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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2016 · Paper 2 · Mathematics · Sequences and Series

Let b _ i > 1 for i = 1, 2, . , 101 . Suppose _ e b _ 1 , _ e b _ 2...

Let bi>1{b}_{i} > 1 for i=1,2,….,101.i = 1, 2, \ldots . , 101 . Suppose log⁡eb1,log⁡eb2,…..,log⁡eb101\log_{e} {b}_{1}, \log_{e} {b}_{2}, \ldots . . , \log_{e} {b}_{101} are in Arithmetic Progression (A.P.) with the common difference log⁡e2.\log_{e} 2 . Suppose a1,a2,….,a101{a}_{1}, {a}_{2}, \ldots . , {a}_{101} are in A.P. such that a1=b1{a}_{1} = {b}_{1} and a51=b51.{a}_{51} = {b}_{51} . If t=b1+b2+…+b51t = {b}_{1} + {b}_{2} + \ldots + {b}_{51} and s=a1+a2+…+a51s = {a}_{1} + {a}_{2} + \ldots + {a}_{51} then

Options

  1. As>tanda101>b101s > tand{a}_{101} > {b}_{101}
  2. Bs>tanda101<b101s > tand{a}_{101} < {b}_{101}
  3. Cs<tanda101>b101s < tand{a}_{101} > {b}_{101}
    s<tanda101>b101s < tand{a}_{101} > {b}_{101}
  4. Ds<tanda101<b101s < tand{a}_{101} < {b}_{101}
    s<tanda101<b101s < tand{a}_{101} < {b}_{101}

Correct answer

B. s>tanda101<b101s > tand{a}_{101} < {b}_{101}

Step-by-step solution

If log⁡eb1,log⁡eb2….log⁡eb101→AP;difference(d)=log⁡e2\log_{e} {b}_{1}, \log_{e} {b}_{2} \ldots . \log_{e} {b}_{101} \to AP; difference(d) = \log_{e} 2
⇒b1,b2,b3.......b101→GP;r=2⇒ {b}_{1} , {b}_{2} , {b}_{3} .......{b}_{101} \to GP; r = 2
∴b1,2b1,22b1.........,2100b1→GP∴ {b}_{1} , 2{b}_{1} , {2}^{2} {b}_{1} ........., {2}^{100} {b}_{1} \to GP
a1,a2,a3..........a101→AP{a}_{1} , {a}_{2} , {a}_{3} ..........{a}_{101} \to AP Let Common Difference = D
Given, a1=b1{a}_{1} = {b}_{1} and a51=b51{a}_{51} = {b}_{51}
⇒⇒ a1+50D=250b1{a}_{1} + 50D = {2}^{50}{b}_{1}
∴a1+50D=250a1(Asb1=a1)⇒D=250a1−a150∴ {a}_{1} + 50D = {2}^{50} {a}_{1} ( As{b}_{1} = {a}_{1} ) ⇒ D = \dfrac{{2}^{50} {a}_{1} - {a}_{1}}{50}
t=b1+b2+.......b51=b1+2b1+22b1+.......250b1⇒t=b1(251−1);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=a1+a2+.......a51⇒s=512(2a1+50D)s = {a}_{1} + {a}_{2} + .......{a}_{51} ⇒ s = \dfrac{51}{2} ( 2{a}_{1} + 50D )
t=a1.251−a1⇒t<a1.251t = {a}_{1} . {2}^{51} - {a}_{1} ⇒ t < {a}_{1} . {2}^{51} .......(i) (asbi>1 and a1=b1soa1>1)( as{b}_{i} > 1 \text{ and } {a}_{1} = {b}_{1} so{a}_{1} > 1 )
s=512(a1+a1+50D)s = \dfrac{51}{2}\left({a}_{1} + {a}_{1} + 50D\right)
s=512(a1+250a1)s = \dfrac{51}{2}\left({a}_{1} + {2}^{50}{a}_{1}\right)
s=51a12+512.250a1s = \dfrac{51{a}_{1}}{2} + \dfrac{51}{2} . {2}^{50}{a}_{1}
⇒⇒ s>a1.251s > {a}_{1} . {2}^{51} ......(ii)
Clearly s>ts > t (from equation (i) and (ii))
Also a101=a1+100D;b101=b1.2100{a}_{101} = {a}_{1} + 100D; {b}_{101} = {b}_{1} . {2}^{100}
∴∴ a101=a1+100(250a1−a150);b101=2100a1{a}_{101} = {a}_{1} + 100\left(\dfrac{{2}^{50}{a}_{1} - {a}_{1}}{50}\right); {b}_{101} = {2}^{100}{a}_{1} .......(ii)
a101=a1+251a1−2a1⇒a101=251a1−a1⇒a101<251a1{a}_{101} = {a}_{1} + {2}^{51}{a}_{1} - 2{a}_{1} ⇒ {a}_{101} = {2}^{51}{a}_{1} - {a}_{1} ⇒ {a}_{101} < {2}^{51}{a}_{1} .......(iv)
Clearly b101>a101{b}_{101} > {a}_{101} (from equation (iii) and (iv))
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