Geometric Progression MCQ Quiz - Objective Question with Answer for Geometric Progression - Download Free PDF

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

⇒ \(\frac{ar(1-r^2)}{{a(r^2-1)}}\) = \(\frac{18}{{9}}\)

⇒ -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 Used:

The Geometric Mean (G.M) of a series containing n observations is the nth root of the product of the values.

\(\begin{array}{l}G. M = \sqrt[n]{x_{1}× x_{2}× …x_{n}}\end{array}\)

Calculation:

Using the above formula -

⇒ 8⁵ = 32 × 4 × 8 × X × 2

⇒ X =  \(\frac{8^{5}}{32\times 4\times 8\times 2}\)= 16

∴ The correct answer is 16

Given:

The geometric mean of squares of two positive integers is 10.

Formula used:

Geometric Mean = √(a² × b²)

Where a and b are positive integers.

Calculation:

√(a² × b²) = 10

⇒ (a² × b²) = 100

⇒ a × b = 10

To minimize the sum (a + b), choose factors of 10:

a = 2, b = 5

⇒ a + b = 2 + 5 = 7

∴ The correct answer is option (1).

Given:

The fifth term of a geometric progression (GP) = 27

The eighth term of the GP = 729

Formula used:

General term of GP: T_n = a × r^n-1

Where, a = first term, r = common ratio, n = term number

Calculation:

Fifth term: T₅ = a × r⁴ = 27

Eighth term: T₈ = a × r⁷ = 729

Dividing both equations:

⇒ (a × r⁷) / (a × r⁴) = 729 / 27

⇒ r³ = 27

⇒ r = 3

Substitute r = 3 into T₅:

⇒ a × 3⁴ = 27

⇒ a × 81 = 27 

⇒ a = 27 / 81 = 1/3

T₁₁ = (1/3) × 3¹⁰

⇒ T₁₁ = (1/3) × 59049

⇒ T₁₁ = 19683

∴ The correct answer is option (1).

Given:

Sum of first 13 terms of a GP = Sum of first 11 terms of the same GP

Sum of first 15 terms of the GP = 1200

Formula used:

Sum of n terms of a GP: S_n = a(1 - rⁿ) / (1 - r)

To find the 21st term: T₂₁ = a × r²⁰

Calculation:

⇒ S₁₃ = S₁₁

⇒ a(1 - r¹³) / (1 - r) = a(1 - r¹¹) / (1 - r)

⇒ 1 - r¹³ = 1 - r¹¹

⇒ r¹³ = r¹¹

⇒ r² = 1

⇒ r = ±1

⇒ r = -1 (since r = 1 not valid for equal sums)

Now using S₁₅ = 1200

⇒ a(1 - (-1)¹⁵) / (1 - (-1)) = 1200

⇒ a(1 + 1) / 2 = 1200

⇒ 2a / 2 = 1200

⇒ a = 1200

T₂₁ = a × r²⁰

⇒ T₂₁ = 1200 × (-1)²⁰

⇒ T₂₁ = 1200 × 1

⇒ T₂₁ = 1200

∴ The correct answer is option (3).

Formula used:

Sum of geometric progression (S_n) = {a × (rⁿ - 1)}/(r - 1)

Where, a = first term ; r = common ratio ; n = number of term

Calculation:

3 +32 + 33 +...+ 38.

Here, a = 3 ; r = 3 ; n = 8

Sum of the series (S₈) = {a × (r8 - 1)}/(r - 1)

⇒ {3 × (3⁸ - 1)}/(3 - 1)

⇒ (3 × 6560)/2 = 3280 × 3 

⇒ 9840

∴ The correct answer is 9840.

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 \( = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\)

Formula Used:

\({S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\)

n = 10

Calculation:

Considering the given series

16, 32, 64, 128, ......

\({S_{10}} = \frac{{16\left( {{2^{10}} - 1} \right)}}{{2 - 1}}\)

⇒ S₁₀ = 16 × 1023 = 16368

Given:

First term of GP = 15

Common ratio of GP = 4

Formula used:

\(S_n=a_1\times \dfrac{(r^n-1)}{(r-1)}\)

Calculation:

\(S_4=15\times \dfrac{(4^4-1)}{(4-1)}\) \(= \dfrac{15\times 255}{3} = 1275\)

∴ The answer is 1275 .

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:

⇒ Tn = arn-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:

The series is 4, 12, 36, 108

Concept used:

In geometric progression

a_n = a₁{r^{(n - 1)}}

Here,

a_n = nth term

a₁ = 1st term

r = common ratio

n = number of term

Calculation:

According to the question:

a₁ = 4

a₂ = 12

So, common ratio of a₁ : a₂ = 4 : 12

⇒ 1 : 3

So,

a₆ = 4 × 3^{(6 - 1)}

⇒ a₆ = 4 × 3⁵

⇒ a₆ = 4 × 243

⇒ a₆ = 972

So, 6^th term of the series = 972

∴ The 6^th term of this series is 972.

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.

Alternate Method

(20 + 22 + 24 +........+28) × 3

⇒ (1 + 4 + 16 + 64 + 256) × 3

⇒ 341 × 3 = 1023

∴ The Sum of the series is 1023.

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:

Concept:

• Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}\)
• nth  term of the G.P. is an = arn−1
• Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1
• Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {1 - {\rm{\;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{\;r}}}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{\;}}\frac{{\rm{a}}}{{1{\rm{\;}} - {\rm{\;r}}}}{\rm{\;}}\) ; |r| < 1

Calculation:

Given series is 5, 10, 20, ...

Here, a = 5, r = 2

Sum of n numbers = sn = 1275

As we know that, Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1

sn = \(\frac{{{\rm{5\;}}\left( {{{\rm{2}}^{\rm{n}}} - 1} \right)}}{{{\rm{2}} - {\rm{\;}}1}}\)

1275 = 5 × (2n - 1)

⇒ 255 = (2n - 1)

⇒ 2n = 256

⇒ 2n = 28

⇒ n = 8

Thus the correct answer is 8.

Formula:

GM of a and b = √ab

Given:

a = 9, b = 81, x = Geometric Mean

Calculation:

x = √729

⇒ x = 27