Geometric Progressions - 2 - Free MCQ Practice Test with solutions, SSC

Concept Used:
if a is the first term and r is the common ratio of the G. P. then nth term, tn = arⁿ-1

Calculation:
let a and r be the first term and common ratio of the required G.P.
Given, 4th term = 2, 5th term = 8
⇒ ar³ = 2      ----(1)
⇒ ar⁴ = 8      ----(2)
Divide equation 2 by 1
⇒ r = 4
To find: Product of find 8 term = a⋅ar⋅ar²⋅ar³⋅ . . . ar⁷
⇒ a⁸r^{(1+2+3+. . . +7)} =  = a⁸r²⁸
⇒(ar³)⁴(ar⁴)⁴
⇒ 2⁴ × 8⁴
⇒ 16 × 64 × 64
⇒ 4⁸
∴ The product of the 1st 8 terms is 4⁸.

Given:
The required series: 
Formula used:
The formula used for summation for G.P: (r < 1 and series is of infinite terms)

Where, 
S, is the summation of G.P
a, is the first term
r, is the common ratio

Calculation:
Let S be the sum of the given series.
The required series can be written as:

Now, divide the above equation by 20.

Now, subtract equation (2) from equation (1).

Now, the above series becomes Geometric series with a common ratio of 10/20.
The value for the common ratio, r = 10/20 = 1/2
The value of the first term in the series = 1/20
Now, 

∴ The value of the required series is 2/19.

Given:
The third term is greater than the first by 9.
The second term is greater than the fourth by 18.

Concept:
n^th Term of GP whose first term & common ratio are 'a' & 'r', is given by T_n = ar^n-1

Calculation:
Let a be the first term and r be the common ratio of the G.P.
a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³
According to the question
a₃ = a₁ + 9
⇒ ar² = a + 9 ....(1)
Now, a₂ = a₄ + 8
⇒ ar = ar³ + 18 ....(2)
From equation (1) and (2),
⇒ a(r² - 1) = 9 ....(3)
⇒ ar(1 - r²) = 18 ...(4)
Dividing (4) and (3), we get

⇒ -r = 2
⇒ r = -2
Substituting the value of r in equation (1), we get
⇒ 4a = a + 9
⇒ 3a = 9
⇒ a = 3
So. the first term of G.P. a1 = a = 3
∴ The required value is 3.

Concept:
If a, ar, ar², ar³,.....,ar^n-1 are in GP then the nth term of GP is given by
T_n = ar^n-1

Given:
First term a = 125
Common ratio r = 2/5
Term n = 4 th

Calculation:
⇒ T_n = ar^n-1
⇒ T₄ = 125 × (2/5)^4-1
⇒ T₄ = 125 × (2/5)³
⇒ T₄ = 125 × (8/125)
⇒ T₄ = 8
Hence, the 4th term of the G is "8".

Given:
(1/2¹) + (1/2²) + (1/2³) + ……… + (1/2¹⁰) = 1/k

Formula used:
Sum of G.P = a(1 – rⁿ)/(1 – r)

Calculation:
(1/2¹) + (1/2²) + (1/2³) + ……… + (1/2¹⁰) = 1/k
Left hand side of the above series in GP and common ratio = ½
∴ Sum of the given series = [1(1 – rⁿ)/(1- r)]
⇒ ½ × [(1 – (½)¹⁰]/ (1-1/2)
⇒ ½ × [1 – 1/1024]/ (1/2)
⇒ 1023/1024 = 1/k
⇒ k = 1024/1023
∴ The correct answer is 1024/1023.

Given:
(2⁰ + 2² + 2⁴ +........+2⁸) × 3

Formula Used:
It is a Geometric Progression. 
a = first term, r = common ratio
Sum of the Geometric Progression = [a(rⁿ - 1)/(r - 1)]

Calculation:
a = 1
r = (2²/2⁰) = 4/1 = 4
⇒ Sum of the series = [1 × (4⁵ - 1)/(4 - 1)]  × 3
⇒ Sum of the series = [1 × (2¹⁰ - 1)/(3)]  × 3
⇒ Sum of the series = [1 × (1024 - 1)]  
⇒ Sum of the series = [1 × (1023)]  
⇒ Sum of the series = 1023
∴ The Sum of the series is 1023.

