Let b i > 1 for i = 1,2,….,101. Suppose log e b 1 ,log e b 2 ,…,log e b 101 are in Arithmetic progression (A.P.) with the common difference log e 2. Suppose a 1 , a 2 ,…,a 101 are in A.P. such that a 1 = b 1 and a 51 = b 51 . If t = b 1 + b 2 + …. + b 51 and s = a 1 + a 2 + … + a 51 , then
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Sol. log e b 1 , log e b 2 , log e b 3 , ...... log e b 101 are in A.P.
b 1 , b 2 , b 3 , ..........., b 101 are in G.P.
Given : log e (b 2 ) – log e (b 1 ) = log e ⇒ 
= 2 = r (common ratio of G.P. )
a 1 , a 2 , a 3 , ......... a 101 are in A.P.
a 1 = b 1 = a
b 1 + b 2 + b 3 + ........ b 51 = t ,
S = a 1 +a 2 + ...... + a 51
t = sum of 51 terms of G.P.= b 1 = 
= 
a(2 51 –1)
s = sum of 51 terms of A.P. = 
[2a 1 + (n–1)d] = 
(2a + 50d)
Given a 51 = b 51
a + 50d = a(b) 50
50d = a(2 50 – 1)
Hence s = 
[2 50 + 1] ⇒ s = a 
s = 
⇒ s = 
s – t = a 
Clearly s > t
a 101 = a 1 + 100d = a + 2a.2 50 – 2a = a (2 51 – 1)
b 101 = b 1 r 100 = a.2 100 Hence b 101 > a 101
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