# Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9

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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9

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
$$\sum_{1}^{n} n^{2}=\frac{n(n+1)(2 n+1)}{6}$$

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

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 = $$\sqrt { 44100 }$$
= 210

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
$$\left(\frac{n(n+1)}{2}\right)^{2}$$ = 14400 = (120)2
$$\frac{n(n+1)}{2}$$ = $$\sqrt { 14400 }$$ = 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

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

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.

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