Geometric Progression — Exercise 11(A)

If the first term of a G.P. is 8 and its common ratio is -2. The 3^rd term of this G.P. is :

1. -32

2. 2

3. 32

4. -2

Answer

Given,

First term of G.P. = 8

Common ratio = -2

By formula,

⇒ a_n = ar^{n - 1}

Substituting values we get :

⇒ a₃ = 8 × (-2)^{3 - 1}

⇒ a₃ = 8 × (-2)²

⇒ a₃ = 8 × 4

⇒ a₃ = 32.

Hence, Option 3 is the correct option.

The 4^th term of a G.P. is 16 and the 7^th term is 128, then its common ratio is equal to :

1. 2

2. -2

3. 1

4. -1

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 a G.P. is 16.

⇒ a₄ = 16

⇒ ar^{4 - 1} = 16

⇒ ar³ = 16 .........(1)

Given,

7^th term of a G.P. is 128.

⇒ a₇ = 128

⇒ ar^{7 - 1} = 128

⇒ ar⁶ = 128 .........(2)

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

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{128}{16} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2.$

Hence, Option 1 is the correct option.

(2x + 2) and (3x + 3) are two consecutive terms of a G.P. The common ratio of the G.P. is :

1. 2

2. 3

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

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

Answer

By formula,

Common ratio = $\dfrac{a_{n + 1}}{a_n}$

Given,

(2x + 2) and (3x + 3) are two consecutive terms of a G.P.

$\therefore \text{Common ratio } = \dfrac{3x + 3}{2x + 2} \\[1em] = \dfrac{3(x + 1)}{2(x + 1)} \\[1em] = \dfrac{3}{2}.$

Hence, Option 3 is the correct option.

The third term of a G.P. is 3. The product of its first five terms is :

1. 15

2. $\dfrac{3 \times 5}{2}$

3. 3⁵

4. 5³

Answer

Let first five terms of G.P. be

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

Given,

3rd term of G.P. is 3.

∴ a = 3

Product of first five terms are

$\Rightarrow \dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 \\[1em] \Rightarrow a^5 \\[1em] \Rightarrow 3^5.$

Hence, Option 3 is the correct option.

The 8^th term of a G.P. is 192 and its common ratio is 2, then the first term of the G.P. is :

1. 3

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

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

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

Answer

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

8^th term of a G.P. is 192.

⇒ a₈ = 192

⇒ ar^{8 - 1} = 192

⇒ ar⁷ = 192

Substituting value of common ratio (r) = 2 in above equation, we get :

⇒ a(2)⁷ = 192

⇒ 128a = 192

⇒ a = $\dfrac{192}{128} = \dfrac{3}{2}$.

Hence, Option 4 is the correct option.

Find the 9^th term of the series :

1, 4, 16, 64, ........

Answer

Since,

$\dfrac{4}{1} = \dfrac{16}{4} = 4$.

Hence, the above sequence is a G.P. with r = 4 and a = 1.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₉ = 1.(4)^{9 - 1}

= (4)⁸

= 65536.

Hence, 9^th term of the series = 65536.

Find the seventh term of the G.P. :

1, $\sqrt{3}, 3, 3\sqrt{3}, ........$

Answer

Since,

$\dfrac{\sqrt{3}}{1} = \dfrac{3\sqrt{3}}{3} = \sqrt{3}$

Hence, the above sequence is a G.P. with r = $\sqrt{3}$ and a = 1.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₇ = 1.$(\sqrt{3})^{7 - 1}$

= $(\sqrt{3})^6$

= 27.

Hence, 7^th term of the G.P. = 27.

Find the 8^th term of the sequence :

$\dfrac{3}{4}, 1\dfrac{1}{2}, 3, .........$

Answer

Sequence = $\dfrac{3}{4}, \dfrac{3}{2}, 3, ........$

Calculating ratio between terms,

$\dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3 \times 4}{2 \times 3} = 2, \dfrac{3}{\dfrac{3}{2}} = 2$.

Since,

$\dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3}{\dfrac{3}{2}} = 2$.

