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

**10th Maths Exercise 2.7 Samacheer Kalvi 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

**10th Maths Exercise 2.7 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, ……………

**Ex 2.7 Class 10 Samacheer Question 3.**

In a G.P. 729, 243, 81,… find t_{7}.

Solution:

G.P = 729, 243, 81 ……

t_{7} = ?

**Ex 2.7 Class 10 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

**Exercise 2.7 Class 10 Maths 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

**10th Maths 2.7 Exercise 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)

**10th Math Exercise 2.7 Solution 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

**Class 10 Maths Exercise 2.7 Solutions 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 .

**Exercise 2.7 Class 10 Maths Solution 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 \(\frac { 57 }{ 2 } \)

**2.7 Exercise Class 10 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

**Exercise 2.7 Class 10 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.

**10 Class Math Exercise 2.7 Solution Pdf 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