Geometric Progression — Test Yourself

-1, k and -1 are three consecutive terms of a G.P., then

(i) k = 1

(ii) k = -1

Which of the following is valid ?

1. only 1

2. only 2

3. both 1 and 2

4. either 1 or -1

Answer

Given,

-1, k and -1 are three consecutive terms of a G.P.

$\therefore \dfrac{k}{-1} = \dfrac{-1}{k} \\[1em] \Rightarrow k \times k = -1 \times -1 \\[1em] \Rightarrow k^2 = 1 \\[1em] \Rightarrow k = \sqrt{1} = \pm 1.$

Hence, Option 4 is the correct option.

x + 9, 10 and 4 are in G.P. The value of x is :

1. 16

2. 8

3. -16

4. 0

Answer

Given,

x + 9, 10 and 4 are in G.P.

$\therefore \dfrac{10}{x + 9} = \dfrac{4}{10} \\[1em] \Rightarrow 4(x + 9) = 10 \times 10 \\[1em] \Rightarrow 4x + 36 = 100 \\[1em] \Rightarrow 4x = 100 - 36 \\[1em] \Rightarrow 4x = 64 \\[1em] \Rightarrow x = \dfrac{64}{4} = 16.$

Hence, Option 1 is the correct option.

The common ratio of a G.P. is 2 and its 6^th term is 48. The first term is :

1. $\dfrac{3}{2}$

2. $\dfrac{2}{3}$

3. 1

4. 2

Answer

Given,

Common ratio of G.P. (r) = 2

By formula,

⇒ a_n = ar^{n - 1}

Given,

6^th term of G.P. is 48.

⇒ a₆ = 48

⇒ a × (2)^{6 - 1} = 48

⇒ a × 2⁵ = 48

⇒ 32a = 48

⇒ a = $\dfrac{48}{32} = \dfrac{3}{2}$.

Hence, Option 1 is the correct option.

Three terms are in G.P., whose product is 27. The middle term is :

1. -3

2. 3

3. -3 and 3

4. -3 or 3

Answer

Let three terms be $\dfrac{a}{r}$, a and ar.

Given,

Product of three terms = 27

$\therefore \dfrac{a}{r} \times a \times ar = 27 \\[1em] \Rightarrow a^3 = 3^3 \\[1em] \Rightarrow a = 3.$

Hence, Option 2 is the correct option.

The common ratio of a G.P., whose 4^th term is 27 and 6^th term is 243; is :

1. 9

2. 3

3. $\dfrac{1}{3}$

4. $\dfrac{1}{9}$

Answer

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

4^th term of G.P. is 27.

⇒ a₄ = 27

⇒ ar^{4 - 1} = 27

⇒ ar³ = 27 ........(1)

Given,

6^th term of G.P. is 243.

⇒ a₆ = 243

⇒ ar^{6 - 1} = 243

⇒ ar⁵ = 243 ........(2)

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

$\Rightarrow \dfrac{ar^5}{ar^3} = \dfrac{243}{27} \\[1em] \Rightarrow r^2 = 9 \\[1em] \Rightarrow r = \sqrt{9} = 3.$

Hence, Option 2 is the correct option.

The third term of a G.P. = 18, the product of its first five terms is :

1. 18

2. 18⁵

3. 9

4. $\sqrt{18}$

Answer

Let first five terms of G.P. be

$\dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2$

Given,

Third term of a G.P. = 18.

∴ a = 18

Product of first five terms = $\dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2$

= a⁵

= 18⁵.

Hence, Option 2 is the correct option.

G.P. : $\dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2},..............$

Assertion (A): 5^th term of the given G.P. is $1\dfrac{1}{8}$.

Reason (R): If for a G.P., the first term is a, the common ratio is r and the number of terms = n, then sum of the first n terms S_n = $\dfrac{a(r^n - 1)}{r - 1}$ for all r.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, the sequence = $\dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2},..............$

First term (a) = $\dfrac{2}{9}$

Common ratio (r) = $\dfrac{\dfrac{1}{3}}{\dfrac{2}{9}} = \dfrac{1 \times 9}{2 \times 3} = \dfrac{3}{2}$

Using the formula; T_n = a.r^{n - 1}

$T_5 = \dfrac{2}{9} × \Big(\dfrac{3}{2}\Big)^{5 - 1}\\[1em] = \dfrac{2}{9} × \Big(\dfrac{3}{2}\Big)^4\\[1em] = \dfrac{2}{9} × \dfrac{81}{16}\\[1em] = \dfrac{9}{8}\\[1em] = 1\dfrac{1}{8}.$

So, assertion (A) is true.