Hence, the sequence is a G.P. with r = 2 and a = $\dfrac{3}{4}$

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₈ = $\dfrac{3}{4} \times (2)^{8 - 1}$

= $\dfrac{3}{4} \times 2^7$

= $\dfrac{3}{4} \times 128$

= 3 × 32

= 96.

Hence, a₈ = 96.

Find the next three terms of the sequence :

$\sqrt{5}, 5, 5\sqrt{5}, .......$

Answer

Since,

$\dfrac{5}{\sqrt{5}} = \dfrac{5\sqrt{5}}{5} = \sqrt{5}$

Hence, the sequence $\sqrt{5}, 5, 5\sqrt{5}, .......$ is a G.P. with r = $\sqrt{5} \text{ and } a = \sqrt{5}.$

Next three terms are = 4^th, 5^th and 6^th.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

⇒ a₄ = ar^{(4 - 1)}

= ar³

= $\sqrt{5}(\sqrt{5})^3 = \sqrt{5}(5\sqrt{5})$

= 25.

⇒ a₅ = ar^{(5 - 1)}

= ar⁴

= $\sqrt{5}(\sqrt{5})^4 = \sqrt{5}(25)$

= $25\sqrt{5}$.

⇒ a₆ = ar^{(6 - 1)}

= ar⁵

= $\sqrt{5}(\sqrt{5})^5 = \sqrt{5}(25\sqrt{5})$

= 125.

Hence, next three terms of the G.P. are = 25, 25√5 and 125.

Find the seventh term of the G.P. :

$\sqrt{3} + 1, 1, \dfrac{\sqrt{3} - 1}{2},$ ...........

Answer

Rationalising the term, $\dfrac{\sqrt{3} - 1}{2}$ we get,

$\Rightarrow \dfrac{\sqrt{3} - 1}{2} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] = \dfrac{\sqrt{3}^2 - 1^2}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{3 - 1}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{2}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{1}{\sqrt{3} + 1}.$

So, Sequence = $\sqrt{3} + 1, 1, \dfrac{1}{\sqrt{3} + 1},$ ...........

Common ratio(r) = $\dfrac{1}{\sqrt{3} + 1}$.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

$\Rightarrow a_7 = (\sqrt{3} + 1)\Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{7 - 1} \\[1em] = (\sqrt{3} + 1)\Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{6} \\[1em] = \Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{5} \\[1em] = \Big(\dfrac{1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\Big)^{5} \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{\sqrt{3}^2 - (1)^2}\Big)^5 \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{3 - 1}\Big)^5 \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{2}\Big)^5 \\[1em] = \dfrac{1}{32}(\sqrt{3} - 1)^5.$

Hence, seventh term of the G.P. = $\dfrac{1}{32}(\sqrt{3} - 1)^5.$

Find the next two terms of the series :

2 - 6 + 18 - 54 .............

Answer

Since,

$\dfrac{-6}{2} = \dfrac{18}{-6}$ = -3.

Hence, the above series is a G.P. with r = -3 and a = 2.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

Next two terms of the series are 5^th and 6^th.

⇒ a₅ = ar⁴

= 2.(-3)⁴

= 2 × 81

= 162.

⇒ a₆ = ar⁵

= 2.(-3)⁵

= 2 × -243

= -486.

Hence, the next two terms are 162 and -486.

Which term of the G.P. :

$-10, \dfrac{5}{\sqrt{3}}, -\dfrac{5}{6}, ........ \text{ is } -\dfrac{5}{72}?$

Answer

Common ratio (r) = $\dfrac{\dfrac{5}{\sqrt{3}}}{-10} = -\dfrac{5}{10\sqrt{3}} = -\dfrac{1}{2\sqrt{3}}$.

Let n^th term of G.P. be $-\dfrac{5}{72}$.

