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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) 6^{2} + 7^{2} + 8^{2} + … + 21^{2}

(vi) 10^{3} + 11^{3} + 12^{3} + … + 20^{3}

(vii) 1 + 3 + 5 + … + 71

Solution:

(i) 1 + 2 + 3 + ……… + 60

= 4278 – 1275 = 3003

(iv) 1 + 4 + 9 + 16 + … + 225

= 1^{2} + 2^{2} + 3^{2} + 4^{2} + ……… + 15^{2}

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

Question 2.

If 1 + 2 + 3 + … + k = 325, then find 1^{3} + 2^{3} + 3^{3} + …………. + K^{3}.

Solution:

1 + 2 + 3 + … + K = 325

If 1 + 2 + 3 … + k = 325

1^{3} + 2^{3} + 3^{3} + … + K^{3} = (325)^{2} = 105625

Question 3.

If 1^{3} + 2^{3} + 3^{3} + … + K^{3} = 44100 then find 1 + 2 + 3 + … + k.

Solution:

If 1^{3} + 2^{3} + 3^{3} + … + K^{3} = 44100

1 + 2 + 3 + … + K = \(\sqrt { 44100 }\)

= 210

Question 4.

How many terms of the series 1^{3} + 2^{3} + 3^{3} + … should be taken to get the sum 14400?

Solution:

1^{3} + 2^{3} + 3^{3} + ……… + n^{3} = 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:

n^{2} + n – 240 = 0

n^{2} + 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:

n^{2} + 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:

1^{2} + 2^{2} + 3^{2} + …… + n^{2} = 285

1^{3} + 2^{3} + 3^{3} + …… + n^{3} = 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:

10^{2} + 11^{2} + 12^{2} + … + 24^{2}

= (1^{2} + 2^{2} + … + 24^{2}) – (1^{2} + 2^{2} + … + 9^{2})

∴ Rekha has 4615 cm^{2} colour papers. She can decorate 4615 cm^{2} area with these colour papers.

Question 7.

Find the sum of the series (2^{3} – 1) + (4^{3} – 3^{3}) + (6^{3} – 15^{3}) +… to (i) n terms (ii) 8 terms.

Solution:

(2^{3} – 1) + (4^{3} – 3^{3}) + (6^{3} – 15^{3}) + ……… n

= 4n^{3} + 3n^{2} = sum of ‘n’ terms.

When n = 8

Sum = 4 × 8^{3} + 3 × 8^{2}

= 2048 + 192 = 2240