Question

# Let bi>1 for i=1,2,...,101. Suppose logeb1.logeb2,......,logeb101 are in Arithmetic Progression (A.P) with the common diffrence 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+......+a53. 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 B s>t and a101<b101logc,b1,logcb2,......logcb101 are in A.P. ⇒b1,b2,.....b101 are in G.P. Also a1,a2,...a101 are in A.P. where a1=b1 are a51=b51. ∴b2,b3,...b50 and GM′s and a2,a3,...,a50 are AM's between b1 and b51. ∴GM<AM⇒b2<a2,b3<a3,...b50<a50∴b1+b2+...+b51<a1+a2+...+a51⇒t<sAlso a1,a51,a101 are in APb1,b51,b101 are in GP∵a1=b1 and a51=b51∴b101>a101

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