23 Nov 2020 9:12 am

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.8,**

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) 13,16,112, ………….

(v) 1, -5, 25,-125,…

(vi) 120, 60, 30, 18,…

(vii) 16, 4, 1, 14, ……….

Answer:

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 = t2t1=t3t2

x+12x+6=x+15x+12

(x + 12)^{2} = (x + 6) (x + 5)

x^{2} + 24x + 144 = x^{2} + 6x + 15x + 90

24x – 21x = 90 – 144

3x = -54

x = −543 = -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 = 84 = 2

t_{n} = 8192

a . r^{n-1} = 8192 ⇒ 4 × 2^{n-1} = 8192

2^{n-1} = 81924 = 2048

2^{n-1} = 2^{11} ⇒ n – 1 = 11

n = 11 + 1 ⇒ n = 12

Number of terms = 12

(ii) 13, 19, 127, ……………, 12187

Answer:

a = 13 ; r = 19 ÷ 13 = 19 × 31 = 13

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} = 12187

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)

ar8ar5 = 328051215 ⇒ r^{3} = 6561243

= 218781 = 72927 = 2439 = 813

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 = 1215243 = 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 572. Find the three terms.

Answer:

Let the three terms of the G.P. be ar, a, ar

Product of three terms = 27

ar × a × ar = 27

a^{3} = 27 ⇒ a^{3} = 3^{3}

a = 3

Sum of the product of two terms taken at a time is 572

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 = 32 (or) r = 23

∴ The three terms are 2, 3 and 92 or 92, 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

= 5100 × 60000

= ₹ 3000

Starting salary for the 2nd year = 60000 + 3000

= ₹ 63000

Year increase = 5% of 63000

= 5100 × 63000

= ₹ 3150

Starting salary for the 3^{rd} year = 63000 + 3150

= ₹ 66150

60000, 63000, 66150,…. form a G.P.

a = 60000; r = 6300060000 = 6360 = 2120

t_{n} = an^{n-1}

t_{5} = (60000) (2120)^{4}

= 60000 × 2120 × 2120 × 2120 × 2120

= 6×21×21×21×212×2×2×2

= 72930.38

5% increase = 5100 × 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

= 5100 × 20000

= ₹ 1200

Salary for the 2nd year = ₹ 20000 + 1200

= ₹ 21200

Here a = 20,000; r = 2120020000 = 212200 = 106100 = 5350

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

= 3100 × 22000

= ₹ 660

Salary for the 2nd year = ₹ 22000 + ₹ 660

= ₹ 22,660

Here a = 22000; r = 2266022000

= 22662200 = 11331100 = 103100

Salary at the end of 4th year = 22000 × (103100)^{4-1}

= 22000 × (103100)^{3}

= 22000 × 103100 × 103100 × 103100

= 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