Let b i > 1 for i = 1 ,   2 , … . , 101 . Suppose log e ⁡ b 1 , log e ⁡ b 2 ,…

Let bi>1 for i=1, 2,….,101. Suppose loge⁡b1,loge⁡b2,…..,loge⁡b101 are in Arithmetic Progression (A.P.) with the common difference loge⁡2.  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
  1. s>t and a101>b101
  2. s>t and a101<b101
  3. s<t and a101>b101
    s<t and a101>b101
  4. s<t and a101<b101
    s<t and a101<b101

Solution

If loge⁡b1,loge⁡b2….loge⁡b101→AP;     difference(d)=loge⁡2
⇒ b 1 , b 2 , b 3 ....... b 101 →GP;r=2
∴ b 1 ,2 b 1 , 2 2 b 1 ........., 2 100 b 1 →GP
a 1 , a 2 , a 3 .......... a 101 →AP        Let Common Difference = D
Given, a1=b1 and a51=b51
⇒ a1+50D=250b1
∴ a 1 +50D= 2 50 a 1 ( As  b 1 = a 1 )⇒D= 2 50 a 1 − a 1 50
t= b 1 + b 2 +....... b 51 = b 1 +2 b 1 + 2 2 b 1 + .......2 50 b 1 ⇒t= b 1 ( 2 51 −1 );
s= a 1 + a 2 +....... a 51 ⇒s= 51 2 ( 2 a 1 +50D ) 
t=a1. 251-a1⇒t<a1. 251 .......(i)  ( as bi>1 &  a 1 = b 1   so  a1>1 )
s=512 a1+a1+50D
s=512 a1+250 a1
s=51a12+512.250 a1
⇒ s>a1. 251 ......(ii)
Clearly s>t (from equation (i) and (ii))
Also a101=a1+100D;  b101=b1. 2100
∴ a101=a1+100 250 a1-a150;  b101=2100 a1 .......(ii)
a101=a1+251 a1-2a1 ⇒  a101=251 a1-a1 ⇒a101<251 a1 .......(iv)
Clearly b101>a101 (from equation (iii) and (iv))

Asked in: JEE Advanced 2016 (Paper 2)

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