Let b_i > 1 for i = 1, 2, , 101. Suppose _e b₁ , _e b₂ , ,_e b₁01} are in Arithmetic Progression (A.P.) with the common difference _e 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 [JEE Adv 2016 P2]

Correct Answer is : s>ts > ts>t and a101<b101a₁01} < b₁01}a101<b101

Correct Answer is : s>ts > ts>t and a101<b101a₁01} < b₁01}a101<b101

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Sequences and Series

Section: Mathematics

Question : 10 of 22

Marks: +1, -0

Let

b_i > 1

i = 1, 2, \dots, 101

\log_e b_1 , \log_e b_2 , \dots ,\log_e b_{101}

\log_e 2

a_1, a_2, \dots, a_{101}

a_1 = b_1

a_{51} = b_{51}

t = b_1 + b_2 + \dots + b_{51}

s = a_1 + a2 + \dots + a_{51}

Solution:

\log_e b_1,\log_e b_2\dots\log_e b_{101}\rightarrow AP; D=\log_e 2

\Rightarrow b_1 b_2 b_3 \dots b_{101}\rightarrow GP; r=2

\therefore b_{1}, 2b_1, 2^2 b_1, \dots, 2^{100} b_1, \dots GP

a_1 a_2 a_3 \dots a_{101} \dots AP

Given,

a_1 = b_1 \;\&\; a_{51} = b_{51}

\Rightarrow a_1 + 50D = 2^{50} b_1

\therefore a_1 = 50D = 2^{50} a_1 (\text{As } b_1 = a_1)

Now,

t = b_1(2^{51} - 1); s = \frac{51}{2}(2a_1 + 50D)

t = a_{1} \cdot 2^{51} - a_1 \Rightarrow t < a_{1} \cdot 2^{51} \dots (i)\;;\; s = \frac{51}{2}(a_1 + a_1 + 50D)

s = \frac{51}{2}(a_1 + 2^{50} a_1)

s = \frac{51 a_1}{2} + \frac{51}{2} \cdot 2^{50} a_1

s > a_{1} \cdot 2^{51}

.......(ii)

clearly s > t (from equation (i) and (ii))

Also

a_{101} = a_1 + 100D; b_{101} = b_{1} \cdot 2^{100}

\therefore a_{101} = a_1 + 100\left(\frac{2^{50} a_1 - a_1}{50}\right); b_{101} = 2^{100} a_1

.....(iii)

a_{101} = a_1 + 2^{51} a_1 - 2a_1 \Rightarrow a_{101} = 2^{51} a_1 - a_1 \Rightarrow a_{101} < 2^{51} a_1

......(iv)

clearly

b_{101} > a_{101}

(from equation (iii) and (iv))

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