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Sequences & Series: Consider Value

JEE Maths question with a full step-by-step solution.

Question

Consider

\alpha_0, \alpha_1, \alpha_2, \ldots

with

\alpha_0=17.23

\alpha_1=33.23

and

\alpha_{r+2}=\frac{\alpha_r+\alpha_{r+1}}{2}

for all

r\ge 0

. The value of

|\alpha_{10}-\alpha_9|

\frac{41}{50}

\frac{28}{61}

\frac{1}{32}

correct

\frac{33}{277}

Solution

Step 1: From the recurrence,

\alpha_{r+2}-\alpha_{r+1}=\frac{\alpha_r+\alpha_{r+1}}{2}-\alpha_{r+1}=-\frac12(\alpha_{r+1}-\alpha_r).

Step 2: The differences form a G.P. with ratio

-\frac12

and first difference

\alpha_1-\alpha_0=16

\alpha_{10}-\alpha_9=\left(-\frac12\right)^9(\alpha_1-\alpha_0)=-\frac{16}{2^9}=-\frac{1}{2^5}.

Step 3:

|\alpha_{10}-\alpha_9|=\dfrac{1}{32}.

Correct answer: (3)

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