This exercise mainly deals with problems based on geometric progression (G.P.) and the general term of a G.P. For students who face difficulty in understanding the concepts during class hours, the subject experts at BYJUâ€™S have formulated the solutions in a simple language based on the studentâ€™s ability. Students can practice the examples and can solve exercise wise problems at ease, which boosts their confidence level before appearing for the board exam. RD Sharma Class 11 Maths Solutions is readily available in the pdf format for students to download easily and start practising offline.

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**1. Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:**

**(i) 4, -2, 1, -1/2, â€¦. **

**(ii) -2/3, -6, -54, â€¦.**

**(iii) a, 3a ^{2}/4, 9a^{3}/16, â€¦.**

**(iv) Â½, 1/3, 2/9, 4/27, â€¦**

**Solution:**

**(i) **4, -2, 1, -1/2, â€¦.

Let a = 4, b = -2, c = 1

In GP,

b^{2 }= ac

(-2)^{2}Â = 4(1)

4 = 4

So, the Common ratio = r = -2/4 = -1/2

**(ii) **-2/3, -6, -54, â€¦.

Let a = -2/3, b = -6, c = -54

In GP,

b^{2 }= ac

(-6)^{2} = -2/3 Ã— (-54)

36 = 36

So, the Common ratio = r = -6/(-2/3) = -6Â Ã— 3/-2 = 9

**(iii) **a, 3a^{2}/4, 9a^{3}/16, â€¦.

Let a = a, b = 3a^{2}/4, c = 9a^{3}/16

In GP,

b^{2 }= ac

(3a^{2}/4)^{2} = 9a^{3}/16 Ã— a

9a^{4}/4 = 9a^{4}/16

So, the Common ratio = r = (3a^{2}/4)/a = 3a^{2}/4a = 3a/4

**(iv) **Â½, 1/3, 2/9, 4/27, â€¦

Let a = 1/2, b = 1/3, c = 2/9

In GP,

b^{2 }= ac

(1/3)^{2} = 1/2 Ã— (2/9)

1/9 = 1/9

So, the Common ratio = r = (1/3)/(1/2) = (1/3)Â Ã— 2 = 2/3

**2. Show that the sequence defined by a _{n} = 2/3^{n}, nÂ âˆˆÂ N is a G.P.**

**Solution:**

Given:

a_{n} = 2/3^{n}

Let us consider n = 1, 2, 3, 4, â€¦ since n is a natural number.

So,

a_{1} = 2/3

a_{2} = 2/3^{2} = 2/9

a_{3} = 2/3^{3} = 2/27

a_{4} = 2/3^{4} = 2/81

In GP,

a_{3}/a_{2} = (2/27) / (2/9)

= 2/27 Ã— 9/2

= 1/3

a_{2}/a_{1} = (2/9) / (2/3)

= 2/9 Ã— 3/2

= 1/3

âˆ´ Common ratio of consecutive term is 1/3. Hence nÂ âˆˆÂ N is a G.P.

