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

Given:

The numbers are 27, 60, 108, 150, and 225, and the geometric mean is denoted by x. We need to find the harmonic mean of x and 60.

Formula used:

Geometric mean (x) = (Product of numbers)^(1/n), where n is the number of terms, and the numbers are the given values.

Harmonic mean (H) of two numbers a and b is given by:

H = 2ab / (a + b)

Calculations:

Find the geometric mean (x) of the numbers 27, 60, 108, 150, and 225.

x = (27 × 60 × 108 × 150 × 225)^(1/5)

x = (3 × 3 × 3 × 3 × 2 × 2 × 5 × 3 × 3 × 3 × 2 × 2 × 3 × 5 × 2 × 5 × 3 × 5 × 3 × 5)(1/5)

x = (3¹⁰ × 2⁵ × 5⁵)(1/5)

x = 32 × 2 × 5

x = 90

Find the harmonic mean of x and 60.

Now that we know x = 90, we can find the harmonic mean of x and 60:

H = 2 × 90 × 60 / (90 + 60)

H = 10800 / 150

H = 72

∴ The harmonic mean of x and 60 is 72.

Concept:

​Geometric Progression (G.P.): A sequence where each term is obtained by multiplying the previous term by a constant called the common ratio.

Sum of n terms (S_n) of G.P. is given by: \( S_n = \frac{a(r^n - 1)}{r - 1} \)

• a: First term of the G.P.
• r: Common ratio of the G.P.
• n: Number of terms in the series

Calculation:

Given, f(1) = 3

Using the property: \( f(x + y) = f(x) \times f(y) \)

⇒ f(2) = f(1 + 1) = f(1) × f(1) = 3 × 3 = 9

⇒ f(3) = f(1 + 2) = f(1) × f(2) = 3 × 9 = 27

⇒ f(4) = f(1 + 3) = f(1) × f(3) = 3 × 27 = 81

So, \( f(1), f(2), f(3), \dots \) = 3, 9, 27, 81, … forms a G.P. with a = 3 and r = 3

Given, \( \sum_{x=1}^{n} f(x) = 120 \)

⇒ \( S_n = \frac{3(3^n - 1)}{3 - 1} \)

⇒ \( 120 = \frac{3(3^n - 1)}{2} \)

⇒ \( 120 = \frac{3}{2}(3^n - 1) \)

⇒ \( 240 = 3(3^n - 1) \)

⇒ \( 80 = 3^n - 1 \)

⇒ \( 3^n = 81 \)

⇒ \( 3^n = 3^4 \Rightarrow n = 4 \)

∴ The value of n is 4.

Concept:

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

• 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

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

Explanation:

Given a = 3

Let common ratio be r

Then GP is a, ar, ar², ar³, ....(2n th term)

Then for the GP taking odd places term is

a, ar², ar⁴, ... (n the term)

i.e, common ratio is r²

By given condition

S_2n = 3 × S_n

\({a(r^{2n}-1)\over r-1}=3\times {a({(r^2)}^n-1)\over r^2-1}\)

\({(r^{2n}-1)\over r-1}=3\times {({r}^{2n}-1)\over r^2-1}\)

\(1=\frac3{r+1}\)

r + 1 = 3

r = 2

Common ratio of the given GP is 2

Option (1) is true.

Concept:

Arithmetic Progression (AP):

• The terms are in AP if the difference between consecutive terms is constant.
• The common difference is given by: \(d = q - p = r - q\)

Geometric Progression (GP):

• The terms are in GP if the ratio between consecutive terms is constant.
• The common ratio is given by: \(r = q \times p\)

Calculation:

Given

p = ln(x),

q = ln(x3)  = 3 lnx

and r = ln(x5) = 5 lnx

Clearly, p - q  = r - q = \(2lnx\)

⇒ p, q, r are in AP

Also \(\frac{q}{p} ≠ \frac{r}{p} \because \frac{q}{p} = 3 , \frac{r}{p} = \frac{5}{3}\)

∴ p, q, r can never be in GP

∴ Option (c) is correct.

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 \(\rm \dfrac{a}{r^2},\dfrac{a}{r},a,ar,ar^2\) .

Calculation:

Let us consider a general geometric progression with common ratio r.

Assume that the five terms in the GP are \(\rm \dfrac{a}{r^2},\dfrac{a}{r},a,ar,ar^2\).

It is given that third term is 9.

Therefore, a = 9.

Now the product of the five terms is given as follows:

\(\rm \dfrac{a}{r^2}\times\dfrac{a}{r}\times a\times ar \times ar^2 = a^5\)

But we know that a = 9.

Thus, the product is \(9^5=3^{10}\).

