General Term MCQ Quiz - Objective Question with Answer for General Term - Download Free PDF

Concept:

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

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

Calculation:

Given: Third term of GP = 4

So, a₃ = ar² = 4

Now, product of five time term is given by  = a ×  ar × ar²  × ar³ ×ar⁴   

= a⁵r¹⁰  

= (ar²)⁵ 

= 4⁵

Concept Used:

In a G.P, the n^th term is given by a*r^(n-1), where a is the first term and r is the common ratio.

Calculation

Let the first term of the G.P. be 'a' and the common ratio be 'r'.

Given, the (m+n)^th term is p.

⇒ ...(1)

Also, the (m-n)^th term is q.

⇒ ...(2)

We need to find the m^th term, which is .

Multiplying equations (1) and (2), we get

⇒

⇒

⇒

⇒

⇒

Thus, the m^th term is .

∴ The m^th term is .

Hence option 3 is correct

Calculation

⇒ 

2^nd G.P.  T₁ = a, T₅ = ar⁴ = b

⇒ 

Hence option (3) is correct

Concept :

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

• Common ratio , r =  
•  term of G.P  is  a_n = ar^n-1 
• Sum of n terms = s =  ; where r >1
• Sum of n terms = s = ; where r
• Sum of infinite GP =  ; |r|

Calculation :  

Here 5^th term of G.P is 96 

i.e  a₅ = ar^5-1 

⇒ a₅ = ar⁴ 

⇒ 96 = ar⁴        ____( i ) 

Given: 8th 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 ,  term of G.P , a_n = ar^n-1 

So, a₃ = 6× 2^3-1 

⇒ a₃ = 24 . 

The correct option is 3. 

Concept:

Common Ratio of G.P.  

r 

Where

• a_n is nth term of G.P.
• a_n-1 is (n-1)th term of G.P.

General term of a G.P. is T_n = ar^n-1

Where a is the first term of G.P.

Solution:

Statement I: If a, b, c are in A.P. & a2, b2, c2 are in G.P. then the common ratio of G.P. is -1.

Given: a, b, c are in A.P.

So, 2b = a + c

and a², b², c²  are in G.P.

So, b² = ra²

c² = r²a²

⇒ 4b² = a² + c² + 2ac

⇒ 4ra² = a² + r²a² + 2a²r

⇒ r²a² + 2a²r - 4a²r + a² = 0

⇒ a²(r² - 2r + 1) = 0

⇒ r² -2r + 1 = 0

⇒ r = 1

Hence the common ratio of G.P. is 1.

∴ Statement I is incorrect.

Statement II: If the 3rd and the 8th term of G.P. are 4 and 128 respectively, then the G.P. is 1, 2, 4, 8....

Let a be the first term and r be the common ratio of the G.P., then

The 3^rd term of G.P. is 4

ar² = 4 ....(1)

The 8^th term of G.P. is 128

ar⁷ = 128 .....(2)

Dividing (2) by (1)

r⁵ = 32

r = 2

Putting r = 2 in equation (1)

we get a = 1.

So the required G.P. = 1, 2, 4, 8....

∴ Statement II is correct.

So, the correct option is (2)

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 .

Concept:

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

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

Sin18o =   

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 r =    = 2 × = 2 Sin18° 

Concept:

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

• Common ratio = r = 
• n^th  term of the G.P. is a_n = arⁿ⁻¹

Calculation:

Given: The sequence 3, 9, 27, 81, ..... is a GP.

Here, we have to find which term of the given sequence is 6561.

As we know that the general term of a GP is given by: 

Here, a = 3, r = 3 and let an = 6561

⇒ 6561 = (3) ⋅ (3)n - 1

⇒ 3n = 3⁸ 

∴ n = 8

Hence, 6561 is the 8^th term of the given sequence.

Additional Information

• Sum of n terms of GP = s_n = ; where r >1
• Sum of n terms of GP = s_n = ; where r
• Sum of infinite GP =  ; |r|

Concept:

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

Common ratio = r = a2/a1 = a3/a2 = .......= an/an-1

nth  term of the G.P. is an = arn−1

Calculation:

Given: The sequence √3, 3, 3√3....... 

a = √3, r = √3 and a_n = 729

an = arn−1 

⇒ 729 = √3 ⋅ (√3)^{n - 1}

⇒ 729 = (√3)ⁿ 

⇒ 3⁶ = (3)^n/2

⇒ 

⇒ n = 12

Concept:

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

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

Calculation:

Here 6^th term = a₆ = 48 

⇒ ar⁵ = 48        ....(1)

And 12^th term = a₁₂ = 384  

⇒ ar¹¹ = 384    ....(2)

Dividing eq. (2) by eq.(1) , we get

r⁶ = 8 

Now 18^th term of GP = a₁₈ = ar^{18 - 1} = ar¹⁷  = ar¹¹ × r⁶ 

= 384 × 8 = 3072

Concept:

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

Common ratio = r = a2/a1 = a3/a2 = .......= an/an-1

nth  term of the G.P. is an = arn−1

Calculation:

Given: The sequence √5, 5, 5√5.......

a = √5, r = √5 and a_n = 125

an = arn−1 

⇒ 125 = √5 ⋅ (√5)^{n - 1}

⇒ 125 = (√5)ⁿ 

⇒ 5³ = (5)^n/2

⇒ 

⇒ n = 6 

Concept:

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

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

Calculation:

Given: Third term of GP = 4

So, a₃ = ar² = 4

Now, product of five time term is given by  = a ×  ar × ar²  × ar³ ×ar⁴   

= a⁵r¹⁰  

= (ar²)⁵ 

= 4⁵

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.

Calculation:

Given: the 3rd term is 24 and the 6th term is 192

T₃ = ar² = 24          ...(1)

T₆ = ar⁵ = 192          ...(2)

Dividing equation (2) by (1), we get:

r³ = 8

⇒ r = 2

Using equation (1), a = 24/2² = 6.

The 12th term = T₁₂ = 6 × 2¹¹ = 12288.

CONCEPT:

Let us suppose a be the first term and r be the common ratio of a GP. Then the general term of a GP is given by:

CALCULATION:

Given: The sequence 5, 10, 20, 40, ..... is a GP.

Here, we have to find which term of the given sequence is 5120.

As we know that, the the general term of a GP is given by: 

Here, a = 5, r = 2 and let a_n = 5120

⇒ 5120 = (5) ⋅ (2)^{n - 1}

⇒ 2^{(n - 1)} = 1024 = 2¹⁰

⇒ n - 1 = 10

⇒ n = 11

Hence, 5120 is the 11th term of the given sequence.

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.

Calculation:

Given: Sum of the first two terms of GP is -2 and the fifth term of GP is 4 times the third term of GP

According to the question:

a + ar = -2           ...(1)

ar⁴ = 4 × ar²

r = ±2         ...(2)

Using equation (1), we get:

a = 

Now, the sixth term of the GP = ar⁵ = .