$\therefore ar^{n - 1} = -\dfrac{5}{72} \\[1em] \Rightarrow -10 \times \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = -\dfrac{5}{72} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = -\dfrac{5}{72} \times -\dfrac{1}{10} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = \dfrac{1}{144} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = \Big(-\dfrac{1}{2\sqrt{3}}\Big)^4 \\[1em] \Rightarrow n - 1 = 4 \\[1em] \Rightarrow n = 5.$

Hence, 5^th term of the G.P. is $-\dfrac{5}{72}.$

The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.

Answer

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

Given,

⇒ a₅ = 81

⇒ ar⁴ = 81 ........(i)

Also,

⇒ a₂ = 24

⇒ ar = 24 ........(ii)

Dividing (i) by (ii) we get,

$\Rightarrow \dfrac{ar^4}{ar} = \dfrac{81}{24} \\[1em] \Rightarrow r^3 = \dfrac{27}{8} \\[1em] \Rightarrow r^3 = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow r = \dfrac{3}{2}.$

Substituting value of r in (ii) we get,

$\Rightarrow a \times \dfrac{3}{2} = 24 \\[1em] \Rightarrow a = \dfrac{2}{3} \times 24 \\[1em] \Rightarrow a = 16.$

⇒ a₃ = ar²

= $16 \times \Big(\dfrac{3}{2}\Big)^2$

= $16 \times \dfrac{9}{4}$

= 4 × 9

= 36.

⇒ a₄ = ar³

= $16 \times \Big(\dfrac{3}{2}\Big)^3$

= $16 \times \dfrac{27}{8}$

= 2 × 27

= 54.

G.P. = 16, 24, 36, 54, 81, ...........

Hence, G.P. = 16, 24, 36, 54, 81, ...........

Fourth and seventh terms of a G.P. are $\dfrac{1}{18} \text{ and } -\dfrac{1}{486}$ respectively. Find the G.P.

Answer

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

Given,

⇒ a₄ = $\dfrac{1}{18}$

⇒ ar³ = $\dfrac{1}{18}$ ........(i)

Also,

⇒ a₇ = $-\dfrac{1}{486}$

⇒ ar⁶ = $-\dfrac{1}{486}$ ........(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{-\dfrac{1}{486}}{\dfrac{1}{18}} \\[1em] \Rightarrow r^3 = -\dfrac{18}{486} \\[1em] \Rightarrow r^3 = -\dfrac{1}{27} \\[1em] \Rightarrow r^3 = \Big(-\dfrac{1}{3}\Big)^3 \\[1em] \Rightarrow r = -\dfrac{1}{3}.$

Substituting value of r in (i) we get,

$\Rightarrow a \times \Big(-\dfrac{1}{3}\Big)^3 = \dfrac{1}{18} \\[1em] \Rightarrow a \times -\dfrac{1}{27} = \dfrac{1}{18} \\[1em] \Rightarrow a = -\dfrac{27}{18} = -\dfrac{3}{2}.$

⇒ a₂ = ar

= $-\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big)$

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

⇒ a₃ = ar²

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

= $-\dfrac{3}{2} \times \dfrac{1}{9}$

= $-\dfrac{1}{6}.$

G.P. = $-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} ...........$

Hence, G.P. = $-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} ...........$

If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.

Answer

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

Given,

⇒ a = 2

⇒ a₃ = 8

⇒ ar² = 8

⇒ 2r² = 8

⇒ r² = 4

⇒ r =√4 = ±2

a₂ = ar

Let r = -2

⇒ a₂ = 2(-2) = -4

Let r = 2

⇒ a₂ = 2(2) = 4.

Hence, a₂ = 4 or -4.

The product of 3^rd term and 8^th terms of a G.P. is 243. If its 4^th term is 3, find its 7^th term.

