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Relation between A.M., G.M. and H.M. Let there are two numbers ‘a’ and ‘b’, a, b > 0 then AM = a+b/2 GM =√ab HM =2ab/a+b ∴ AM × HM =a+b/2 × 2ab/a+b = ab = (√ab)2 = (GM)2 Note that these means are in G.P. Hence AM.GM.HM follows the rules of G.P. i.e. G.M. =√A.M. × H.M. Now, let us see the difference between AM and GM AM – GM =a+b/2 – √ab =(√a2)+(√b)–2√a√b/2 i.e. AM > GM Similarly, G.M. – H.M. = √ab –2ab/a+b =√ab/a+b (√a – √b)2 > 0 So. GM > HM Combining both results, we get AM > GM > HM …….. (12) All sequences of numbers cannot be put into A.P./G.P./H.P. Let us study these. Important Points r3 (r – 1)3 = 3 r2 – 3r + 1 r = 1 : 13 – 0 = 3 . 12 – 3 . 1 + 1 r = 2 : 23 – 13 = 3 . 22 – 3 . 2 + 1 r = 3 : 33 – 23 = 3 . 32 – 3 . 3 + 1 r = n : n3 – (n–1)3 = 3.(n2) – 3(n) + 1 Adding n3 = 3 (12 + 22 +…+ n2) –3 (1 + 2 + 3 +…+ n) + (1 + 1 +…+ n times) n3 = 3 Σnr=1 r2 – 3 (n(n+1))/2 + n ⇒ 3 Σnr=1 r2 = n3 + 3n(n+1)/2 – n = n/2 (2n2 + 3n + 1) ⇒ Σnr=1 r2 = n(n+1)(2n+1)/6 To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
Let there are two numbers ‘a’ and ‘b’, a, b > 0
then AM = a+b/2
GM =√ab
HM =2ab/a+b
∴ AM × HM =a+b/2 × 2ab/a+b = ab = (√ab)2 = (GM)2
Note that these means are in G.P.
Hence AM.GM.HM follows the rules of G.P.
i.e. G.M. =√A.M. × H.M.
Now, let us see the difference between AM and GM
AM – GM =a+b/2 – √ab
=(√a2)+(√b)–2√a√b/2
i.e. AM > GM
Similarly,
G.M. – H.M. = √ab –2ab/a+b
=√ab/a+b (√a – √b)2 > 0
So. GM > HM
Combining both results, we get
AM > GM > HM …….. (12)
All sequences of numbers cannot be put into A.P./G.P./H.P. Let us study these.
r3 (r – 1)3 = 3 r2 – 3r + 1
r = 1 : 13 – 0 = 3 . 12 – 3 . 1 + 1
r = 2 : 23 – 13 = 3 . 22 – 3 . 2 + 1
r = 3 : 33 – 23 = 3 . 32 – 3 . 3 + 1
r = n : n3 – (n–1)3 = 3.(n2) – 3(n) + 1
Adding
n3 = 3 (12 + 22 +…+ n2) –3 (1 + 2 + 3 +…+ n) + (1 + 1 +…+ n times)
n3 = 3 Σnr=1 r2 – 3 (n(n+1))/2 + n
⇒ 3 Σnr=1 r2 = n3 + 3n(n+1)/2 – n
= n/2 (2n2 + 3n + 1)
⇒ Σnr=1 r2 = n(n+1)(2n+1)/6
To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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