Given:
A ball dropped from a height of 120 m

Formula:

Calculation:
After dropping 120 m height the ball bounce = 120 × 4/5 = 96 m
⇒ First term (a) = 120 + 96 = 216 m
⇒ Common ratio (r) = 4/5
∴ Total distance that it travels before coming to rest = 216/(1 – 4/5) = 216/(1/5) = 216 × 5 = 1080 m

Given:
7, 7², 7³, _________ 7ⁿ

Concept used:
Gm of two no a & b = √ab

Calculation:
GM = ⁿ√(a¹.aⁿ)
​⇒ GM of ⁿ√(7, 7² , .....7ⁿ)
​⇒ ⁿ√7^{n(n + 1)/2}
⇒ (7n(n + 1)/2)^1/n
​⇒ 
∴ Required answer is ​​​​​​​

Given:
The ratio of 10th term to 7th term of a G.P. = 1 ∶ 8.

Calculation:
Let the common ratio of the G.P. be r
Now, According to the question
⇒ 10^th term/7^th term = 1/8
⇒ 8 × 10^th term = 7^th term
⇒ 8 × (7^th term × r³) = 7^th term
⇒ r³ = 18
⇒ r = 1/2
∴ The common ratio is 1/2.

Given:
The third term of a G.P. is 16. 

Concept used:
The general form of G.P. is
a, ar, ar², ar³, ar⁴,.....ar^n-1
Where, a = First term, r = common ratio, ar^n-1 = n^th term

Calculation:
Third term of a GP, ar² = 16
Product of first five terms, P = a(ar)(ar²)(ar³)(ar⁴)
⇒ P = a⁵r¹⁰
⇒ P = (ar²)⁵
⇒ P = 16⁵ = (2⁴)⁵ = 2²⁰
∴  The product of its first five terms is 2²⁰.

Given:
A Geometric Progression with first term ‘a’ = 16 and common ratio ‘r’ = 2.

Concept Used:
In this type of question, where ‘r’ > 1, then the sum of n terms of a G.P = S_n 

Formula Used:

n = 10

Calculation:
Considering the given series
16, 32, 64, 128, ......

⇒ S₁₀ = 16 × 1023 = 16368
∴ Sum of first 10 terms of given series is 16368

Concept:
Geometric mean: The value which indicates the central tendency by using the product of the values of a set of numbers.
Formula: GM = (Product of all numbers in the set)^1/n
Where, n = total numbers in the set
Example: 2, 3, 4, 5
GM = (2 × 3 × 4 × 5)^1/4 = (120)^1/4

Calculation:
G = G₁ × G₂ × G₃ ……………….. × G_r

Given∶
The 3 ages are in G.P. such that their sum is 57 and their product is 5832.

Formula Used∶
3 terms of a G.P. = a/r, a, ar.

Calculation∶
Let the ages of Alpha, Beta and Gamma be a/r, a, ar.
Also, Product of their ages = 5832
(a/r) × a × ar = 5832
⇒ a³ = 5832
⇒ a = 18
and Sum = 57
⇒ a/r + a + ar = 57
⇒ a( 1/r + 1 + r) = 57
⇒ 18(1 + r + r²) = 57
⇒ 6 + 6r + 6r² = 19r
⇒ 6r² - 13r + 6 = 0
⇒ 6r² - 9r - 4r + 6 = 0
⇒ (3r - 2) (2r - 3) = 0
⇒ r = 2/3 or r = 3/2
Putting the values of a and r in a/r, a, ar, we find that the required ages are 12, 18, 27 or 27, 18, 12.
∴ The required ages are 12, 18, 27 or 27, 18, 12.

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 5, 10, 20, ...
Here, a = 5, r = 2
Sum of n numbers = s_n = 1275
As we know that, Sum of n terms of GP 

1275 = 5 × (2ⁿ - 1)
⇒ 255 = (2ⁿ - 1)
⇒ 2ⁿ = 256
⇒ 2ⁿ = 28
⇒ n = 8
Thus the correct answer is 8.

Given:

Formula used:
Sum of infinite G.P. = a/(1 – r)

Calculation:

Now, multiply both sides by 1/13, we get
⇒ 1/13s = 5/13² + 55/13³ ....(2)
Subtracting (2) from (1)
⇒ s – 1/13s = [5/13 + 55/13² + 555/13³ + ....] – [5/13² + 55/13³ +....]
⇒ 12s/13 = 5/13 + [55/13² – 5/13²] + (555/13³ – 55/13³) + .....
⇒ 12s/13 = 5/13 + 50/13² + 500/13³ + ....
Now,
Here R.H.S is an infinite G.P. with first term a = 5/13 and common ratio (r) = 10/13
So,  
⇒ 12s/13 = 5/13/(1 – 10/13)
⇒ 12s/13 = 5/13/(13 – 10)/13
⇒ 12s/13 = (5/13)/(3/13)
⇒ 12s/13 = (5/13 × 13/3)
⇒ 12s/13 = 5/3
⇒ s = (13 × 5)/(12 × 3)
⇒ s = 65/36
∴ The required value is 65/36