Answer

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

Given,

⇒ a₃.a₈ = 243

⇒ ar².ar⁷ = 243

⇒ a²r⁹ = 243 ..........(i)

Also,

⇒ a₄ = 3

⇒ ar³ = 3

⇒ a = $\dfrac{3}{r^3}$ ........(ii)

Substituting value of a from (ii) in (i) we get,

$\Rightarrow \Big(\dfrac{3}{r^3}\Big)^2 \times r^9 = 243 \\[1em] \Rightarrow \dfrac{9}{r^6} \times r^9 = 243 \\[1em] \Rightarrow 9r^3 = 243 \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3.$

Substituting value of r in (ii),

$\Rightarrow a = \dfrac{3}{r^3} \\[1em] = \dfrac{3}{3^3} \\[1em] = \dfrac{1}{3^2} \\[1em] = \dfrac{1}{9}.$

a₇ = ar⁶

= $\dfrac{1}{9} \times 3^6$

= $\dfrac{1}{9} \times 729$

= 81.

Hence, 7^th term = 81.

Find the geometric progression with fourth term = 54 and seventh term = 1458.

Answer

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

Given,

⇒ a₄ = 54

⇒ ar³ = 54 ........(i)

Also,

⇒ a₇ = 1458

⇒ ar⁶ = 1458 ........(ii)

Dividing (ii) by (i) we get

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3.$

Substituting value of r in (i) we get,

⇒ a(3)³ = 54

⇒ 27a = 54

⇒ a = $\dfrac{54}{27}$ = 2.

a₂ = ar

= 2.(3) = 6.

a₃ = ar²

= 2.(3)²

= 2.(9) = 18.

G.P. = 2, 6, 18, 54, ......

Hence, G.P. = 2, 6, 18, 54, ......

Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.

Answer

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

Given,

⇒ a₂ = 6

⇒ ar = 6 ........(i)

Also,

⇒ a₅ = 9a₃

⇒ ar⁴ = 9ar²

⇒ $\dfrac{\text{ar}^4}{\text{ar}^2}$ = 9

⇒ r² = 9

⇒ r = √9

⇒ r = ±3

As all terms of G.P. are positive so, r ≠ -3

∴ r = 3

Substituting r in (i),

⇒ 3a = 6

⇒ a = 2.

G.P. = a, ar, ar², ar³, ......

= 2, 6, 18, 54, .......

Hence, G.P. = 2, 6, 18, 54, .......

The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

Answer

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

Given,

⇒ a₄ = 10

⇒ ar³ = 10 ........(i)

Also,

⇒ a₇ = 80

⇒ ar⁶ = 80 .........(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{80}{10} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} \\[1em] \Rightarrow r = 2.$

Substituting r in (i) we get,

$\Rightarrow a(2)^3 = 10 \\[1em] \Rightarrow 8a = 10 \\[1em] \Rightarrow a = \dfrac{10}{8} \\[1em] \Rightarrow a = \dfrac{5}{4}.$

Let n be no. of terms,

ar^{n - 1} = 2560

$\Rightarrow \dfrac{5}{4}\times (2)^{n - 1} = 2560 \\[1em] \Rightarrow (2)^{n - 1} = \dfrac{4}{5} \times 2560 \\[1em] \Rightarrow (2)^{n - 1} = 4 \times 512 \\[1em] \Rightarrow (2)^{n - 1} = 2048 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{11} \\[1em] \Rightarrow n - 1 = 11 \\[1em] \Rightarrow n = 12.$

Hence, first term = $\dfrac{5}{4}$, common ratio = 2 and number of terms = 12.

If the fourth and ninth terms of a G.P. are 54 and 13122 respectively, find the G.P. Also, find its general term.

Answer

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

Given,

⇒ a₄ = 54

⇒ ar³ = 54 ........(i)

Also,

⇒ a₉ = 13122

⇒ ar⁸ = 13122 .........(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{ar^8}{ar^3} = \dfrac{13122}{54} \\[1em] \Rightarrow r^5 = 243 \\[1em] \Rightarrow r^5 = 3^5 \\[1em] \Rightarrow r = 3.$

Substituting r in (i) we get,

$\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2.$

n^th term of a G.P. = ar^{n - 1}

= 2(3)^{n - 1}

= 2 × 3^{n - 1}

2^nd term of a G.P. = 2 × 3^{2 - 1}

= 2 × 3

= 6.

3^rd term of a G.P. = 2 × 3^{3 - 1}

= 2 × 3²

= 18.

