Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.7

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.7

Question 1.
Which of the following sequences are in G.P?
(i) 3,9,27,81,…
(ii) 4,44,444,4444,…
(iii) 0.5,0.05,0.005,
(iv) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \), ………….
(v) 1, -5, 25,-125,…
(vi) 120, 60, 30, 18,…
(vii) 16, 4, 1, \(\frac { 1 }{ 4 } \), ……….
Answer:

Question 2.
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
Answer:
a = 6, r = 3
ar = 6 × 3 = 18,
ar² = 6 × 9 = 54
The three terms are 6, 18 and 54

(ii) a = \(\sqrt { 2 }\), r = \(\sqrt { 2 }\).
Answer:
ar = \(\sqrt { 2 }\) × \(\sqrt { 2 }\) = 2,
ar² = \(\sqrt { 2 }\) × 2 = 2 \(\sqrt { 2 }\)
The three terms are \(\sqrt { 2 }\), 2 and 2\(\sqrt { 2 }\)

(iii) a = 1000, r = \(\frac { 2 }{ 5 } \)
Answer:
ar = 1000 × \(\frac { 2 }{ 5 } \) = 400,
ar² = 1000 × \(\frac { 4 }{ 25 } \) = 40 × 4 = 160
The three terms are 1000,400 and 160.

Question 3.
In a G.P. 729, 243, 81,… find t₇.
Answer:
The G.P. is 729, 243, 81,….

Question 4.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression
Solution:
G.P = x + 6, x + 12, x + 15
In G.P r = \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}\)
\(\frac{x+12}{x+6}=\frac{x+15}{x+12}\)
(x + 12)² = (x + 6) (x + 5)
x² + 24x + 144 = x² + 6x + 15x + 90
24x – 21x = 90 – 144
3x = -54
x = \(\frac { -54 }{ 3 } \) = -18
x = -18

Question 5.
Find the number of terms in the following G.P.

(i) 4,8,16,…,8192?
Answer:
Here a = 4; r = \(\frac { 8 }{ 4 } \) = 2
t_n = 8192

a . r^n-1 = 8192 ⇒ 4 × 2^n-1 = 8192
2^n-1 = \(\frac { 8192 }{ 4 } \) = 2048
2^n-1 = 2¹¹ ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12

(ii) \(\frac { 1 }{ 3 } \), \(\frac { 1 }{ 9 } \), \(\frac { 1 }{ 27 } \), ……………, \(\frac { 1 }{ 2187 } \)
Answer:
a = \(\frac { 1 }{ 3 } \) ; r = \(\frac { 1 }{ 9 } \) ÷ \(\frac { 1 }{ 3 } \) = \(\frac { 1 }{ 9 } \) × \(\frac { 3 }{ 1 } \) = \(\frac { 1 }{ 3 } \)

t_n = \(\frac { 1 }{ 2187 } \)
a. r^n-1 = \(\frac { 1 }{ 2187 } \)

n – 1 = 6 ⇒ n = 6 + 1 = 7
Number of terms = 7

Question 6.
In a G.P. the 9^th term is 32805 and 6^th term is 1215. Find the 12^th term.
Answer:
Given, 9^th term = 32805
a. r^n-1 = \(\frac { 1 }{ 2187 } \)
t₉ = 32805 [t_n = ar^n-1]
a.r⁸ = 32805 …..(1)
6^th term = 1215
a.r⁵ = 1215 …..(2)
Divide (1) by (2)
\(\frac{a r^{8}}{a r^{5}}\) = \(\frac { 32805 }{ 1215 } \) ⇒ r³ = \(\frac { 6561 }{ 243 } \)
= \(\frac { 2187 }{ 81 } \) = \(\frac { 729 }{ 27 } \) = \(\frac { 243 }{ 9 } \) = \(\frac { 81 }{ 3 } \)
r³ = 27 ⇒ r³ = 3³
r = 3
Substitute the value of r = 3 in (2)
a. 3⁵ = 1215
a × 243 = 1215
a = \(\frac { 1215 }{ 243 } \) = 5
Here a = 5, r = 3, n = 12
t₁₂ = 5 × 3^(12-1)
= 5 × 3¹¹
∴ 12^th term of a G.P. = 5 × 3¹¹

Question 7.
Find the 10th term of a G.P. whose 8^th term is 768 and the common ratio is 2.
Solution:
t₈ = 768 = ar⁷
r = 2
t₁₀ = ar⁹ = ar⁷ × r × r
= 768 × 2 × 2 = 3072

Question 8.
If a, b, c are in A.P. then show that 3^a, 3^b, 3^c are in G.P.
Answer:
a, b, c are in A.P.
t₂ – t₁ = t₃ – t₂
b – a = c – b
2b = a + c …..(1)
3^a, 3^b, 3^c are in G.P.

