10th Maths - Book Back Answers - Chapter 2 Excercise 2.7 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Excercise 2.7 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Excercise 2.7: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Excercise 2.7 – Maths from the Tamil Nadu State Board 10th Standard Maths textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 2 Excercise 2.7 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 2 Numbers and Sequences Ex 2.7

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 Key:

 

 

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 Key:
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 Key:
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 = 25
Answer Key:
ar = 1000 × $\frac{2}{5}$= 400,
ar² = 1000 × $\frac{4}{25}$= 40 × 4 = 160
The three terms are 1000,400 and 160.

 

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

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

 

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

(i) 4,8,16,…,8192?
Answer Key:
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 Key:
$a=\frac{1}{3}; r=\frac{1}{9}÷\frac{1}{3}=\frac{1}{9}\times \frac{3}{1}=\frac{1}{3}$

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

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

 

6. In a G.P. the 9^th term is 32805 and 6^th term is 1215. Find the 12^th term.
Answer Key:
Given, 9^th term = 32805
a. r^n-1 =
t₉ = 32805 [t_n = ar^n-1]
a.r⁸ = 32805 …..(1)
6^th term = 1215
a.r⁵ = 1215 …..(2)
Divide (1) by (2)
$\frac{a{r}^8}{a{r}^5}=\frac{32805}{1215}\Rightarrow r^3=\frac{6562}{243}$
$\frac{2187}{81}=\frac{729}{27}=\frac{243}{9}=\frac{81}{3}$
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

 

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

 

8. If a, b, c are in A.P. then show that 3^a, 3^b, 3^c are in G.P.
Answer Key:
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.

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 Key:
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 92 or 92, 3 and 2

 

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 Key:
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 × $\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}$
= $\frac{6\times 21\times 21\times 21\times 21}{2\times 2\times 2\times 2}$
= 72930.38
5% increase = $\frac{5}{100}$ × 72930.38
= ₹ 3646.51
Salary after 5 years = ₹ 72930.38 + 3646.51
= ₹ 76576.90

= ₹ 76577

 

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 Key:
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 = $\frac{22660}{22000}$
= $\frac{22660}{22000}=\frac{1133}{1100}=\frac{103}{100}$
Salary at the end of 4th year = 22000 × ($\frac{3}{100}$)^4-1
= 22000 × ($\frac{3}{100}$)³
= 22000 ×$\frac{103}{100}\times \frac{103}{100}\times \frac{103}{100}$

=$\frac{22\times 103\times 103\times 103}{1000}$

=$\frac{11\times 103\times 103\times 103}{500}$ 

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

 

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 Key:
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