MCQ: Arithmetic Progressions - 3 - SSC CGL MCQ

# MCQ: Arithmetic Progressions - 3 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Arithmetic Progressions - 3

MCQ: Arithmetic Progressions - 3 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Arithmetic Progressions - 3 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Arithmetic Progressions - 3 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Arithmetic Progressions - 3 below.
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MCQ: Arithmetic Progressions - 3 - Question 1

### If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term. find the A.P.

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 1

According to the question,
The 5th term of an A.P. is 31.

And,
The 25th term is 140 more than the 5th term.

Also,

The required A.P. is as follows,
3, 10, 17, 24, 31.......

MCQ: Arithmetic Progressions - 3 - Question 2

### Find the sum of all even numbers from 1 to 100.

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

Formula used:
Sn = n/2(a + l)
l = a + (n − 1)d, where,
Sn = Sum of elements,
n  = number of elements.
a = First element,
n  = number of elements.
d  = difference between two consecutive elements.

Calculation:
Even numbers between 1 to 100 = 2, 4, 6, 8,....98, 100.
l = 100, a = 2, n = ?
According to the question,
l = a + (n − 1)d
⇒ 100 = 2 + (n - 1)2
⇒ n = 50
Now,
Sn = n/2(a + l)
⇒ Sn = 50/2(2 + 100) = 102 × 25 = 2,550
∴ The sum of all even numbers between 1 to 100 is 2550.

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MCQ: Arithmetic Progressions - 3 - Question 3

### Sum of n terms of the series

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

We have to find sum of first n term of series

as we know that sum of first n terms of natural numbers is
hence
So the sum of series is

MCQ: Arithmetic Progressions - 3 - Question 4

Find the sum to n terms of the A.P., whose nth term is 5n + 1

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 4

Concept:
For AP series,

Calculations:
We know that, For AP series,

Given, the nth term of the given series is an = 5n + 1.
Put n = 1, we get
a1 = 5(1) + 1 = 6.
We know that

MCQ: Arithmetic Progressions - 3 - Question 5

The tenth term common to both the A. P. 3, 7, 11, ... and 1, 6, 11, ... is:

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 5

Concept:
Arithmetic Progressions:

• The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.

• If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:

a, a + d, a + 2d, ..., a + (n - 1)d

• Common Terms to two A. P.s form an A. P. themselves, with common difference equal to the LCM of the common difference of the two A. P.s.

Calculation:
For the given two A. P.s 3, 7, 11, ... and 1, 6, 11, ..., the common differences are 4 and 5 respectively and 11 is the first common term.
The common difference of the terms common to both the series will be: LCM of (4 and 5) = 20.
The required 10th term common to both the A. P.s = a + (n - 1)d
= 11 + (10 - 1) × 20
= 11 + 180
191.

MCQ: Arithmetic Progressions - 3 - Question 6

Find the sum of all numbers divisible by 6 in between 100 to 400

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 6

Concept:

Let us consider sequence a1, a2, a3 …. an is an A.P.

• Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
• nth term of the A.P. is given by an = a + (n – 1) d
• Sum of the first n terms =

Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term

Calculation:
Here 1st term = a = 102 (Which is the 1st term greater than 100 that is divisible by 6.)
The last term less than 400, Which is divisible by 6 is 396.
Terms in the AP; 102, 108, 114 … 396
Now
First term = a = 102
Common difference = d = 108 - 102 = 6
nth term = 396
As we know, nth term of AP = an = a + (n – 1) d
⇒ 396 = 102 + (n - 1) × 6
⇒ 294 = (n - 1) × 6
⇒ (n - 1) = 49
∴ n = 50
Now,

MCQ: Arithmetic Progressions - 3 - Question 7

If fourth term of an A.P. is zero, then t25/t11 is, where tn denotes the nth

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 7

Concept:
Let us consider sequence a1, a2, a3 …. an is an A.P.

• Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
• nth term of the A.P. is given by an = a + (n – 1) d
• Sum of the first n terms =

Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term

Calculation:
Let the first term of AP be 'a' and the common difference be 'd'
Given: Fourth term of an A.P. is zero
⇒ a4 = 0
⇒ a + (4 - 1) × d = 0
⇒ a + 3d = 0
∴ a = -3d          .... (1)

MCQ: Arithmetic Progressions - 3 - Question 8

The middle term of arithmatic series 2, 6, 10, ...,146

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 8

Concept:
Let us consider sequence a1, a2, a3 …. an is an A.P.

• Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
• nth term of the A.P. is given by an = a + (n – 1) d

Calculation:
Given series is 2, 6, 10, ...,146
First term, a = 2, last term, an = 146, an
Common difference d = 4, so it is an AP
an = a + (n – 1) d
146 = 2 + (n - 1) (4)
⇒ n - 1 = 144/4
⇒ n = 36 + 1 = 37
So, number of terms in given series = 37
Middle term = (37 + 1)/2 = 19th term
a19 = 2 + (19 - 1) × 4
= 2 + 72
= 74
Hence, option (3) is correct.

MCQ: Arithmetic Progressions - 3 - Question 9

The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 9

According to the question,
The eighth term of an A.P. is half of its second term.
The eighth term of an A.P. = a + 7d
The second term of an A.P = a + d
Now,

The eleventh term exceeds one-third of its fourth term by 1
The eleventh of an A.P = a + 10d
The fourth of an A.P = a + 3d
One-third of a fourth of an A.P
Now,

By solving (i) and (ii) we will get,
2(-13d) = -27d + 3
d = 3
a = -39
Now, the 15th term of the A.P. will be,

Hence, the 15th term of the A.P. is 3

MCQ: Arithmetic Progressions - 3 - Question 10

The common difference of an A.P., the Sum of whose x terms is Sn, is

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 10

Given :- Sum of x terms is
Concept used :- nth term of an A.P. is
As we know that :

MCQ: Arithmetic Progressions - 3 - Question 11

The sum of the series 5 + 9 + 13 + … + 49 is:

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 11

Concept:
Arithmetic Progression (AP):

• The sequence of numbers where the difference of any two consecutive terms is same is called an Arithmetic Progression.
• If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:
• a, a + d, a + 2d, ..., a + (n - 1)d.
• The sum of n terms of the above series is given by:

Calculation:
The given series is 5 + 9 + 13 + … + 49 which is an arithmetic progression with first term a = 5 and common difference d = 4.
Let's say that the last term 49 is the nth term.
∴ a + (n - 1)d = 49
⇒ 5 + 4(n - 1) = 49
⇒ 4(n - 1) = 44
⇒ n = 12.
And, the sum of this AP is:

MCQ: Arithmetic Progressions - 3 - Question 12

If the numbers n - 3, 4n - 2, 5n + 1 are in AP, what is the value of n?

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 12

Concept:
If a, b, c are in A.P then 2b = a + c

Calculation:
Given:
n - 3, 4n - 2, 5n + 1 are in AP
Therefore, 2 × (4n - 2) = (n - 3) + (5n + 1)
⇒ 8n - 4 = 6n - 2
⇒ 2n = 2
∴ n = 1

MCQ: Arithmetic Progressions - 3 - Question 13

The sum of (p + q)th and (p – q)th terms of an AP is equal to

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 13

Concept:
The nth term of an AP is given by: Tn = a + (n - 1) × d, where a = first term and d = common difference.

Calculation:
As we know that, the nth term of an AP is given by: Tn = a + (n - 1) × d, where a = first term and d = common difference.
Let a be the first term and d is the common difference.

By adding (1) and (2), we get

MCQ: Arithmetic Progressions - 3 - Question 14

In an A.P. twenty fifth term is 70 more than to fifteenth term. Find the common differences.

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 14

Concept:
The nth number in A.P. series = Tn = a + (n - 1)d
Where 'a' is the first number of the series and 'd' is the common difference

Calculation:
Let the first element of A.P. is 'a' and common difference is 'd'

Given:
T25 = T15 + 70
According to the question,
T25 = T15 + 70
⇒ a + (25 – 1)d = a + ( 15 – 1)d + 70
⇒ 24d – 14d = 70
⇒ 10d = 70
⇒ d = 7
∴ The common difference is 7.

MCQ: Arithmetic Progressions - 3 - Question 15

If the first term of an AP is 2 and the sum of the first five terms is equal to one-fourth of the sum of the next five terms, then what is the sum of the first ten terms?

Detailed Solution for MCQ: Arithmetic Progressions - 3 - Question 15

Concept:
Let us consider sequence a1, a2, a3 …. an is an A.P.
Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
nth term of the A.P.= an = a + (n – 1) d

Also, S =  (n/2)(a + l)
Where,
a = First term,
d = Common difference,
n = number of terms,
an = nth term,
l = Last term

Calculation:
Given, a1 = 2      ...(1)

[∵ n5 = 5 and n10 = 10]

∴ d = -3a1 = - 3 × 2 = - 6
Hence,

⇒ 5[2a1 + 9d]
⇒ 5[4 - 54] = -250
∴ S10 = 5 × (-50) = - 250.

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