From (1) and (2) we get
3^a, 3^b, 3^c are in G.P.

Question 9.
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is \(\frac { 57 }{ 2 } \). Find the three terms.
Answer:
Let the three terms of the G.P. be \(\frac { a }{ r } \), a, ar
Product of three terms = 27
\(\frac { a }{ r } \) × a × ar = 27
a³ = 27 ⇒ a³ = 3³
a = 3
Sum of the product of two terms taken at a time is \(\frac { 57 }{ 2 } \)

6r² – 13r + 6 = 0
6r² – 9r – 4r + 6 = 0
3r (2r – 3) -2(2r – 3) = 0
(2r – 3) (3r – 2) = 0
2r – 3 = 0 or 3r – 2 = 0
2r = 3 (or) 3r – 2 = 0
r = \(\frac { 3 }{ 2 } \) (or) r = \(\frac { 2 }{ 3 } \)

∴ The three terms are 2, 3 and \(\frac { 9 }{ 2 } \) or \(\frac { 9 }{ 2 } \), 3 and 2

Question 10.
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Answer:
Starting salary (a) = ₹ 60000
Increased salary = 5% of starting salary
= \(\frac { 5 }{ 100 } \) × 60000
= ₹ 3000
Starting salary for the 2nd year = 60000 + 3000
= ₹ 63000
Year increase = 5% of 63000
= \(\frac { 5 }{ 100 } \) × 63000
= ₹ 3150
Starting salary for the 3^rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = \(\frac { 63000 }{ 60000 } \) = \(\frac { 63 }{ 60 } \) = \(\frac { 21 }{ 20 } \)
t_n = an^n-1
t₅ = (60000) (\(\frac { 21 }{ 20 } \))⁴
= 60000 × \(\frac { 21 }{ 20 } \) × \(\frac { 21 }{ 20 } \) × \(\frac { 21 }{ 20 } \) × \(\frac { 21 }{ 20 } \)
= \(\frac{6 \times 21 \times 21 \times 21 \times 21}{2 \times 2 \times 2 \times 2}\)
= 72930.38
5% increase = \(\frac { 5 }{ 100 } \) × 72930.38
= ₹ 3646.51
Salary after 5 years = ₹ 72930.38 + 3646.51
= ₹ 76576.90
= ₹ 76577

Question 11.
Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.
Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4^th year with respect to the offers A and B?
Answer:
Starting salary (a) = ₹ 20,000
Annual increase = 6% of 20000
= \(\frac { 5 }{ 100 } \) × 20000
= ₹ 1200
Salary for the 2nd year = ₹ 20000 + 1200
= ₹ 21200
Here a = 20,000; r = \(\frac { 21200 }{ 20000 } \) = \(\frac { 212 }{ 200 } \) = \(\frac { 106 }{ 100 } \) = \(\frac { 53 }{ 50 } \)
n = 4 years
t_n = ar^n-1

Salary at the end of 4^th year = 23820

For B
Starting salary = ₹ 22000
(a) = 22000
Annual increase = 3% of 22000
= \(\frac { 3 }{ 100 } \) × 22000
= ₹ 660
Salary for the 2nd year = ₹ 22000 + ₹ 660
= ₹ 22,660
Here a = 22000; r = \(\frac { 22660 }{ 22000 } \)
= \(\frac { 2266 }{ 2200 } \) = \(\frac { 1133 }{ 1100 } \) = \(\frac { 103 }{ 100 } \)
Salary at the end of 4th year = 22000 × (\(\frac { 103 }{ 100 } \))^4-1
= 22000 × (\(\frac { 103 }{ 100 } \))³
= 22000 × \(\frac { 103 }{ 100 } \) × \(\frac { 103 }{ 100 } \) × \(\frac { 103 }{ 100 } \)

= 24039.99 = 24040
4^th year Salary for A = ₹ 23820 and 4^th year Salary for B = ₹ 24040

Question 12.
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x^b-c × y^c-a × z^a-b = 1
Answer:
a, b, c are three consecutive terms of an A.P
∴ a = a, b = a + dand c = a + 2d respectively ….(1)
x, y, z are three consecutive terms of a G.P
∴ x = x, y = xr, z = xr² respective ……(2)
L.H.S = x^b-c × y^c-a × z^a-b ( Substitute the values from 1 and 2 we get)

L.H.S = R.H.S
Hence it is proved