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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,

ar^{2} = 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,

ar^{2} = \(\sqrt { 2 }\) × 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,

ar^{2} = 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 t_{7}.

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)

x^{2} + 24x + 144 = x^{2} + 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} = 2^{11} ⇒ 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} = \(\frac { 1 }{ 2187 } \)

t_{9} = 32805 [t_{n} = ar^{n-1}]

a.r^{8} = 32805 …..(1)

6^{th} term = 1215

a.r^{5} = 1215 …..(2)

Divide (1) by (2)

\(\frac{a r^{8}}{a r^{5}}\) = \(\frac { 32805 }{ 1215 } \) ⇒ r^{3} = \(\frac { 6561 }{ 243 } \)

= \(\frac { 2187 }{ 81 } \) = \(\frac { 729 }{ 27 } \) = \(\frac { 243 }{ 9 } \) = \(\frac { 81 }{ 3 } \)

r^{3} = 27 ⇒ r^{3} = 3^{3}

r = 3

Substitute the value of r = 3 in (2)

a. 3^{5} = 1215

a × 243 = 1215

a = \(\frac { 1215 }{ 243 } \) = 5

Here a = 5, r = 3, n = 12

t_{12} = 5 × 3^{(12-1)}

= 5 × 3^{11}

∴ 12^{th} term of a G.P. = 5 × 3^{11}

Question 7.

Find the 10th term of a G.P. whose 8^{th} term is 768 and the common ratio is 2.

Solution:

t_{8} = 768 = ar^{7}

r = 2

t_{10} = ar^{9} = ar^{7} × 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.

t_{2} – t_{1} = t_{3} – t_{2}

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

a^{3} = 27 ⇒ a^{3} = 3^{3}

a = 3

Sum of the product of two terms taken at a time is \(\frac { 57 }{ 2 } \)

6r^{2} – 13r + 6 = 0

6r^{2} – 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}

t_{5} = (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 = xr^{2} 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