# ICSE Solutions for Chapter 11 Geometric Progression Class 10 Mathematics

**Question 1:**

**Find, which of the following sequence form a G.P. :**

**(i) 8, 24, 72, 216, …**

**(ii) 1/8, 1/24, 1/72, 1/216, ...**

**(iii) 9, 12, 16, 24, …**

**Solution 1:**

(i) Given sequence: 8, 24, 72, 216 ....

Now,

24/8 = 3, 72/24 = 3, 216/72 = 3

Since 24/8 = 72/24 = 216/72 = .... = 3, the given sequence is a G.P.

With common ratio 3.

(ii) Given sequence: 1/8, 1/24, 1/72, 1/216 ….

Now,

(1/24)/(1/8) = 1/3, (1/72)/(1/24) = 1/3, (1/216)/(1/72) = 1/3

Since (1/24)/(1/8) = (1/72)/(1/24) = (1/216)/(1/72) = ….1/3, the given sequence is a G.P.

With common ratio 1/3.

(iii) Given sequence: 9, 12, 16, 24

Now,

12/9 = 4/3, 16/12 = 4/3, 24/16 = 3/2

Since 24/8 = 72/24 ≠ 216/72, the given sequence is not a G.P.

**Question 2:**

**Find the 9th term of the series :**

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

**Solution 2:**

Given sequence: 1, 4, 16, 64……

Now,

4/1 = 4, 16/4 = 4, 64/16 = 4

Since 4/1 = 16/4 = 64/16 = .….= 4, the given sequence is a G.P.

With first term, a = 1 and common ratio, r = 4

Now, t

_{n}= ar

^{n - 1}

⇒ t

**= 1×48 = 65536**

_{9}

**Question 3:**

**Find the seventh term of the G.P. :**

**1, √3, 3, 3√**3

**…**

**Solution 3:**

Given G.P.: 1, √3, 3, 3√3,…

Here,

First term, a = 1

Common ratio, r = √3/1 = √3

Now, t

_{n}= ar

^{n - 1}

⇒ t

_{7}= 1×(√3)

^{6}= 27

**Question 4:**

**Find the 8**

^{th}

**term of the sequence :**

**3/4, 1 1/2, 3, …….**

**Solution 4:**

Given sequence: 3/4., 1½, 3, ……

i.e. 3/4, 3/2, 3, ..…

Now,

(3/2)/(3/4) = 2, 3/(3/2) = 2,

Since (3/2)/(3/4) = 3/(3/2) = …….= 2, the given sequence is a G.P.

With first term, a = ¾ and common ratio, r = 2

Now, t

_{n}= ar

^{n - 1}

⇒ t

_{8}= ¾ × 2

^{7}= ¾ × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 3 × 2

^{5}= 96

**Question 5:**

**Find the 10**

^{th}

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

**Solution 5:**

Given G.P.: 12, 4, 1 1/3,…….

Here,

First term, a = 12

Common ratio, r = 4/12 = 1/3

Now, t

_{n}= ar

^{n - 1}

⇒ t

_{10}= 12 × (1/3)

^{9}= 12 × 1/19683 = 4/6561

**Question 6:**

**Find the n**

^{th}term of the series :

**Solution 6:**

Given series: 1, 2, 4, 8, ……

Now,

2/1 = 2, 4/2 = 2, 8/4 = 2

Since 2/1 = 4/2 = 8/4 = ….. = 2, the given sequence is a G.P.

With first term, a = 1 and common ratio, r = 2.

Now, t

_{n}= ar

^{n - 1}

⇒ t

_{n}= 1 × 2

^{n - 1}= 2

^{n - 1}

**Question 7:**

**Find the next three terms of the sequence :**

**√5, 5, 5√5, …**

**Solution 7:**

Given sequence: √5, 5, 5√5,……

Now,

5/√5 = √5, (5√5)/5 = √5

Since 5/√5 = (5√5)/52 = ……. = √5, the given sequence is a G.P.

With first term, a = √5 and common ratio, r = √5.

Now, t

_{n}= ar

^{n - 1}

∴ Next three terms :

4

^{th}term = √5×(√5)

^{3}= √5×5√5 = 25

5

^{th}term = √5×(√5)

^{4}= √5×25 = 25√5

6

^{th}term = √5×(√5)

^{5}= √5×25√5 = 125

**Question 8:**

**Find the sixth term of the series :**

**2**

^{2}

**, 2**

^{3}

**, 2**

^{4}

**, …**

**Solution 8:**

Given sequence : 2

^{2}, 2

^{3}, 2

^{4},…..

Now, 2

^{3}/2

^{2}= 2, 2

^{4}/2

^{3}= 2

Since 2

^{3}/2

^{2}= 2

^{4}/2

^{3}= ……= 2, the given sequence is a G.P.

With first term, a = 2

^{2}= 4 and common ratio, r = 2.

Now, t

_{n}= ar

^{n - 1}

∴ t

_{6}= 4 × (2)

^{5}= 4×32 = 128

**Question 9:**

Find the seventh term of the G.P. :

3 + 1, 1, (√3 – 1)/2, …

Find the seventh term of the G.P. :

3 + 1, 1, (√3 – 1)/2, …

**Solution 9:**

Given G.P.: √3 + 1, 1, (√3 – 1)/2, …

Here,

First term, a = √3 + 1

Common ratio, r = 1/(√3 + 1)

Now, t

_{n}= ar

^{n - 1}

⇒ t

_{7}= {(√3 + 1)×(1/(√3 + 1)}

^{6}

= (1/(√3 + 1)

^{5}

= {1/(√3 + 1) × (√3 – 1)/(√3 – 1)}

^{5}

= {(√3 – 1)/2}

^{5}

= 1/32(√3 – 1)

^{5}

**Question 10:**

**Find the G.P. whose first term is 64 and next term is 32.**

**Solution 10:**

First term , a = 64

Second term, t

_{2}= 32

⇒ ar = 32

⇒ 64 × r = 32

⇒ r = 32/64 = ½

∴ Required G.P. = a, ar, ar

^{n-1}, ar

^{n-2}, …

= 64, 32, 64 × (1/2)

^{2}, 64 × (1/2)

^{3},…

= 64, 32, 16, 8, ……