MCQ: Arithmetic Progressions - 3 - SSC CGL MCQ

The 5^th term is given as 31, and the 25^th term is 140 more than the 5^th term, which makes it 31 + 140 = 171. Using the formula for the nth term of an A.P., a + (n-1)d, we have:

a + 4d = 31
a + 24d = 171

Solving these equations:

From the first equation: a + 4d = 31 => a = 31 - 4d

Substituting in the second equation: 31 - 4d + 24d = 171

20d = 140

d = 7

Substitute d = 7 back into a = 31 - 4d:

a = 31 - 28 = 3

Thus, the A.P. is 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.