10th Maths - Book Back Answers - Chapter 2 Excercise 2.9 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Excercise 2.9 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Excercise 2.9: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Excercise 2.9 – 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 2.9 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 2.9

1. Find the sum of the following series
(i) 1 + 2 + 3 + …….. + 60
(ii) 3 + 6 + 9 + …….. +96
(iii) 51 + 52 + 53 + …….. + 92
(iv) 1 + 4 + 9 + 16 + …….. + 225
(v) 6² + 7² + 8² + …….. + 21²
(vi) 10³ + 11³ + 12³ + …….. + 20³
(vii) 1 + 3 + 5 + …… + 71
Answer key:
(i) 1 + 2 + 3 + …….. + 60 =$\frac{60\times 61}{2}$
[Using $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$formula]
= 1830

 

(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
=$\frac{3\times 32\times 33}{2}$
= 1584

 

(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)

[$\frac{92\times 93}{2}$]-[$\frac{50\times 51}{2}$]

= 4278 – 1275
= 3003

 

(iv) 1 + 4 + 9 + 16 + …….. + 225 = 1² + 2² + 3² + 4² + ………… + 15²
$\frac{15\times 16\times 31}{6}$
[using $\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}$] formula
= 1240

 

(v) 6² + 7² + 8² + …….. + 21² = 1 + 2² + 3² + 4² + ………… + 21² – (1 + 2² + ………… + 5²)
=$\frac{21\times 22\times 43}{6}-\frac{5\times 6\times 11}{6}$
= 3311 – 55
= 3256

 

(vi) 10³ = 11³ + 12³ + …….. + 20³ = 1³ + 2³+ 3³ + ………… + 20³ – (1³ + 2³ + 3³ + …………. + 9³)

[$\frac{20\times 21}{2}$]²-[$\frac{9\times 10}{2}$]² 

[Using ($\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}$)² formula]
= 210² – 45² = 44100 – 2025
= 42075

 

(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71

$\mathrm{n}=\frac{\mathrm{l}-\mathrm{a}}{\mathrm{d}}+1$ 

$n=\frac{71-1}{2}+1=35+1=36$

1+3+5+......+71=(36)²=1296

1+3+5+......+71= [$\frac{71+1}{2}$]² 

Using ($\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}$)²

=1296 [Using1+3+5+......+71=(number of terms)²]

  

2. If 1 + 2 + 3 + …. + k = 325 , then find 1³ + 2³ + 3³ + …………. + k³
Answer key:
1 + 2 + 3 + …. + k = 325
${\frac{\mathrm{k}(\mathrm{k}+1)}{2}}^{}$= 325 ……(1)

1 + 2 + 3 + …. + k³=${\left[\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right]}^2$

= 325² (From 1)
= 105625

 

3. If 1³ + 2³ + 3³ + ………… + K³ = 44100 then find 1 + 2 + 3 + ……. + k
Answer key:
1³ + 2³ + 3³ + ………….. + k³ = 44100

${\left[\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right]}^2$= 44100

${\frac{\mathrm{k}(\mathrm{k}+1)}{2}}^{}$ = $\sqrt{44100}$= 210
1 + 2 + 3 + …… + k =${\frac{\mathrm{k}(\mathrm{k}+1)}{2}}^{}$ 
= 210

 

4. How many terms of the series 1³ + 2³ + 3³ + …………… should be taken to get the sum 14400?
Answer key:
1³ + 2³ + 3³ + ……. + n³ = 14400

${\left[\frac{\mathrm{n}(n+1)}{2}\right]}^2$= 14400

${\frac{\mathrm{n}(\mathrm{n}+1)}{2}}^{}$  = $\sqrt{14400}$
${\frac{\mathrm{n}(\mathrm{n}+1)}{2}}^{}$ = 120 ⇒ n² + n = 240

 n² + n – 240 = 0
(n + 16) (n – 15) = 0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15

 

5. The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Answer key:
1² + 2² + 3² + …. + n² = 285

$\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}=285....\left(1\right)$ 

$1^3+2^3+3^3+......+n^3=2025$

${\left[\frac{n(n+1)}{2}\right]}^2=2025$ 

$\frac{n(n+1)}{2}=\sqrt{2025}\Rightarrow \frac{n\left(n+1\right)}{2}=45$

$substitute the value of\frac{n\left(n+1\right)}{2}in\left(1\right)$

$\frac{45\left(2n+1\right)}{3}=285$

$(2n+1)=\frac{285\times 3}{45}=\frac{285}{15}=19$

$2n+1=19\Rightarrow 2n=19-1=18$

$n=\frac{18}{2}=9\Rightarrow n=9$ 

 

6. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?
Answer key:
Area of 15 square colour papers
= 10² + 11² + 12² + …. + 24²
= (1² + 2² + 3² + …. + 24²) – (1² + 2² + 9²)
$=\frac{24\times 25\times 49}{6}-\frac{9\times 10\times 19}{6}$
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm²

 

7. Find the sum of the series (2³ – 1)+(4³ – 3³) + (6³ – 15³) + …….. to
(i) n terms
(ii) 8 terms
Answer key:
Sum of the series = (2³ – 1) + (4³ – 3³) + (6³ – 15³) + …. n terms
= 2³ + 4³ + 6³ + …. n terms – (1³ + 3³ + 5³ + …. n terms) …….(1)
2³ + 4³ + 6³ + …. n = ∑(2³ + 4³ + 6³ + ….(2n)³]
∑ 2³ (1³ + 2³ + 3³ + …. n³)
=$8{\left(\frac{n(n+1)}{2}\right)}^2$
= 2[n (n + 1)]²
1³ + 3³ + 5³ + ……….(2n – 1)³ [sum of first 2n cubes – sum of first n even cubes]

${\left[\frac{\mathrm{2n}(\mathrm{2n}+1)}{2}\right]}^2$-8${\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]}^2$

$\frac{4n^2(2n+1{)}^2}{4}-\frac{8{\left[n^2(n+1)\right]}^2}{4}$

$n^2(2n+1{)}^2-2n^2(n+1{)}^2$ 

Substituting (2) and (3) in (1)
Sum of the series = 2n² (n + 1)² – n² (2n + 1)² + 2n²(n + 1)²
= 4n² (n + 1)² – n² (2n + 1)²
= n² [(4(n + 1)² – (2n + 1)²]
= n² [4n² + 4 + 8n – 4n² – 1 – 4n]
= n² [4n + 3]
= 4n³ + 3n²

 

(ii) when n = 8 = 4(8)³ + 3(8)²
= 4(512) + 3(64)
= 2240

${\left[\frac{\mathrm{2n}(\mathrm{2n}+1)}{2}\right]}^2$