Let bi>1 for i = 1, 2, .... ,101. Suppose logeb1,logeb2,……logeb101 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 Bs>t and a101<b101 If logb1,logb2,……logb101 are in AP, with common difference loge2, 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+100Dandb101=2100b1 ∴a101=a1+100(250a1−a150) and b101=2100a1 ⇒a101=a1+251a1−2a1=251a1−a1 ⇒a101<251a1andb101>251a1 ⇒b101>a101