G.P. = 2, 6, 18, 54, .........

Hence, G.P. = 2, 6, 18, 54, ......... and general term = 2 × 3^{n - 1}

The fifth, eighth and eleventh terms of a geometric progression are p, q and r respectively. Show that : q² = pr.

Answer

Let first term of the G.P. be A and it's common ratio be R.

Given,

⇒ a₅ = p

⇒ AR⁴ = p ........(i)

Also,

⇒ a₈ = q

⇒ AR⁷ = q .........(ii)

⇒ a₁₁ = r

⇒ AR¹⁰ = r .........(iii)

Multiplying (i) by (iii) we get,

AR⁴ x AR¹⁰ = pr

⇒ A²R¹⁴ = pr .........(iv)

Squaring eq. (ii) we get,

(AR⁷)² = q²

A²R¹⁴ = q² .........(v)

L.H.S. of eq. (iv) and (v) are equal so, R.H.S. will also be equal.

∴ q² = pr.

Hence, proved that q² = pr.

Find the seventh term from the end of the series :

$\sqrt{2}, 2, 2\sqrt{2}, .........., 32$

Answer

Since,

$\dfrac{2}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2}$.

So, the above sequence is a G.P. with r = $\sqrt{2}$.

Let there be n terms.

$\Rightarrow ar^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2} \times (\sqrt{2})^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2}^{n - 1 + 1} = 32 \\[1em] \Rightarrow (\sqrt{2})^n = (\sqrt{2})^{10} \\[1em] \Rightarrow n = 10.$

m^th term from end is (n - m + 1)^th term from starting.

So, 7^th term from end is (10 - 7 + 1) = 4^th term from starting.

a₄ = ar³

= $\sqrt{2}(\sqrt{2})^3$

= $\sqrt{2} \times 2\sqrt{2}$

= 4.

Hence, 7^th term from end is 4.

Find the third term from the end of the G.P.

$\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ........, 162.$

Answer

Common ratio,

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

So, the above sequence is a G.P. with r = 3.

Let there be n terms.

$\Rightarrow ar^{n - 1} = 162 \\[1em] \Rightarrow \dfrac{2}{27} \times (3)^{n - 1} = 162 \\[1em] \Rightarrow \dfrac{2}{3^3} \times 3^{n - 1} = 162 \\[1em] \Rightarrow 3^{n - 1 - 3} = \dfrac{162}{2} \\[1em] \Rightarrow 3^{n - 4} = 81 \\[1em] \Rightarrow 3^{n - 4} = 3^4 \\[1em] \Rightarrow n - 4 = 4 \\[1em] \Rightarrow n = 8.$

m^th term from end is (n - m + 1)^th term from starting.

So, 3^rd term from end is (8 - 3 + 1) = 6^th term from starting.

a₆ = ar⁵

= $\dfrac{2}{27} \times (3)^5$

= $\dfrac{2}{27} \times 243$

= 2 × 9

= 18.

Hence, 3^rd term from end is 18.

For the G.P. $\dfrac{1}{27}, \dfrac{1}{9}, \dfrac{1}{3}, ........., 81$;

find the product of fourth term from the beginning and the fourth term from the end.

Answer

Common ratio,

$\dfrac{\dfrac{1}{9}}{\dfrac{1}{27}} = \dfrac{27}{9} = 3$.

So, the above sequence is a G.P. with r = 3.

Let there be n terms.

$\Rightarrow ar^{n - 1} = 81 \\[1em] \Rightarrow \dfrac{1}{27} \times (3)^{n - 1} = 81 \\[1em] \Rightarrow 3^{n - 1} = 81 \times 27 \\[1em] \Rightarrow 3^{n - 1} = 3^4 \times 3^3 \\[1em] \Rightarrow 3^{n - 1} = 3^7 \\[1em] \Rightarrow n - 1 = 7 \\[1em] \Rightarrow n = 8.$

a₄ = ar³

= $\dfrac{1}{27} \times (3)^3$

= $\dfrac{1}{27} \times 27$

= 1.

m^th term from end is (n - m + 1)^th term from starting.

