Let A = {1, 2, 3, ..., 100} and R be a relation on A such that R = {(a, b) : a = 2b + 1}. Let be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k,

Question

Let A = {1, 2, 3, ..., 100} and R be a relation on A such that R = {(a, b) : a = 2b + 1}. Let $(a_{1}, a_{2}), (a_{2}, a_{3}), (a_{3}, a_{4}), . . ., (a_{k}, a_{k + 1})$ be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k, for which such a sequence exists, is equal to : [2025]

Smart Achievers · Accepted Answer

(2)

Let the smallest value is $a_{k + 1}$ in A = {1, 2, 3, ..., 100}.

Since, $a_{k} = 2 a_{k + 1} + 1 = (2 \times 1) + 1 = 3$

$∴$ $a_{k - 1} = 2 a_{k} + 1 = (2 \times 3) + 1 = 7$

$a_{k - 2} = 2 a_{k - 1} + 1 = (2 \times 7) + 1 = 15$

$a_{k - 3} = 2 a_{k - 2} + 1 = (2 \times 15) + 1 = 31$

$a_{k - 4} = 2 a_{k - 3} + 1 = (2 \times 31) + 1 = 63$

$a_{k - 5} = 2 a_{k - 4} + 1 = (2 \times 63) + 1 = 127 \notin A$

$∴$ The sequence is {(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)}

$∴$ k = 5

Solution

1 8

2 5

3 6

4 7

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