}

d = 7

**5. What is the value of x for which the numbers 3x + 2, 7x − 3, 25 are in A.P.?**

a) 4

b) 2

c) 5

d) 3

**Answer: **d) 3

**Explanation: **We know that if the numbers a, b and c are in A.P., then

b − a = c − b

**2b = a + c**

Given: 3x + 2, 7x − 3, 25 are in A.P.

Thus, a = 3x + 2, b = 7x − 3 and c = 25

2(7x − 3) = (3x + 2) + 25

14x − 6 = 3x + 27

14x − 3x = 27 + 6

11x = 33

x = 33 / 11

x = 3

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