1. Which term of the A.P. 5, 14, 23, 32, ……. will be 198 more than its 32nd term?
a) 52nd
b) 55th
c) 54th
d) 56th
Answer: c) 54th
Explanation: Given A.P.: 5, 14, 23, 32, …….
Thus,
a = 5
d = 14 − 5 = 9
We know that an = a + (n − 1) d
a32 = 5 + (32 − 1) 9
= 5 + (31) 9
= 5 + 279
= 284
Let ak be 198 more than the 32nd term of the A.P.
ak = a32 + 198
= 284 + 198
= 482
Now we will find the value of k using an = a + (n − 1) d
ak = 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, 54th term of the A.P. is 198 more than its 32nd 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 Sn = n ÷ 2 [2a + (n - 1) d]
S55 = 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: a4 = 28 and a9 = 53
We know that an = a + (n − 1) d
a4 = a + 3d
28 = a + 3d … (1)
a9 = 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:
Sn = n ÷ 2[2a + (n - 1) d]
Here, n = 19
S19 = 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 − d2 = 627
676 − d2 = 627
676 − 627 = d2
49 = d2
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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