According to reason :

Sum of first n terms (S_n) = $\dfrac{a(r^n - 1)}{r - 1}$, for all r.

But this is not the case when r = 1.

Thus,

Reason (R) is false.

Hence, option 1 is the correct option.

For a G.P., its fourth term = x, seventh term = y and tenth term = z.

Assertion (A): x, y and z are in G.P.

Reason (R): y² = (ar⁶)² = ar³ × ar⁹ = xz.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

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

By formula :

⇒ T_n = a.r^{n - 1}

Given, fourth term = x, seventh term = y and tenth term = z

⇒ a₄ = x, a₇ = y and a₁₀ = z

⇒ ar^{4 - 1} = x, ar^{7 - 1} = y and ar^{10 - 1} = z

⇒ ar³ = x, ar⁶ = y and ar⁹ = z

If x, y and z are in G.P., then the ratio between the consecutive terms will be equal.

Ratio between y and x :

$\Rightarrow \dfrac{y}{x} = \dfrac{ar^6}{ar^3} = r^3.$

Ratio between z and y :

$\Rightarrow \dfrac{z}{y} = \dfrac{ar^9}{ar^6} = r^3.$

Since, the ratio between the consecutive terms are equal.

Thus, x, y and z are in G.P.

∴ Assertion (A) is true.

⇒ y² = (ar⁶)²

⇒ y² = ar¹²

⇒ y² = ar³ × ar⁹

⇒ y² = xz.

∴ Reason (R) is true.

Hence, option 3 is the correct option.

G.P. : = 3 - 6 + 12 - 24 + ............. - 384

Statement (1): Product of 5^th term from the beginning and 5^th term from the end of the G.P. is -1152.

Statement (2): In an G.P. the product of n^th term from the beginning and n^th term from the end is

1^st term + last term

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, G.P.: = 3 - 6 + 12 - 24 + ............. - 384

Here, a = 3

common ratio, r = $\dfrac{-6}{3}$ = -2

a_n = -384

Using the formula; T_n = a.r^{n - 1}

⇒ 3 x (-2)^{n - 1} = -384

⇒ (-2)^{n - 1} = $-\dfrac{384}{3}$

⇒ (-2)^{n - 1} = -128

⇒ (-2)^{n - 1} = (-2)⁷

⇒ n - 1 = 7

⇒ n = 7 + 1 = 8

5_th term from the beginning,

⇒ T₅ = 3 x (-2)^{5 - 1}

= 3 x (-2)⁴

= 3 x 16 = 48

5_th term from the last = (8 - 5 + 1) = 4_th term from the beginning,

⇒ T₄ = 3 x (-2)^{4 - 1}

= 3 x (-2)³

= 3 x (-8) = -24

Product of 5_thterm from the beginning and 5_thterm from the end = 48 x (-24) = -1152.

So, statement 1 is true.

Let in a G.P.

a be the first term and N be the total number of terms.

n^th term from the beginning,

⇒ T_n = a.r^{n - 1} ........(1)

n^th term from the end,

⇒ T_{N - n + 1} = a.r^{N - n + 1 - 1}

⇒ T_{N - n + 1} = ar^{N - n} .....(2)

Multiplying equation (1) and (2), we get :

⇒ T_n x T_{N - n + 1} = a.r^{n - 1} x a.r^{N - n}

= a².r^{(n - 1) + (N - n)}

= a².r^{N - 1} ....................(3)

Now, first term + last term = a + a.r^{n - 1}

= a(1 + r^{n - 1}) ....................(4)

From equation (3) and (4),

The product of n^th term from the beginning and n^th term from the end is not equal to 1^st term + last term.

So, statement 2 is false.

Hence, option 3 is the correct option.

In a G.P., common ratio = 2, first term = 3 and last term = 96.

Statement (1): The number of terms in this G.P. = 96 - 3.

Statement (2): a : ar^{n - 1} = 3 : 96

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

First term, a = 3

Common ratio, r = 2

Last term = 96

Using the formula;

⇒ T_n = a.r^{n - 1}

⇒ 3 x 2^{n - 1} = 96

⇒ 2^{n - 1} = $\dfrac{96}{3}$

⇒ 2^{n - 1} = 32

⇒ 2^{n - 1} = 2⁵

⇒ n - 1 = 5

⇒ n = 5 + 1 = 6

And, according to statement 1, the number of terms in this G.P. = 96 - 3 = 93, which is not correct.

