MCQ: Geometric Progressions - 3 - SSC CGL MCQ

Concept:
Geometric Progression
The general form of Geometric Progression is:
a, ar, ar², ar³, ar⁴,…, ar^n-1
where, a = First term, r = common ratio, ar^n-1 = nth term.

Calculation:
Let A be first term of GP with common ratio r.
The nth term, A_n = Ar^n-1
Given 4th term = a
Ar³ = a …(i)
Given 7th term = b
Ar⁶ = b …(ii)
Given 10th term = c
Ar⁹ = c …(iii)
Multiply (i) and (iii)
⇒ ac = Ar³(Ar⁹)
⇒ ac = A²r¹²
⇒ ac = (Ar⁶)²
⇒ ac = b²
∴ b² = ac

Concept:
If three terms p, q, r are in GP, then pr = q2.
If three terms a, b, c are in AP, then 2b = a + c

Calculation:

⇒ a, b, c are in AP
∴ The correct option is (1).

Concept :

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

• Common ratio ,  
• nth term of G.P  is  a_n = ar^n-1 
• Sum of n terms =
• Sum of n terms =  
• Sum of infinite

Calculation: 
Here 5^th term of G.P is 96 
i.e  a₅ = ar^5-1 
⇒ a₅ = ar⁴ 
⇒ 96 = ar⁴        ____( i ) 
Given: 8^th term is 768
⇒ a₈ = ar⁷ 
768 = ar⁷       ____(ii) 
Divide eqn. (ii) by eqn. (i) , we get 
8 = r³ 
⇒ r = 2.
Putting this in eqn. (i) , we get 
a = 6.
We know that , n^th term of G.P , a_n = ar^n-1 
So, a₃ = 6× 2^3-1 
⇒ a₃ = 24 . 
The correct option is 3.

Concept:
Let us consider sequence a₁, a₂, a₃ …. a_n is a G.P.

• Common ratio =
• nth  term of the G.P. is a_n = arⁿ⁻¹
• Sum of n terms of GP =
• Sum of n terms of GP =
• Sum of infinite GP = 

Calculation:
Given series is 5, 10, 20, ...
Here, a = 5, r = 2
Sum of n numbers = sn = 1275
To Find: nAs we know that, Sum of n terms of GP where r >1

1275 = 5 × (2ⁿ - 1)
⇒ 255 = (2ⁿ - 1)
⇒ 2ⁿ = 256
⇒ 2ⁿ = 2⁸
∴ n = 8

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

• Common ratio =
• nth  term of the G.P. is a_n = arⁿ⁻¹

Sin18^o  

Calculation:
We know that if the first term of a G.P is 'a' and the common ratio is 'r' then in this case then G.P = a, ar, ar²............ 
Since we have given  a = ar + ar² 
Now, 1= r + r² 
⇒ r² + r - 1 = 0 
After solving we get

Concept:
Geometric Progression (GP):

• The series of numbers where the ratio of any two consecutive terms is the same is called a Geometric Progression.
• A Geometric Progression of n terms with first term a and common ratio r is represented as:
  a, ar, ar², ar³, ..., ar^n-2, ar^n-1.
• The sum of the first n terms of a GP is: 
• The sum to ∞ of a GP, when |r| < 1, is: 

Calculation:

Let us consider the infinite series 

Concept:

Let us consider sequence a₁, a₂, a₃ …. an is a G.P.
Common ratio 

Calculation:
Consider,
(a = 3) be the 3rd term of the G.P series,
So, we can write the five terms as,

So, the product of the five terms (P) will be,

Since,
a = 3,
∴ The product of the first five terms (P) = 3⁵ = 243

Concept:
If a₁, a₂, …., a_n is a GP then the general term is given by: a_n = a × r^{n - 1} where a is the 1^st term and r is the common ratio.

Calculation:
Given: 25, -125, 625, -3125, …….
Here, first term a = 25 and common ratio r = -5.
As we know that, if a₁, a₂, …., a_n is a GP then the general term is given by: a_n = a × r^{n - 1} where a is the first term and r is the common ratio.
⇒ The general term is : a_n = 25 × (-5)^{n – 1} = (-1)^{n – 1} × 5^{n + 1}

Concept:
The general form of Geometric Progression is:
a, ar, ar², ar³, ar⁴,…, ar^n-1
Where,
a = First term
r = common ratio
ar^n-1 = nth term

Calculation:
It is given that each term is equal to the sum of two terms that follows it.
T_n = T_n+1 + T_n+2
⇒ ar^n-1 = arⁿ + arⁿ⁺¹
⇒ r^n-1 = rⁿ + rⁿ⁺¹
⇒ r^-1 = r + 1
⇒ r² + r - 1 = 0

Since each term of the G.P. is positive,

∴ The common ratio is 

Concept:
Sum of n terms of a Geometric Progression, 

Given that a_n = 2ⁿ of the G.P.
Then, a₁ = 2
a₂ = 4
a₃ = 8
i.e. G.P. series is 2, 4, 8, 16, 32, . . .
where first term, a = 2 ;
Common ration, r = 4/2 = 8/ 4 = ... = 2,
Number of terms, n = 6 (given in the question)

⇒ 2(64 - 1)
⇒ 2(63)
⇒ 126 

Concept:
Five terms in a geometric progression:
If a G.P. has first term a and common ratio r then the five consecutive terms in the GP are of the form

Calculation:
Let us consider a general geometric progression with common ratio r.
Assume that the five terms in the GP are  

It is given that third term is 9.
Therefore, a = 9.
Now the product of the five terms is given as follows:

But we know that a = 9.
Thus, the product is 9⁵ = 3¹⁰.

Concept:
If a, b and c are in a GP then b² = ac

Calculation:
Given: The numbers - 2/7, x, - 7/2 are in GP
As we know that, if a, b and c are in GP then b² = ac
Here, a = - 2/7, b = x and c = - 7/2
⇒ x² = (-2/7) × (-7/2) = 1
⇒ x = ± 1
Hence, correct option is 3.

Concept: 

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

• Sum of n terms =
• Sum of n terms = 

Calculation:  
The required no. of ancestors 
= 2 + 4 + 8 +... upto 8 terms 
As we know that sum  of G.P , 

Where , a = 2, r = 2 and n = 8
⇒ No. of ancestors required 
∴ No. of ancestors required is  510. 
The correct option is 4. 

Concept:
Geometric Progression (GP): The series of numbers where the ratio of any two consecutive terms is the same, is called a Geometric Progression.

• A Geometric Progression of n terms with first term a and common ratio r is represented as:
• a, ar, ar², ar³, ..., ar^n-2, ar^n-1.
• The sum of the first n terms of a GP is
  

Calculation:
For the given geometric series 1 + 3 + 3² + ..., we have a = 1 and r = 3.
Let the sum of first n terms be equal to 3280.

⇒ 3ⁿ - 1 = 3280 × 2
⇒ 3ⁿ - 1 = 6560
⇒ 3ⁿ = 6561 = 3⁸
⇒ n = 8.

Concept:

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

• Common ratio =
• nth  term of the G.P. is a_n = arⁿ⁻¹
• Sum of n terms of GP =
• Sum of n terms of GP =
• Sum of infinite GP = 

Calculation:
Given series is 4, 8, 16, ...
Here, a = 4, r = 2
Sum of n numbers = s_n = 2044
To Find: nAs we know that, Sum of n terms of GP 

2044 = 4 × (2ⁿ - 1)
⇒ 511 = (2ⁿ - 1)
⇒ 2ⁿ = 512
⇒ 2n = 2⁹
∴ n = 9