**3. Find:**

**(i) the ninth term of the G.P. 1, 4, 16, 64, â€¦.**

**(ii) the 10 ^{th} term of the G.P. -3/4, Â½, -1/3, 2/9, â€¦.**

**(iii) the 8 ^{th}Â term of the G.P. 0.3, 0.06, 0.012, â€¦.**

**(iv) the 12 ^{th} term of the G.P. 1/a^{3}x^{3} , ax, a^{5}x^{5}, â€¦.**

**(v) nth term of the G.P. âˆš3, 1/âˆš3, 1/3âˆš3, â€¦**

**(vi) the 10 ^{th} term of the G.P. âˆš2, 1/âˆš2, 1/2âˆš2, â€¦.**

**Solution:**

**(i) **the ninth term of the G.P. 1, 4, 16, 64, â€¦.

We know that,

t_{1} = a = 1, r = t_{2}/t_{1} = 4/1 = 4

By using the formula,

T_{n} = ar^{n-1}

T_{9} = 1 (4)^{9-1}

= 1 (4)^{8}

= 4^{8}

**(ii) **the 10^{th} term of the G.P. -3/4, Â½, -1/3, 2/9, â€¦.

We know that,

t_{1} = a = -3/4, r = t_{2}/t_{1} = (1/2) / (-3/4) = Â½ Ã— -4/3 = -2/3

By using the formula,

T_{n} = ar^{n-1}

T_{10} = -3/4 (-2/3)^{10-1}

= -3/4 (-2/3)^{9}

= Â½ (2/3)^{8}

**(iii) **the 8^{th}Â term of the G.P., 0.3, 0.06, 0.012, â€¦.

We know that,

t_{1} = a = 0.3, r = t_{2}/t_{1} = 0.06/0.3 = 0.2

By using the formula,

T_{n} = ar^{n-1}

T_{8} = 0.3 (0.2)^{8-1}

= 0.3 (0.2)^{7}

**(iv) **the 12^{th} term of the G.P. 1/a^{3}x^{3} , ax, a^{5}x^{5}, â€¦.

We know that,

t_{1} = a = 1/a^{3}x^{3}, r = t_{2}/t_{1} = ax/(1/a^{3}x^{3}) = ax (a^{3}x^{3}) = a^{4}x^{4}

By using the formula,

T_{n} = ar^{n-1}

T_{12} = 1/a^{3}x^{3} (a^{4}x^{4})^{12-1}

= 1/a^{3}x^{3} (a^{4}x^{4})^{11}

= (ax)^{41}

**(v) **nth term of the G.P. âˆš3, 1/âˆš3, 1/3âˆš3, â€¦

We know that,

t_{1} = a = âˆš3, r = t_{2}/t_{1} = (1/âˆš3)/âˆš3 = 1/(âˆš3Ã—âˆš3) = 1/3

By using the formula,

T_{n} = ar^{n-1}

T_{n} = âˆš3 (1/3)^{n-1}

**(vi) **the 10^{th} term of the G.P. âˆš2, 1/âˆš2, 1/2âˆš2, â€¦.

We know that,

t_{1} = a = âˆš2, r = t_{2}/t_{1} = (1/âˆš2)/âˆš2 = 1/(âˆš2Ã—âˆš2) = 1/2

By using the formula,

T_{n} = ar^{n-1}

T_{10} = âˆš2 (1/2)^{10-1}

= âˆš2 (1/2)^{9}

= 1/âˆš2 (1/2)^{8}

**4. Find the 4 ^{th} term from the end of the G.P. 2/27, 2/9, 2/3, â€¦., 162.**

**Solution:**

The nth term from the end is given by:

a_{n} = l (1/r)^{n-1} where, l is the last term, r is the common ratio, n is the nth term

Given: last term, l = 162

r = t_{2}/t_{1} = (2/9) / (2/27)

= 2/9 Ã— 27/2

= 3

n = 4

So, a_{n} = l (1/r)^{n-1}

a_{4} = 162 (1/3)^{4-1}

= 162 (1/3)^{3}

= 162 Ã— 1/27

= 6

âˆ´Â 4^{th}Â term from last is 6.

**5. Which term of the progression 0.004, 0.02, 0.1, â€¦. is 12.5?**

**Solution:**

By using the formula,

T_{n} = ar^{n-1}

Given:

a = 0.004

r = t_{2}/t_{1} = (0.02/0.004)

= 5

T_{n} = 12.5

n = ?

So, T_{n} = ar^{n-1}

12.5 = (0.004) (5)^{n-1}

12.5/0.004 = 5^{n-1}

3000 = 5^{n-1}

5^{5} = 5^{n-1}

5 = n-1

n = 5 + 1

= 6

âˆ´Â 6^{th}Â term of the progression 0.004, 0.02, 0.1, â€¦. is 12.5.

**6. Which term of the G.P.:**

**(i) âˆš2, 1/âˆš2, 1/2âˆš2, 1/4âˆš2, â€¦ is 1/512âˆš2 ?**

**(ii) 2, 2âˆš2, 4, â€¦ is 128 ?**

**(iii) âˆš3, 3, 3âˆš3, â€¦ is 729 ?**

**(iv) 1/3, 1/9, 1/27â€¦ is 1/19683 ?**

**Solution:**

**(i) **âˆš2, 1/âˆš2, 1/2âˆš2, 1/4âˆš2, â€¦ is 1/512âˆš2 ?

By using the formula,

T_{n} = ar^{n-1}

a = âˆš2

r = t_{2}/t_{1} = (1/âˆš2) / (âˆš2)

= 1/2

T_{n} = 1/512âˆš2

n = ?

