# 10th Maths- Chapter 2- Numbers and Sequences- Exercise 2.9 (All Sums)

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### 10th Maths- Chapter 2- Numbers and Sequences- Exercise  2.9 (All Sums)

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

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

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 = √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

(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

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 color papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these color papers?

Solution:

102 + 112 + 122 + … + 242

= (12 + 22 + … + 242) – (12 + 22 + … + 92)

∴ Rekha has 4615 cm2 color 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