So, statement 1 is false.

$\Rightarrow \dfrac{a}{ar^{n - 1}} = \dfrac{\text{first term}}{\text{last term}}\\[1em] = \dfrac{3}{96} = 3 : 96.$

So, statement 2 is true.

Hence, option 4 is the correct option.

The 5^th and the 8^th terms of a G.P. are 32 and 256 respectively. Find its first term and the common ratio.

Answer

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

5^th term of G.P. = 32

⇒ a₅ = 32

⇒ ar^{5 - 1} = 32

⇒ ar⁴ = 32 ..........(1)

Given,

8^th term of G.P. = 256

⇒ a₈ = 256

⇒ ar^{8 - 1} = 256

⇒ ar⁷ = 256 ..........(2)

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

$\Rightarrow \dfrac{ar^7}{ar^4} = \dfrac{256}{32} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2.$

Substituting value of r in equation (1), we get :

⇒ a × (2)⁴ = 32

⇒ 16a = 32

⇒ a = $\dfrac{32}{16}$ = 2.

Hence, first term = 2 and common ratio = 2.

The third term of a G.P. is greater than its first term by 9 whereas its second term is greater than the fourth term by 18. Find the G.P.

Answer

Let first term be a and common ratio be r.

By formula,

⇒ a_n = ar^{n - 1}

Given,

The third term of a G.P. is greater than its first term by 9.

∴ a₃ - a = 9

⇒ ar^{3 - 1} - a = 9

⇒ ar² - a = 9

⇒ a(r² - 1) = 9

⇒ a = $\dfrac{9}{r^2 - 1}$ .......(1)

Given,

Second term is greater than the fourth term by 18.

∴ a₂ - a₄ = 18

⇒ ar^{2 - 1} - ar^{4 - 1} = 18

⇒ ar - ar³ = 18

⇒ ar(1 - r²) = 18

Substituting value of a from equation (1) in above equation, we get :

$\Rightarrow \dfrac{9}{r^2 - 1} \times r(1 - r^2) = 18 \\[1em] \Rightarrow \dfrac{9}{r^2 - 1} \times -r(r^2 - 1) = 18 \\[1em] \Rightarrow -9r = 18 \\[1em] \Rightarrow r = -\dfrac{18}{9} = -2.$

Substituting value of r in equation (1), we get :

$\Rightarrow a = \dfrac{9}{(-2)^2 - 1} \\[1em] = \dfrac{9}{4 - 1} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

G.P. = a, ar, ar², ar³, .......

= 3, 3 × -2, 3 × (-2)², 3 × (-2)³, .......

= 3, -6, 3 × 4, 3 × -8, ........

= 3, -6, 12, -24, ........

Hence, required G.P. = 3, -6, 12, -24, ........

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P. Find the value of G.

Answer

Given,

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P.

$\therefore \dfrac{2x + 2}{x} = \dfrac{3x + 3}{2x + 2} \\[1em] \Rightarrow (2x + 2)^2 = x(3x + 3) \\[1em] \Rightarrow (2x)^2 + 2^2 + 2 \times 2x \times 2 = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 + 4 + 8x = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 - 3x^2 + 8x - 3x + 4 = 0 \\[1em] \Rightarrow x^2 + 5x + 4 = 0 \\[1em] \Rightarrow x^2 + 4x + x + 4 = 0 \\[1em] \Rightarrow x(x + 4) + 1(x + 4) = 0 \\[1em] \Rightarrow (x + 1)(x + 4) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x + 4 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = -4.$

Substituting value of x = -1, in terms we get :

Terms : -1, 2(-1) + 2, 3(-1) + 3, G

= -1, -2 + 2, -3 + 3, G

= -1, 0, 0, G

This is not possible as in this case common ratio is different.

Substituting value of x = -4, in terms we get :

Terms : -4, 2(-4) + 2, 3(-4) + 3, G

= -4, -8 + 2, -12 + 3, G

= -4, -6, -9, G

Here, common difference = $\dfrac{-6}{-4} = \dfrac{3}{2}$.

G = $-9 \times \dfrac{3}{2} = -\dfrac{27}{2}$.

Hence, G = $-\dfrac{27}{2}$.

