Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9
Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9
Exercise 2.9 Class 10 Maths Question 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
Solution:
(i) 1 + 2 + 3 + ……… + 60
= 4278 – 1275 = 3003
(iv) 1 + 4 + 9 + 16 + … + 225
= 12 + 22 + 32 + 42 + ……… + 152
∑n1n2=n(n+1)(2n+1)/6
10th Maths Exercise 2.9 Samacheer Kalvi Question 2.
If 1 + 2 + 3 + … + k = 325, then find 13 + 23 + 33 + …………. + K3.
Solution:
1 + 2 + 3 + … + K = 325
If 1 + 2 + 3 … + k = 325
13 + 23 + 33 + … + K3 = (325)2 = 105625
10th Maths Exercise 2.9 Answers Question 3.
If 13 + 23 + 33 + … + K3 = 44100 then find 1 + 2 + 3 + … + k.
Solution:
If 13 + 23 + 33 + … + K3 = 44100
1 + 2 + 3 + … + K = √44100
= 210
10th Maths Exercise 2.9 Question 4.
How many terms of the series 13 + 23 + 33 + … should be taken to get the sum 14400?
Solution:
13 + 23 + 33 + ……… + n3 = 14400
(n(n+1)/2)² = 14400 = (120)2
n(n+1)/2 = √14400= 120
n(n + 1) = 240
Method 1:
n2 + n – 240 = 0
n2 + 16n – 15n – 240 = 0
n(n + 16) – 15(n + 16) = 0
(n + 16)(n – 15) = 0
n = -16, 15
∴ 15 terms to be taken to get the sum 14400.
Method 2:
n2 + n – 240 = 0
Ex 2.9 Class 10 Samacheer Question 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.
Solution:
12 + 22 + 32 + …… + n2 = 285
13 + 23 + 33 + …… + n3 = 2025
Exercise 2.9 Class 10 Maths Samacheer Question 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?
Solution:
102 + 112 + 122 + … + 242
= (12 + 22 + … + 242) – (12 + 22 + … + 92)
∴ Rekha has 4615 cm2 colour papers. She can decorate 4615 cm2 area with these colour papers.
10th Maths Samacheer Question 7.
Find the sum of the series (23 – 1) + (43 – 33) + (63 – 153) +… to (i) n terms (ii) 8 terms.
Solution:
(23 – 1) + (43 – 33) + (63 – 153) + ……… n
= 4n3 + 3n2 = sum of ‘n’ terms.
When n = 8
Sum = 4 × 83 + 3 × 82
= 2048 + 192 = 2240
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