You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## 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, \(\frac { 1 }{ 4 } \), ……

Solution:

(i) 3, 9, 27, 81

r = Common ratio

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

(ii) a = \(\sqrt { 2 }\) , r = \(\sqrt { 2 }\)

(iii) a = 1000, r = \(\frac { 2 }{ 5 } \)

Solution:

(i) a = 6, r = 3

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

t_{1} = ar^{1-1} = ar^{0 }= a = 6

t_{2} = ar^{2-1} = ar^{1} = 6 × 3 = 18

t_{3} = ar^{3-1} = ar^{2} = 6 × 3^{2} = 54

∴ The 3 terms are 6, 18, 54, ….

The 3 terms are 1000, 400, 160, ……………

Question 3.

In a G.P. 729, 243, 81,… find t_{7}.

Solution:

G.P = 729, 243, 81 ……

t_{7} = ?

Question 4.

Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.

Answer:

\(\frac{t_{2}}{t_{1}}\) = \(\frac { x+12 }{ x+6 } \),\(\frac{t_{3}}{t_{2}}\) = \(\frac { x+15 }{ x+12 } \)

Since it is a G.P.

\(\frac { x+12 }{ x+6 } \) = \(\frac { x+15 }{ x+12 } \)

(x + 12)^{2} = (x + 6) (x + 15)

x^{2} + 24x + 144 = x^{2} + 21x + 90

3x = -54 ⇒ x = \(\frac { -54 }{ 3 } \) = -18

Question 5.

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

(i),4, 8, 16, …, 8192

(ii) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 9 } \),\(\frac { 1 }{ 27 } \),……\(\frac { 1 }{ 2187 } \)

Solution:

(i) 4, 8, 16, …… 8192

Question 6.

In a G.P. the 9^{th} term is 32805 and 6^{th} term is 1215. Find the 12th term.

Solution:

In a G.P

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

t_{9} = 32805

t_{6} = 1215

t_{12} = ?

Let

t_{9} = ar^{8} = 32805 ………(1)

t_{6} = ar^{5} = 1215 ………. (2)

Question 7.

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

Answer:

Here r = 2, t_{8} = 768

t_{8} = 768 (t_{n} = ar^{n-1})

a. r^{8-1} = 768

ar^{7} = 768 …..(1)

10^{th} term of a G.P. = a.r 10^{-1}

= ar^{9}

= (ar^{7}) × (r^{2})

= 768 × 2^{2} (from 1)

= 768 × 4 = 3072

∴ 10^{th} term of a G.P. = 3072

Question 8.

If a, b, c are in A.P. then show that 3^{a},3^{b},3^{c} are in G.P.

Solution:

If a, b, c are in A.P

t_{2} – t_{1} = t_{3} – t_{2}

b – a = c – b

2b = c + a

To prove that 3^{a}, 3^{b}, 3^{c} are in G.P

⇒ 3^{2b} = 3^{c+a} + a [Raising the power both sides]

⇒ 3^{b}.3^{b} = 3^{c}.3^{a}

\(\Rightarrow \frac{3^{b}}{3^{a}}=\frac{3^{c}}{3^{b}}\)

\(\Rightarrow \frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{1}}\)

⇒ Common ratio is same for 3^{a},3^{b},3^{c}

⇒ 3^{a}, 3^{b}, 3^{c} forms a G.P

∴ Hence it is proved .

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.

Solution:

Let the three consecutive terms in a G.P are \(\frac { a }{ r } \), a, ar.

Their Product = \(\frac { a }{ r } \) × a × ar = 27

a^{3} = 27 = 3^{3}

a = 3

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

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?

Solution:

Starting salary = ₹ 60,000

Increase per year = 5%

∴ At the end of 1 year the increase

= 60,0,00 × \(\frac { 5 }{ 100 } \)

₹ 3000

∴ At the end of first year his salary

= ₹ 60,000 + 3000

I year salary = ₹ 63,000

II Year increase = 63000 × \(\frac { 5 }{ 100 } \)

At the end of II year, salary

= 63000 + 3150

= ₹ 66150

III Year increase = 66150 × \(\frac { 5 }{ 100 } \)

= 3307.50

At the end of III year, salary = 66150 + 3307.50

= ₹ 69457.50

IV year increase = 69457.50 × \(\frac { 5 }{ 100 } \)

= ₹ 3472.87

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 3% 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 4th year with respect to the offers A and B?

Solution:

Offer A

Starting salary ₹ 20,000

Annual increase 6%

At the end of

III year ,salary = 22472 + 1348 = 23820

∴ IV year salary = ₹ 23820

Offer B

Starting salary = ₹ 22,000

Salary as per Option A = ₹ 23820

Salary as per Option B = ₹ 24040

∴ Option B is better.

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.

Solution:

a, b, c are three consecutive terms of an AP.

∴ Let a, b, c be a, a + d, a + 2d respectively ………… (1)

x, y, z are three consecutive terms of a GP.

∴ Assume x, y, z as x, x.r, x.r^{2} respectively ……… (2)

PT : x^{b-c} , y^{c-a} , z^{a-b} = 1

Substituting (1) and (2) in LHS, we get

LHS = x^{a+d-a-2d} × (xr)^{a+2d-a} × (xr^{2})^{a-a-d}

= (x)^{-d} . (xr)^{2d} (xr^{2})^{-d}

= \(\frac{1}{x^{d}}\) × x^{2d} . r^{2d} × \(\frac{1}{x^{d} r^{2 d}}\) = 1 = RHS