Special Series MCQ Quiz - Objective Question with Answer for Special Series - Download Free PDF

Concept:

If a1, a2, a3,...are in GP with common ratio r, 

If a be the first term, r be the common ratio of a GP then,

Calculation:

Calculation:

Let required sum is S.

⇒ S =

⇒ S =

⇒ S = 

⇒ S = 

Threfore, a = 1/2 and r = 1/2

We know that the sum of the n^th term of GP (r

⇒ S = 

⇒ S =           (∵ 1/aⁿ = a^-n)

⇒ S = n - 1 + 2-n

∴ S = n + 2-n -1

Concept:

1. The nth term of the A.P. is given by:

an = a + (n – 1) d

2. The sum of n terms of an AP with first term a and common difference d is given by: 

      ----(1)

Or

Where,

an = nth term and l = Last term

Calculation:

Given: 

d = 3/2

n = 25

a = 3

From equation (1);

Concept: 

Sum of consecutive numbers from 1 to n: .

Calculation:

The sum of the first 24 terms of the given series can be written as:

.

Calculation:

Given:

Series: 

⇒

Let n = 1/3: 1 = A(3) => A = 1/3

Let n = -2/3: 1 = B(-3) => B = -1/3

⇒

series: 

⇒

⇒

⇒

⇒

⇒

⇒

∴ The sum of the series is

Hence option 4 is correct.

Concept:

Expansion of ex:

Calculation:

Put x = -1,

Calculation:

Calculation:

Given: Let S =  

Now we can re-write S as shown below:

On further simplifying the above equation we get,

⇒ S = 1/6 × [1/4 - 1/34] = 5/136

Concept:

Expansion of ex:

Calculation:

Put x = 1,

Concept:

Arithmetic-Geometric Progression (AGP):

• The first few terms of an Arithmetic-Geometric progression composed of an arithmetic progression with first term a and common difference d, and a geometric progression with first term b and common ratio r are given by:

  ab, (a + d)br, (a + 2d)br², ... [a + (n - 1)d]br^{n - 1}, ...

• The sum to infinity of an AGP, whose |r| ≤ 1, is given by:

  S_∞ = 

Calculation:

The given series  is an AGP with

a = 1, d = 2 and b = , r = .

⇒ S∞ = 

⇒ S∞  = 1 + 

⇒ S∞  = 1 + 2

⇒ S∞  = 3.

Concept:

If a₁, a₂, …., a_n be an AP then  where a is the 1^st term and d is the common difference.

If a₁, a₂, …, a_n be an AP then the general term is given by: a_n = a + (n - 1) × d where a is the 1^st term and d is the common difference.

Calculation:

Here, we have to find the value of 1 - 2 + 3 - 4 + 5 - ______ + 101.

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100)

As, we can see that (1, 3, ……, 101) is an AP with a = 1 and d = 2.

⇒ a_n = 101 = 1 + (n - 1) × 2

⇒ n = 51

Similarly, (2, 4, …, 100) is an AP with a = 2 and d = 2

⇒ a_n = 100 = 2 + (n - 1) × 2

⇒ n = 50

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100) = 2601 - 2550 = 51.

Concept: 

Sum of consecutive numbers from 1 to n: .

Calculation:

The sum of the first 20 terms of the given series can be written as:

.

Formula used:

n! = n × (n - 1) × (n - 2) × ....... × 3 × 2 × 1

Calculation:

a_n = n (n)!

a_n = (n + 1 - n)n!

a_n = (n + 1)n! - n!

a_n = (n + 1)! - n!

Hence, 

a₁ = 2! - 1! 

a₂ = 3! - 2! 

----------------

a₁₀ = 11! - 10!

Now,

 a1 + a2 + a3 +...+ a10 

2! - 1! + 3! - 2! + ......11! - 10! 

= 11! - 1

∴ The value of a1 + a2 + a3 +...+ a10 is 11! - 1.

Concept:​

• 12 + 22 + 32 + ..... + n2 = 

Calculation:

Here we have to find the sum of the series 2 + 8 + 18 + 32 + 50 +......+ 200 

The given series can be re-written as: 2(12 + 22 + 32 + ..... + 102)

As we know that, 12 + 22 + 32 + ..... + n2 = 

Hence, option 1 is the correct answer.

Concept: 

Sum of consecutive numbers from 1 to n: .

Calculation:

The sum of the first 18 terms of the given series can be written as:

3 + 6 + 9 + 12 + 15, ....... 

⇒ 3(1 + 2 + 3 + 4 + 5 + ....)

⇒