10th Maths - Book Back Answers - Chapter 2 Excercise 2.8 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Excercise 2.8 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Excercise 2.8: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Excercise 2.8 – 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 2.8 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 2.8

1. Find the sum of first n terms of the G.P.
(i) 5, -3,$\frac{9}{5},\frac{27}{25}$, ……
(ii) 256,64,16,…….
Answer Key:
(i) 5,-3,$\frac{9}{5},\frac{27}{55}$, ….. n terms

(ii) 256,64,16,…….
Answer Key:
Here a = 256, r =$\frac{64}{256}$ = 14

2. Find the sum of first six terms of the G.P. 5,15,45,…
Answer Key:
Here a = 5, r = $\frac{15}{3}$= 3, n = 6

3. Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Answer Key:
Common ratio (r) = 5

S₆ = 46872

The first term of the G.P. is 12.

 

4. Find the sum to infinity of (i) 9 + 3 + 1 + ….(ii) 21 + 14 +$\frac{28}{3}$……
Answer Key:
(i) 9 + 3 + 1 + ….
a = 9, r = $\frac{3}{9}=\frac{1}{3}$
Sum of infinity term =$\frac{a}{1-r}=\frac{9}{1-\frac{1}{3}}$

5. If the first term of an infinite G.P. is 8 and its sum to infinity is $\frac{32}{3}$ then find the common ratio.
Answer Key:
Here a = 8, S∞ = $\frac{32}{3}$
$\frac{a}{1-r}=\frac{32}{3}$
$\frac{\mathit{8}}{1-r}=\frac{32}{3}$
32 – 32 r = 24 ⇒ 32 r = 8
r = $\frac{8}{32}=\frac{1}{4}$ 
Common ration =$\frac{1}{4}$

 

6. Find the sum to n terms of the series
(i) 0.4 + 0.44 + 0.444 + …… to n terms

Answer Key:

(ii) 3 + 33 + 333 + ………… to n terms
Answer Key:
S_n = 3 + 33 + 333 + …. to n terms
= 3[1 + 11 + 111 + …. to n terms]
= $\frac{3}{9}$[9 + 99 + 999 + …. n terms]
= $\frac{1}{3}$[(10 – 1) + (100 – 1) + (1000 – 1) + …… n terms]
= $\frac{1}{3}$[10 + 100 + 1000 + ….. n terms – (1 + 1 + 1 ….. n terms)] 

($a=10,r=10,s_n=\frac{a(r^n-1)}{r-1}$) 

=$\frac{1}{3}$ 

=$\frac{1}{3}$ 

$s_n=\frac{10}{27}\left({10}^n-1\right)-\frac{n}{3}$

 

7. Find the sum of the Geometric series 3 + 6 + 12 + …….. + 1536
Answer Key:
3 + 6 + 12 …. +1536
a = 3, r = $\frac{6}{3}$ = 2
t_n = 1536

$\mathrm{a}.{\mathrm{r}}^{\mathrm{n}-1}=1536\Rightarrow 3\left(2^{\mathrm{n}-1}\right)=1536$

$2^{\mathrm{n}-1}=\frac{1536}{3}\Rightarrow 2^{\mathrm{n}-1}=512$ 

$2^{\mathrm{n}-1}=2^9$

$n-1=9$ 

$n=9+1=10$ 

$Number of terms=10$

$s_n=\frac{a(r^n-1)}{r-1}$ 

$s_n=\frac{3\left(2^{10}-1\right)}{2-1}=\frac{3(1024-1)}{1}$

$=3\times 1023=3069$    

∴ Sum of the series is 3069

 

8. Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs ₹2 to mail ong letter, find the amount spent on postage when 8th set of letters is mailed.
Answer Key:
When kumar writes a letter to his friend.Friend writes a letter to another person.
It form a G.P
The G.P is 4, 16, 64,………
Here a = 4, r = 4
The last term is 4 (4)^8-1 = 4(4)⁷

$S_8=\frac{4(4^8-1)}{4-1}=\frac{4\left(65536-1\right)}{3}$

^{$\frac{4\times 65535}{3}=87380$}

^{cost to mail the leeter = ₹2(87380)}

                                      ^=₹174760

9. Find the rational form of the number 0.123 .
Answer Key:
Let x =$\overline{0.123}$
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P

Here a=0.123,$\mathrm{r}=\frac{0.000123}{0.123}$=0.001 

$S_n=\frac{a}{1-r}=\frac{0.123}{1-0.001}=\frac{0.123}{0.999}$

$S_n=\frac{41}{333}$ 

 

10. If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ………… n terms then prove that
Answer Key:
S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …….. n
Multiply by x
x S_n = x(x + y) + x(x² + xy + y²) + x(x³ + x²y + xy² + y³) + ……….. n
= x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …… n terms ……(1)
Multiply by y
yS_n = y(x + y) + y(x² + xy + y²) + y(x³ + x²y + xy² + y³) + ….. n
= xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ….. n terms
Subtract (1) and (2)
x S_n – y S_n = x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …….
– xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ……
(x – y) S_n = (x² + x³ + x⁴ + ……) – (y² + y³ + y⁴ + ……)
[ a = x²; r = x and a = y²; r = y, S_n = ]

(x-y)S_n=

Hence it is proved.