MCQ: Arithmetic Progressions - 3 - SSC CGL MCQ

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.......

Formula used:
S_n = n/2(a + l)
l = a + (n − 1)d, where,
S_n = 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.

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 

Concept:
For AP series, 

Calculations:
We know that, For AP series, 

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

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.

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 a_n = a + (n – 1) d
• Sum of the first n terms =

Where, a = First term, d = Common difference, n = number of terms, a_n = n^th 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
n^th term = 396
As we know, n^th term of AP = a_n = a + (n – 1) d
⇒ 396 = 102 + (n - 1) × 6
⇒ 294 = (n - 1) × 6
⇒ (n - 1) = 49
∴ n = 50
Now,

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, a_n = n^th 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)

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

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

Calculation:
Given series is 2, 6, 10, ...,146
First term, a = 2, last term, a_n = 146, an
Common difference d = 4, so it is an AP
a_n = 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.

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

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

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:

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

Concept:
The nth term of an AP is given by: T_n = 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: T_n = 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

Concept:
The nth number in A.P. series = T_n = 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.

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

Also, S =  (n/2)(a + l)
Where,
a = First term,
d = Common difference,
n = number of terms,
a_n = n^th 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.