OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(c)

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S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(c)

Question 1.
Determine whether the following sequences are geometric progressions or not? If yes then find the common ratio.
(i) 27, 9, 3, 1, …
(ii) -1, 2, 4, 8, …
(iii) 2, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 8 }\), \(\frac { 1 }{ 32 }\), …
(iv) -12, -6, 0, 6, …
Solution:
(i) 27, 9, 3, 1, …
Here, a = 27, r = \(\frac { 9 }{ 27 }\) = \(\frac { 1 }{ 3 }\), \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\), ….
∴ It is a G.P. and r = \(\frac { 1 }{ 3 }\)

(ii) -1, 2, 4, 8, …
Here, a = – 1, r = \(\frac { 2 }{ -1 }\) = – 2, \(\frac { 4 }{ 2 }\) = 2, \(\frac { 4 }{ 2 }\) = 2, \(\frac { 8 }{ 4 }\) = 2
∵ differs
∴ It is not a G.P.

(iii) 2, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 8 }\), \(\frac { 1 }{ 32 }\), …
Here, a = 2
r = \(\frac{1}{2} \div 2,=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
\(\frac{1}{8} \div \frac{1}{2}=\frac{1}{8} \times \frac{2}{1}=\frac{1}{4}\)
\(\frac{1}{32} \div \frac{1}{8}=\frac{1}{32} \times \frac{8}{1}=\frac{1}{4}\) …
∴ It is G.P. and r = \(\frac { 1 }{ 4 }\)

(iv) – 12, – 6, 0, 6, …
a = – 12, r = \(\frac { -6 }{ -12 }\) = \(\frac { 1 }{ 2 }\)
= \(\frac { 0 }{ -6 }\) = 0
∴ It is not G.P.

Question 2.
Write the next three terms in each of the GPs given below:
(i) 2, 6, …
(ii) \(\frac { 1 }{ 16 }\), – \(\frac { 1 }{ 8 }\), …
(iii) 0.3, 0.06, …
Solution:
Next 3 terms of G.P.
(i) 2, 6, … (r = \(\frac { 6 }{ 2 }\) = 3)
2, 6, 18, 54, 162

(ii) \(\frac { 1 }{ 16 }\), – \(\frac { 1 }{ 8 }\), …. (r = \(-\frac{1}{8} \div \frac{1}{16}=-\frac{1}{8} \times \frac{16}{1}\) = – 2)
\(\frac{1}{16},-\frac{1}{8}, \frac{1}{4},-\frac{1}{2}\), 1

(iii) 0.3, 0.06, …
r = \(\frac { 0.06 }{ 0.3 }\) = \(\frac { 0.06 }{ 0.30 }\) = \(\frac { 1 }{ 5 }\) = -.2
0.3, 0.03, 0.012, 0.0024, 0.00048

Question 3.
Find the:
(i) 6th term of the G.P. 2, 10, 50 …
(ii) 11th term of the GP. 4, 12, 36 …
Solution:
(i) 6th term of the G.P. 2, 10, 50 …
Here, a = 2, r = \(\frac { 10 }{ 2 }\) = 5
∴ T₆ = ar^n-1 = 2 x 5^{6 – 1}
= 2 x 55 = 2 x 3125
= 6250

(ii) 11th term of the G.P. 4, 12, 36 …
Here, a = 4, r = \(\frac { 12 }{ 4 }\) = 3
T₁₁ = ar^n-1 = 4 x (3)^11-1 = 4 x 3¹⁰
= 4 x 243 x 243
= 4 x 59049
= 236196

Question 4.
Write the first five terms of the G.P. where nth term is given as:
(i) 4.3^n-1
(ii) \(\frac{5^{n-1}}{2^{n+1}}\)
Solution:
(i) T_n = 4.3^n-1.
∴ T₁ = 4.3¹ = 4.3° = 4 x l = 4
T₂= 4.3² = 4.3¹ = 4 x 3 = 12
T₃ = 4.33³ = 4.3² = 4 x 9 = 36
T₄ = 4.3^4-1 = 4.3³ = 4 x 27 = 108
T₅ = 4.3^5-1 = 4.3⁴ = 4 x 81 = 324
∴ Terms are 4, 12, 36, 108, 324

(ii) T_n = \(\frac{5^{n-1}}{2^{n+1}}\)

∴ 5 terms are , \(\frac{1}{4}, \frac{5}{8}, \frac{25}{16}, \frac{125}{32}, \frac{625}{64}\)

Question 5.
Write down the nth term of each of the following GPs whose first two terms are given as follows. Also find the term stated besides each G.P.
(i) 12, -36, … sixth term
(ii) 3, – \(\frac { 1 }{ 3 }\), …, 8th term
(iii) b²c³, b³c², …, 5th term
Solution:
(i) 12, -36, … sixth term -36
Here, a = 12, r = \(\frac { -36 }{ 12 }\) = – 3
∴ T₆ = ar^n-1 = 12 x (- 3)^6-1
= 12(- 3)⁵ = 12 x (- 243) = – 2916