The third term of a G.P. is 2. Find the product of the first five terms of this G.P.

Answer

Let first five terms of G.P. be $\dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2$.

Given,

Third term of the G.P. is 2.

∴ a = 2

Product of terms = $\dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 = a^5$

= 2⁵

= 32.

Hence, product of the first five terms of this G.P. = 32.

The 10^th, 16^th and 22^nd terms of a G.P. are x, y and z respectively. Show that x, y and z are in G.P.

Answer

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

The 10^th, 16^th and 22^nd terms of a G.P. are x, y and z respectively.

⇒ a₁₀ = x

⇒ x = ar^{10 - 1}

⇒ x = ar⁹ ...........(1)

⇒ a₁₆ = y

⇒ y = ar^{16 - 1}

⇒ y = ar¹⁵ ...........(2)

⇒ a₂₂ = z

⇒ z = ar^{22 - 1}

⇒ z = ar²¹ ...........(3)

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

$\Rightarrow \dfrac{y}{x} = \dfrac{ar^{15}}{ar^9} \\[1em] \Rightarrow \dfrac{y}{x} = \dfrac{r^{15}}{r^9} \\[1em] \Rightarrow \dfrac{y}{x} = r^{15 - 9} \\[1em] \Rightarrow \dfrac{y}{x} = r^6.$

Dividing equation (3) by (2), we get :

$\Rightarrow \dfrac{z}{y} = \dfrac{ar^{21}}{ar^{15}} \\[1em] \Rightarrow \dfrac{z}{y} = \dfrac{r^{21}}{r^{15}} \\[1em] \Rightarrow \dfrac{z}{y} = r^{21 - 15} \\[1em] \Rightarrow \dfrac{z}{y} = r^6.$

Since, $\dfrac{y}{x} = \dfrac{z}{y}$ = r⁶.

Hence, proved that x, y and z are in G.P.

Which term of the G.P. $2, 2\sqrt{2}, 4, ........ \text{ is } 128\sqrt{2} ?$

Answer

Given,

G.P. : $2, 2\sqrt{2}, 4, ........$

In above G.P.,

First term (a) = 2

Common ratio (r) = $\dfrac{2\sqrt{2}}{2} = \sqrt{2}$

Let n^th term of G.P. be $128\sqrt{2}$.

$\Rightarrow ar^{n-1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (\sqrt{2})^{n - 1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (2^{\dfrac{1}{2}})^{n - 1} = (2)^7.\sqrt{2} \\[1em] \Rightarrow 2 \times (2)^{\dfrac{n - 1}{2}} = 2^7.2^{\dfrac{1}{2}} \\[1em] \Rightarrow (2)^{1 + \dfrac{n - 1}{2}} = (2)^{7 + \dfrac{1}{2}} \\[1em] \Rightarrow (2)^{\dfrac{n - 1 + 2}{2}} = (2)^{\dfrac{14 + 1}{2}} \\[1em] \Rightarrow (2)^{\dfrac{n + 1}{2}} = (2)^{\dfrac{15}{2}} \\[1em] \Rightarrow \dfrac{n + 1}{2} = \dfrac{15}{2} \\[1em] \Rightarrow n + 1 = 15 \\[1em] \Rightarrow n = 15 - 1 = 14.$

Hence, 14^th term of G.P. = $128\sqrt{2}$.

Find the 8^th term of a G.P., if its common ratio is 2 and 10^th term is 768.

Answer

Given,

Common ratio (r) = 2

By formula,

⇒ a_n = ar^{n - 1}

Given,

10^th term of G.P. = 768

⇒ a₁₀ = 768

⇒ ar^{10 - 1} = 768

⇒ ar⁹ = 768

⇒ a × 2⁹ = 768

⇒ a × 512 = 768

⇒ a = $\dfrac{768}{512} = \dfrac{3}{2}$.

8th term of G.P. is given by :

⇒ a₈ = ar^{8 - 1}

⇒ a₈ = ar⁷

⇒ a₈ = $\dfrac{3}{2} \times 2^7$

⇒ a₈ = 3 × 2⁶

⇒ a₈ = 3 × 64 = 192.

Hence, 8^th term of a G.P. is given by 192.

In a G.P., the 4^th term is 48 and 7^th term is 384. Find its 6^th term.