So, 4^th term from end is (8 - 4 + 1) = 5^th term from starting.

a₅ = ar⁴

= $\dfrac{1}{27} \times (3)^4$

= $\dfrac{1}{27} \times 81$

= 3.

a₄.a₅ = 1 × 3 = 3.

Hence, product of fourth term from the beginning and the fourth term from the end = 3.

If a, b and c are in G.P. and a, x, b, y, c are in A.P. prove that :

(i) $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$

(ii) $\dfrac{a}{x} + \dfrac{c}{y} = 2.$

Answer

Given,

a, b and c are in G.P.

⇒ b² = ac .............(i)

a, x, b, y, c are in A.P.

⇒ 2x = a + b and 2y = b + c.

⇒ $x = \dfrac{a + b}{2} \text{ and } y = \dfrac{b + c}{2}$ ........(ii)

(i) Substituting value of x and y from (ii) in L.H.S. of $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$,

$\text{L.H.S}. = \dfrac{1}{\dfrac{a + b}{2}} + \dfrac{1}{\dfrac{b + c}{2}} \\[1em] = \dfrac{2}{a + b} + \dfrac{2}{b + c} \\[1em] = \dfrac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\[1em] = \dfrac{2b + 2c + 2a + 2b}{(a + b)(b + c)}\\[1em] = \dfrac{2a + 2c + 4b}{ab + ac + b^2 + bc} \\[1em] = \dfrac{2(a + c + 2b)}{ab + b^2 + b^2 + bc} \text{from (i)} \\[1em] = \dfrac{2(a + c + 2b)}{b(a + b + b + c)} \\[1em] = \dfrac{2(a + c + 2b)}{b(a + c + 2b)} \\[1em] = \dfrac{2}{b} = \text{R.H.S.}$

Hence, proved that $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$.

(ii) Substituting value of x and y from (ii) in L.H.S. of $\dfrac{a}{x} + \dfrac{c}{y} = 2$,

$\text{L.H.S.} = \dfrac{a}{\dfrac{a + b}{2}} + \dfrac{c}{\dfrac{b + c}{2}} \\[1em] = \dfrac{2a}{a + b} + \dfrac{2c}{b + c} \\[1em] = 2\Big(\dfrac{a}{a + b} + \dfrac{c}{b + c}\Big) \\[1em] = 2\Big[\dfrac{a(b + c) + c(a + b)}{(a + b)(b + c)}\Big] \\[1em] = 2\Big(\dfrac{ab + ac + ac + bc}{ab + ac + b^2 + bc}\Big) \\[1em] = 2\Big(\dfrac{ab + b^2 + b^2 + bc}{ab + b^2 + b^2 + bc}\Big) ......\text{from (i)} \\[1em] = 2 = \text{R.H.S.}$

Hence, proved that $\dfrac{a}{x} + \dfrac{c}{y} = 2$.

If a, b and c are in A.P. and also in G.P., show that : a = b = c.

Answer

Since, a, b and c are in A.P.

⇒ 2b = a + c

⇒ b = $\dfrac{a + c}{2}$ .......(i)

Since, a, b and c are also in G.P.

⇒ b² = ac

Substituting value of b from (i) in above equation we get,

$\Rightarrow \Big(\dfrac{a + c}{2}\Big)^2 = ac \\[1em] \Rightarrow \dfrac{a^2 + c^2 + 2ac}{4} = ac \\[1em] \Rightarrow a^2 + c^2 + 2ac = 4ac \\[1em] \Rightarrow a^2 + c^2 - 2ac = 0 \\[1em] \Rightarrow (a - c)^2 = 0\\[1em] \Rightarrow a - c = 0 \\[1em] \Rightarrow a = c .......(ii)$

Substituting value of a from (ii) in (i) we get,

b = $\dfrac{c + c}{2} = \dfrac{2c}{2}$ = c .......(iii)

From (ii) and (iii) we get,

a = b = c.

Hence, proved that a = b = c.