11Let the first term be a1 d ≠0
an=a1 + (n - 1)d
an=a1 + nd - d
Therefore Sn =
an + dn(n+1)/2 -nd
n(a+ d (n + 1)/2 - d)
sn=n(a + d(n/2 + 1/2 -1)
sn=n(a + d((n-1)/2)
Therefore the a) is false
if a1,a2,a3..............an ar n postive numbers in arthmetic progression with common difference d≠0 and
sn= \Sigma r=1 to n ar,
then
a) sn>n x nth root of a1a2an
b)sn>n√a1an
c)sn>n√2a1(n-1)d
d)sn>2[√a1a2+√a3a4+√an-1an] if n is even
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12 Answers
11
11answer is a,b,c,d
Sn=n/2(a1+an)
applying Am Gm inequality for (a1+an)
we get Sn≥n/2(2√a1an)
so b is true
11for c
we write Sn as n/2(2a1+(n-1)d)
again applying Am Gm for the 2 terms gives us the answer
11a is direct by writing Sn as a1+a2+....
for d
taking Sn as (a1+a2)+(a3+a3)+.............
we get Sn≥2√a1a2+2√a3a4
and so on so we get a,b,c,d
11
Some from my side :
e. Sn __ n - 2(√a1 + √a2 +...+√an) (>,<,=,≤ or ≥)
f. 2Sn > (np . a1 . an)
Find maximum value of p.
11@rkrish why is b true for only even values what i proved was for all values
11and if n is not even in d how will the grouping be done so n should be even
11yup...youre rite....I'm sry...I was thinking of something else!!!![200][200]
