    Question

# Let bi>1 for i=1,2,....,101. Suppose logeb1logeb2,...,logeb101 are in Arithmetic Progression (A.P) with the common difference loge2. Suppose a1,a2,...a101 are in A.P. such that a1=b1 and a51=b51. If t=b1+b2+....+b51 and s=a1+a2+....+a51, then

A
s>t and a101>b101
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B
s>t and a101<b101
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C
s<t and a101>b101
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D
s<t and a101<b101
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Solution

## The correct option is D s>t and a101<b101logeb1,logeb2,logeb3,......logeb101 are in A.P.b1,b2,b3,...........,b101 are in G.P.Given : loge(b2)−loge(b1)=loge(2)⇒b2b1=2=r (common ratio of G.P. )a1,a2,a3,.........a101 are in A.P. a1=b1=ab1+b2+b3+........b51=t ,S=a1+a2+......+a51t= sum of 51 terms of G.P. =b1(r51−1)r−1=a(251−1)2−1=a(251−1)s= sum of 51 terms of A.P =512]2a1+(n−1)d]=512(2a+50d)Given a51=b51a+50d=a(2)5050d=a(250−1)Hence, ⇒s=a(51.249+512)s=2(4.249+47.249+512)⇒s=a((251−1)+47.249+532)s−t=a(47.249+532)Clearly s>ta101=a1+100d=a+2a.2502a=a(251−1)b101=b1r100=a.2100 Hence : b101>a101  Suggest Corrections  0      Similar questions  Related Videos   Introduction to Direct Proportions
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