MCQ: Arithmetic Progressions - 2 - SSC CGL MCQ

# MCQ: Arithmetic Progressions - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Arithmetic Progressions - 2

MCQ: Arithmetic Progressions - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Arithmetic Progressions - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Arithmetic Progressions - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Arithmetic Progressions - 2 below.
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MCQ: Arithmetic Progressions - 2 - Question 1

### The first four terms of an AP whose a is 10 and d is 10 will be?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 1

a = 10, d = 10
a(1) = a = 10
a(2) = a1 + d = 10 + 10 = 20
a(3) = a2 + d = 20 + 10 = 30
a(4) = a3 + d = 30 + 10 = 40

MCQ: Arithmetic Progressions - 2 - Question 2

### The missing terms in AP: __, 13, __, 3 are:

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 2

a2 = 13 and
a(4) = 3
The nth term of an AP;
a(n) = a + (n − 1) x d
a(2) = a + (2 - 1) x d
13 = a + d ………………. (i)
a(4) = a + (4 - 1) x d
3 = a + 3d ………….. (ii)
Subtracting equation (i) from (ii), we get
– 10 = 2 x d
d = – 5
put value of d in eq 1
13 = a + (-5)
a = 18
a(3) = 18 + (3 - 1) x (-5)
= 18 + 2 x (-5)
= 18 - 10 = 8

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MCQ: Arithmetic Progressions - 2 - Question 3

### If the a(2) is 13 and a(5) is 25, then it’s a(7) is

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 3

a(2) = 13
a + d = 13
a = 13 – d…. (i)
a(5) = 25
a + 4d = 25….(ii)
Substituting the value of (i) in (ii),
13 – d + 4 x d = 25
3 x d = 12
d = 4
So, a = 13 – 4 = 9
a(7) = a + 6d = 9 + 6(4) = 9 + 24 = 33

MCQ: Arithmetic Progressions - 2 - Question 4

In an AP, if d = -4, a = 28, n = 7, then an is:

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 4

a(n) = a + (n - 1) x d
= 28 + (7 - 1) x (-4)
= 28 + 6 x (-4)
= 28 - 24
a(n) = 4

MCQ: Arithmetic Progressions - 2 - Question 5

Starting from the last, the 20th term of the A.P. 3, 8, 13, …, 253 is:

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 5

A.P. is 3, 8, 13, …, 253
d= 5.
Starting from the end of the series,
253, 248, 243, …, 13, 8, 5
a = 253
d = 248 − 253 = −5
n = 20
a(20) = a + (20 − 1) x d
a(20) = 253 + (19) x (−5)
a(20) = 253 − 95
a(20) = 158

MCQ: Arithmetic Progressions - 2 - Question 6

What will be the nth term of the AP 1, 5, 9, 13, 17…….?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 6

Here a = 1 and d = 4
The nth term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3

MCQ: Arithmetic Progressions - 2 - Question 7

If the nth term of the AP is 7n – 1, then the first term is ________

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 7

nth term of the AP is 7n – 9
Tn = 7n – 1
T1 = 7(1) – 1 = 6
The first term of AP is 6.

MCQ: Arithmetic Progressions - 2 - Question 8

What will be the 99th term from the end of the AP 500, 489, 478, 467… – 1139?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 8

In this case since we have to find the 99th term from the end.
We will consider the first term to be -1139 and the common difference will be 11
Now, a = -1139, d = 11 and n = 99
T99 = a + (n – 1)d
T99 = -1139 + (99 – 1)11
T99 = -1139 + 1078 = -61
The value of 99th term from the end is -61.

MCQ: Arithmetic Progressions - 2 - Question 9

If 5 times the 5th term of an AP is equal to 15 times its 15th term, then the value of its 20th term will be _______

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 9

5(5th term) = 15(15th term)
5th term = a + (5 – 1)d = a + 4d
15th term = a + (15 – 1)d = a + 14d
5(a + 4d) = 15(a + 14d)
5a + 20d = 15a + 210d
20d – 210d = 15a – 5a
-190d = 10a
a = -19d
Now, the 20th term = a + (20 – 1)d = a + 19d
But a = -19d
Hence, -19d + 19d = 0

MCQ: Arithmetic Progressions - 2 - Question 10

If the 7th term of an AP is 20 and its 11th term is 40 then, what will be the common difference?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 10

Here 7th term = 20
11thterm = 40
Let the first term of the AP be a and common difference be d
T7 = a + (n – 1)d = 20
T7 = a + (7 – 1)d = 20
T7 = a + 6d = 20     (1)
T11 = a + (n – 1)d = 40
T11 = a + (11 – 1)d = 40
T11 = a + 10d = 40     (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5

MCQ: Arithmetic Progressions - 2 - Question 11

If the nth term of the AP is 2n + 7, then the common difference will be _____

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 11

nth term of the AP is 2n + 7
Tn = 2n + 7
Common difference = T2 – T1 = (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2

MCQ: Arithmetic Progressions - 2 - Question 12

If the nth term of the AP is 8n + 1, then the 20th term will be ______

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 12

nth term of the AP is 8n + 1
Tn = 8n + 1
T20 = 8(20) + 1
T20 = 161

MCQ: Arithmetic Progressions - 2 - Question 13

If the pth term of an AP is q and its qth term is p, then what will be the value of its (p + q)th term?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 13

pth term = q
a + (p – 1)d = q
a + pd – d = q     (1)
qth term = p
a + (q – 1)d = p
a + qd – d = p     (2)
Subtracting (2) from (1) we get,
a + qd – d – (a + pd – d) = p – q
qd – pd = p – q
d = -1
Substituting in equation 1, we get,
a = p + q – 1
(p + q)th term = a + (n – 1)d = p + q – 1 + (p + q – 1)(-1) = p + q – 1 – p – q + 1 = 0

MCQ: Arithmetic Progressions - 2 - Question 14

What will be the 14th term of the AP 5, 8, 11, 14, 17…….?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 14

Here a = 5, d = 8 – 5 = 3 and n = 14
T14 = a + (n – 1)d
T14 = 5 + (14 – 1)3
T14 = 5 + 13 × 3
T14 = 44

MCQ: Arithmetic Progressions - 2 - Question 15

If the 11th term of an AP is 1/13 and its 13th term is 1/11, then what will be the value of 143th term?

Detailed Solution for MCQ: Arithmetic Progressions - 2 - Question 15

Here 11th term = 1/13
13th term = 1/11
Let the first term of the AP be a and common difference be d

Subtracting (1) from (2)
We get,

Now, substituting value of d in equation 1
We get,

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests