Numbers and Sequences Ex 2.9-10th Std Maths-Book Back Question And Answer
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 = 60×612
[Using n(n+1)2 formula]
= 1830
(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
= 3×32×332
= 1584
(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)
= 4278 – 1275
= 3003
(iv) 1 + 4 + 9 + 16 + …….. + 225 = 12 + 22 + 32 + 42 + ………… + 152
15×16×316
[using n(n+1)(2n+1)6] formula
= 1240
(v) 62 + 72 + 82 + …….. + 212 = 1 + 22 + 32 + 42 + ………… + 212 – (1 + 22 + ………… + 52)
= 21×22×436 – 5×6×116
= 3311 – 55
= 3256
(vi) 103 = 113 + 123 + …….. + 203 = 13 + 23+ 33 + ………… + 203 – (13 + 23 + 33 + …………. + 93)
[Using (n(n+1)2)2 formula]
= 2102 – 452 = 44100 – 2025
= 42075
(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71
Question 2.
If 1 + 2 + 3 + …. + k = 325 , then find 13 + 23 + 33 + …………. + k3
Answer
1 + 2 + 3 + …. + k = 325
k(k+1)2 = 325 ……(1)
= 3252 (From 1)
= 105625
Question 3.
If 13 + 23 + 33 + ………… + K3 = 44100 then find 1 + 2 + 3 + ……. + k
Answer
13 + 23 + 33 + ………….. + k3 = 44100
k(k+1)2 = 44100−−−−−√ = 210
1 + 2 + 3 + …… + k = k(k+1)2
= 210
Question 4.
How many terms of the series 13 + 23 + 33 + …………… should be taken to get the sum 14400?
Answer
13 + 23 + 33 + ……. + n3 = 14400
n(n+1)2 = 14400−−−−−√
n(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
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.
Answer
12 + 22 + 32 + …. + n2 = 285
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?
Answer
Area of 15 square colour papers
= 102 + 112 + 122 + …. + 242
= (12 + 22 + 32 + …. + 242) – (12 + 22 + 92)
= 24×25×496−9×10×196
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm2
Question 7.
Find the sum of the series (23 – 1)+(43 – 33) + (63 – 153) + …….. to
(i) n terms
(ii) 8 terms
Answer
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 (n(n+1)2)2
= 2[n (n + 1)]2
13 + 33 + 53 + ……….(2n – 1)3 [sum of first 2n cubes – sum of first n even cubes]
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