T_{n} = ar^{n-1}

1/512âˆš2 = (âˆš2) (1/2)^{n-1}

1/512âˆš2Ã—âˆš2 = (1/2)^{n-1}

1/512Ã—2 = (1/2)^{n-1}

1/1024 = (1/2)^{n-1}

(1/2)^{10} = (1/2)^{n-1}

10 = n â€“ 1

n = 10 + 1

= 11

âˆ´Â 11^{th}Â term of the G.P is 1/512âˆš2

**(ii) **2, 2âˆš2, 4, â€¦ is 128 ?

By using the formula,

T_{n} = ar^{n-1}

a = 2

r = t_{2}/t_{1} = (2âˆš2/2)

= âˆš2

T_{n} = 128

n = ?

T_{n} = ar^{n-1}

128 = 2 (âˆš2)^{n-1}

128/2 = (âˆš2)^{n-1 }

64 = (âˆš2)^{n-1 }

2^{6} = (âˆš2)^{n-1}

12 = n â€“ 1

n = 12 + 1

= 13

âˆ´Â 13^{th}Â term of the G.P is 128

**(iii) **âˆš3, 3, 3âˆš3, â€¦ is 729 ?

By using the formula,

T_{n} = ar^{n-1}

a = âˆš3

r = t_{2}/t_{1} = (3/âˆš3)

= âˆš3

T_{n} = 729

n = ?

T_{n} = ar^{n-1}

729 = âˆš3 (âˆš3)^{n-1}

729 = (âˆš3)^{n}

3^{6} = (âˆš3)^{n}

(âˆš3)^{12} = (âˆš3)^{n}

n = 12

âˆ´Â 12^{th}Â term of the G.P is 729

**(iv) **1/3, 1/9, 1/27â€¦ is 1/19683 ?

By using the formula,

T_{n} = ar^{n-1}

a = 1/3

r = t_{2}/t_{1} = (1/9) / (1/3)

= 1/9 Ã— 3/1

= 1/3

T_{n} = 1/19683

n = ?

T_{n} = ar^{n-1}

1/19683 = (1/3) (1/3)^{n-1}

1/19683 = (1/3)^{n}

(1/3)^{9} = (1/3)^{n}

n = 9

âˆ´Â 9^{th}Â term of the G.P is 1/19683

**7. Which term of the progression 18, -12, 8, â€¦ is 512/729 ?**

**Solution:**

By using the formula,

T_{n} = ar^{n-1}

a = 18

r = t_{2}/t_{1} = (-12/18)

= -2/3

T_{n} = 512/729

n = ?

T_{n} = ar^{n-1}

512/729 = 18 (-2/3)^{n-1}

2^{9}/(729 Ã— 18) = (-2/3)^{n-1}

2^{9}/36 Ã— 1/2Ã—3^{2} = (-2/3)^{n-1}

(2/3)^{8} = (-1)^{n-1} (2/3)^{n-1}

8 = n â€“ 1

n = 8 + 1

= 9

âˆ´Â 9^{th}Â term of the Progression is 512/729

**8. Find the 4th term from the end of the G.P. Â½, 1/6, 1/18, 1/54, â€¦ , 1/4374**

**Solution:**

The nth term from the end is given by:

a_{n} = l (1/r)^{n-1} where, l is the last term, r is the common ratio, n is the nth term

Given: last term, l = 1/4374

r = t_{2}/t_{1} = (1/6) / (1/2)

= 1/6 Ã— 2/1

= 1/3

n = 4

So, a_{n} = l (1/r)^{n-1}

a_{4} = 1/4374 (1/(1/3))^{4-1}

= 1/4374 (3/1)^{3}

= 1/4374 Ã— 3^{3}

= 1/4374 Ã— 27

= 1/162

âˆ´Â 4^{th}Â term from last is 1/162.