(ii) 3, – \(\frac { 1 }{ 3 }\), …, 8th term
Here, a = 3, r = – \(\frac { 1 }{ 3 }\) ÷ 3 = – \(\frac { 1 }{ 3 }\) x \(\frac { 1 }{ 3 }\) = – \(\frac { 1 }{ 9 }\)
∴ T₈ = ar^n-1 = 3(-\(\frac { 1 }{ 9 }\))^8-1 = 3(-\(\frac { 1 }{ 9 }\))⁷

(iii) b²c³, b³c², …, 5th term

Question 6.
Which term of the G.P. 27, -18,12, -8, … is \(\frac { 1024 }{ 2187 }\).
Solution:
G.P. is 27,-18, 12, -8, … is \(\frac { 1 }{ 2 }\)
Here, a = 27, r = \(\frac { -18 }{ 27 }\) = \(\frac { -2 }{ 3 }\)
Let \(\frac { 1027 }{ 2187 }\) be the nth term, then

Comparing, we get n – 1 = 10 ⇒ n = 10 + 1 = 11
It is 11th term.

Question 7.
Write the GP. whose 4th term is 54 and the 7th term is 1458.
Solution:
In a G.P.
T₄ = 54, T₇ = 1458
Let a be the first term and r be the common ratio.
∴ T₄ = ar^n-1 = ar³ = 54 … (i)
and T₇ = ar^7-1 = ar⁶ = 1458 … (ii)
Dividing, we get
\(\frac{a r^6}{a r^3}=\frac{1458}{54}\) ⇒ r³ = 27 = (3)³
∴ r = 3
Now, T₄ = ar^4-1
54 = a(3)³
∴ r = 3
54 = 27a
∴ a = \(\frac { 54 }{ 27 }\) = 2
∴ a = 2, r = 3
∴ G.P. will be 2, 6, 18, 54, …

Question 8.
The last term of the G.P. : 3, 3\(\sqrt{3}\) , 9,… is 2187. How many terms in all there in the GP.?
Solution:
In a G.P. 3, 3\(\sqrt{3}\), 9, …
Last term (l) = 2187
Here, a – 3, r = \(\frac{3 \sqrt{3}}{3}\) = \(\sqrt{3}\)

Comparing, n – 1 = 12
B = 12 + 1 = 13
∴ It is 13th term.

Question 9.
Find the value of x + y + z if 1, x, y, z, 16 are in GP.
Solution:
1, x, y, z, 16 are in G.P.
∴ a (first term) = 1
Common ratio (r) = \(\frac { x }{ 1 }\)
T_s = 16
T₅ = ar^{(n – 1)}
= ar^(5-1)
16 = ar⁴
16 = 1(r)⁴
16 = (2)⁴
r⁴ = (2)⁴
By comparing,
r = 2
i.e. the common ratio = 2
Now, r = \(\frac { x }{ 1 }\) = \(\frac { 2 }{ 1 }\)
i.e. x = 2 … (i)
Also, common ratio
\(\frac { 16 }{ z }\) = 2
z = \(\frac { 16 }{ 2 }\)
z = 8 … (ii)
Also, common ratio
\(\frac { z }{ y }\) = 2
\(\frac { 8 }{ y }\) = 2
y = \(\frac { 8 }{ 2 }\)
y = 4 … (iii)
As per condition,
From (i), (ii) and (iii)
⇒ x + y + z = 2 + 8 + 4 = 14

Question 10.
The third term of a G.P. is 18 and its seventh term is 3\(\frac { 5 }{ 9 }\). Find the tenth term of the G.P.
Solution:
In a G.P.
T₃ = 18, T₇ = 3\(\frac { 5 }{ 9 }\) = \(\frac { 32 }{ 9 }\)
Let a be the first term and r be the common ratio, then
T_n = ar^n-1
T₃ = ar^3-1 = ar² = 18 … (i)
T₇ = ar^7-1 = ar⁶ = \(\frac { 50 }{ 9 }\) … (ii)
Dividing, we get

Question 11.
The 5th, 8th and 11th terms of a G.P. are P, Q and S respectively. Show that Q² = PS.
Solution:
In a G.P.
T₅ = P, T₈ = Q, T₁₁ = S
To prove : Q² = PS
Let a be the first term and r be the common ratio
∴ T₅ = ar^5-1 = ar⁴ = P
T₈ = ar^8-1 = ar⁷ = Q
T₁₁ = ar^11-1= ar¹⁰ = S
Q² = (ar⁷)² = a²r¹⁴
and P x S = ar⁴ x ar¹⁰ = a²r⁴⁺¹⁰ = a²r¹⁴
Hence, Q² = P x 5