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

To Find: nAs 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

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 a G.P.

• 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

Sin18o =  \(\frac{\sqrt{5}- 1}{4}\) 

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 = \(\frac{\sqrt{5}- 1}{2}\)   = 2 ×\( \frac{\sqrt{5}- 1}{4}\) = 2 Sin18° 

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: S_n = \(\rm a\left(\dfrac{r^n-1}{r-1}\right)\).
• The sum to ∞ of a GP, when |r| < 1, is: S_∞ = \(\rm \dfrac{a}{1-r}\).

Calculation:

Let us consider the infinite series \(\rm \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ ... \infty\).

Here, a = \(\rm \dfrac{1}{3}\) and r = \(\rm \dfrac{\tfrac{1}{9}}{\tfrac{1}{3}}=\dfrac{1}{3}\).

∴ S∞ = \(\rm \dfrac{a}{1-r}=\dfrac{\tfrac{1}{3}}{1-\tfrac{1}{3}}=\dfrac{\tfrac{1}{3}}{\tfrac{2}{3}}=\dfrac{1}{2}\).

Now, let P = \(\rm 9^{\tfrac{1}{3}} 9^{\tfrac{1}{9}} 9^{\tfrac{1}{27}}\ ...\ \infty\).

∴ P = \(\rm 9^{\tfrac{1}{3}+\tfrac{1}{9}+\tfrac{1}{27}+\ ... \infty}=9^{\tfrac{1}{2}}=\sqrt9=3\).

Concept: 

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

• Sum of n terms = s = \(\rm \frac{a\left ( r^{n}-1 \right )}{r-1}\); where r >1
• Sum of n terms = s = \(\rm \frac{a\left ( 1-r^{n} \right )}{1-r}\) ; where r <1  

Calculation:  

The required no. of ancestors 

= 2 + 4 + 8 +... upto 8 terms 

As we know that sum  of G.P , 

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

Where , a = 2, r = 2 and n = 8

⇒ No. of ancestors required = \(\rm \frac{2\times \left ( 2^{8}-1\right )}{2-1}\) =  \(\rm 2 \times \left ( 2^{8} -1\right )\) = 510 

∴ No. of ancestors required is  510. 

The correct option is 4. 

Concept:

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

Common ratio = r = \(\rm a_2\over a_1 \) = \(\rm a_3\over a_2 \) = \(\rm a_n\over a_{n-1} \)

Calculation:

Consider,

(a = 3) be the 3^rd term of the G.P series,

So, we can write the five terms as,

\(\rm a\over r^2\), \(\rm a\over r\), a, ar, ar²

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

P = \(\rm a\over r^2\) × \(\rm a\over r\) × a × ar × ar2 = a⁵

Since,

a = 3,

∴ The product of the first five terms (P) = 3⁵ = 243

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:

  Sn = \(\frac{a(r^n-1)}{r-1}\) if r > 1  or Sn = \(\frac{a(1 -r^n)}{1-r}\) if r < 1

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.

∴ Sn = \(\rm 1\left(\frac{3^n\ -\ 1}{3\ -\ 1}\right)\) = 3280

⇒ \(\rm \left(\frac{3^n\ -\ 1}{2}\right)\) = 3280

⇒ 3ⁿ - 1 = 3280 × 2

⇒ 3ⁿ - 1 = 6560

⇒ 3ⁿ = 6561 = 3⁸

⇒ n = 8.

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 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 = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1

∴ sn = \(\frac{{{\rm{4\;}}\left( {{{\rm{2}}^{\rm{n}}} - 1} \right)}}{{{\rm{2}} - {\rm{\;}}1}}\)

2044 = 4 × (2ⁿ - 1)

⇒ 511 = (2n - 1)

⇒ 2ⁿ = 512

⇒ 2n = 2⁹

∴ n = 9

Concept:

The geometric mean is defined as the nth root of the product of n numbers.

The geometric mean of a data set \(\rm {\textstyle \left\{a_{1},a_{2},\,\ldots ,\,a_{n}\right\}}\) is given by: 

\(\rm GM = \rm {\textstyle \left\{a_{1}\times a_{2}\times\,\ldots \times\,a_{n}\right\}}^{\frac{1}{n}}\)

Calculation:

To Find: Geometric mean of 6, 8, 16 and 27

Here n = 4

Now,

 \(\rm GM = \rm ({{6 \times 8 \times 16 \times 27}})^{\frac{1}{4}}\)

\(= \rm ({{2 \times 3 \times 2^3 \times 2^4 \times 3^3}})^{\frac{1}{4}} \\ =(2^8 \times 3^4)^{\frac{1}{4}}\\ = (2^2 \times 3)\\=12\)