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# Let bi>1 for i = 1, 2, .... ,101. Suppose loge b1,loge b2,……loge b101 are in AP with the common difference loge2. Suppose a1,a2,……a101 are in AP, 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 B s>t and a101<b101If log b1, log b2,……log b101 are in AP, with common difference loge 2, then b1,b2,……b101 are in GP, with common ratio 2. ∴b1=20b1,b2=21b1, b3=22b1……,b101=2100b1 ……(i) Also, a1,a2……a101 are in AP Given a1=b1 and a51=b51 ⇒a1+50D=250b1 ⇒a1+50D=250a1 [∵a1=b1]……(ii) Now, t=b1+b2+……+b51 ⇒t=b1(251−1)2−1 ……(iii) and s=a1+a2+……+a51 =512(2a1+50D) ……(iv) ∴t=a1(251−1) [∵a1=b1] or t=251a1−a1<251a1 ……(v) and s=512[a1+(a1+50D)] [fromEq.(ii)] =512[a1+250a1]=512a1+512250a1∴s>251a1……(vi) From Eqs. (v) and (vi), we get s > t Also, a101=a1+100D and b101=2100b1 ∴a101=a1+100(250a1−a150) and b101=2100a1 ⇒a101=a1+251a1−2a1=251a1−a1 ⇒a101<251a1 and b101>251a1 ⇒b101>a101  Suggest Corrections  0      Similar questions
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