Answer

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

4^th term of G.P. = 48

⇒ a₄ = 48

⇒ ar^{4 - 1} = 48

⇒ ar³ = 48 ..........(1)

Given,

7^th term of G.P. = 384

⇒ a₇ = 48

⇒ ar^{7 - 1} = 384

⇒ ar⁶ = 384 ..........(2)

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

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{384}{48} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2.$

Substituting value of r in equation (1), we get :

⇒ a(2)³ = 48

⇒ a × 8 = 48

⇒ a = $\dfrac{48}{8}$ = 6.

⇒ a₆ = ar^{6 - 1}

⇒ a₆ = ar⁵

⇒ a₆ = 6 × 2⁵

⇒ a₆ = 6 × 32

⇒ a₆ = 192.

Hence, 6^th term of G.P. = 192.

$-\dfrac{5}{3}, x \text{ and } -\dfrac{3}{5}$ are three consecutive terms of a G.P. Find the value(s) of x.

Answer

Given,

$-\dfrac{5}{3}, x \text{ and } -\dfrac{3}{5}$ are three consecutive terms of a G.P.

$\therefore \dfrac{x}{-\dfrac{5}{3}} = \dfrac{-\dfrac{3}{5}}{x} \\[1em] \Rightarrow x^2= -\dfrac{3}{5} \times -\dfrac{5}{3} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.$

Hence, x = +1 or -1.

Rohit writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instructions that they move the chain similarly. Assuming that the chain is not broken and that it costs ₹ 10 to mail one letter, determine the amount spent on postage when the fourth set of letters is mailed.

Answer

Rohit writes a letter to four of his friends, and each of them copies the letter and mail it to four different persons and the chain continues. Thus, the above sequence is :

4, 16, 64, .......

The above sequence is a G.P., with first term (a) = 4 and r = 4.

By formula,

Sum of G.P. = $\dfrac{a(r^n - 1)}{(r - 1)}$

Thus,

$S_4 = \dfrac{4 \times (4^4 - 1)}{4 - 1} \\[1em] = \dfrac{4 \times (256 - 1)}{3} \\[1em] = \dfrac{4 \times 255}{3} \\[1em] = 4 \times 85 \\[1em] = 340.$

Thus, total 340 letters are mailed till fourth set.

If, it costs ₹ 10 to mail one letter.

Total cost = No. of letters × 10 = 340 × ₹ 10 = ₹ 3,400.

Hence, amount spent on postage = ₹ 3,400.

The fourth term of a G.P. is eight times its seventh term. The fifth term of then G.P. is $\dfrac{3}{16}$, then find its 12^th term.

Answer

By formula,

a_n = ar^{n - 1}

Given,

The fourth term of a G.P. is eight times its seventh term.

a₄ = 8a₇

ar^{4 - 1} = 8ar^{7 - 1}

ar³ = 8ar⁶

$\dfrac{r^6}{r^3} = \dfrac{a}{8a}$

$r^3 = \dfrac{1}{8}$

r = $\sqrt[3]{\dfrac{1}{8}} = \dfrac{1}{2}$.

Given,

Fifth term = $\dfrac{3}{16}$

$\Rightarrow ar^4 = \dfrac{3}{16} \\[1em] \Rightarrow a \times \Big(\dfrac{1}{2}\Big)^4 = \dfrac{3}{16} \\[1em] \Rightarrow a \times \dfrac{1}{16} = \dfrac{3}{16} \\[1em] \Rightarrow a = 3.$

12th term :

a₁₂ = ar^{12 - 1}

= ar¹¹

= $3 \times \Big(\dfrac{1}{2}\Big)^{11}$.

Hence, 12^th term = $3 \times \Big(\dfrac{1}{2}\Big)^{11}$.

Richard borrows ₹ 10,000 from his friend Ramesh. Ramesh asks him to return the money just by repaying ₹ 10 at first day and double the amount in the subsequent days than the each previous day continuously for 10 days. Richard thanks Ramesh for his help. Determine whether Ramesh was in loss or profit.

Answer

Sequence :

10 + 20 + 40 + ......... upto 10 terms.

The above is a G.P. with first term (a) = 10 and common ratio (r) = 2.

By formula,

S_n = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{10 \times (2^{10} - 1)}{2 - 1} \\[1em] = 10 \times (1024 - 1) \\[1em] = 10 \times 1023 \\[1em] = 10,230.$

Since, Ramesh lent ₹ 10,000 but received ₹ 10,230 thus he is in profit of ₹ 230.

Hence, Ramesh makes a profit of ₹ 230.