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ISI 2018 Objective Problem 8 | A Problem from Sequence

Try this beautiful Objective Sequence Problem appeared in ISI Entrance - 2018.

Problem

Consider the real valued function \(h:\{0,1,2, \ldots, 100\} \longrightarrow \mathbb{R}\) such that \(h(0)=5, h(100)=20\) and satisfying \(h(i)=\frac{1}{2}(h(i+1)+h(i-1))\), for every \(i=1,2, \ldots, 99\). Then the value of \(h(1)\) is:

(A) \(5.15\)
(B) \(5.5\)
(C) \(6\)
(D) \(6.15.\)


Key Concepts


Sequence

Arithmetic Progression

Suggested Book | Source | Answer


IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2018 , Objective problem number - 8

(B) \(5.15\)

Try with Hints


Hint 1

Observe the following,

\(2 h(i)=h(i-1)+h(i+1)\)

\(2h(1) = h(0) + h(2) \)

\(2h(2) = h(1) + h(3) \)

and so on.

Hint 2

Therefore we have the following,

\(h(i+1)-h(i)=h(i)-h(i-1).\)

Hint 3

Means

\(h(0) , h(1) , h(2) , \ldots \ldots ,h(100) \) are in Arithmetic Progression.

Hint 4

\(h(0) \) and \( h(100)\) are the first and last terms of the AP.

Conclusion

Common difference

\[=\frac{h(100) - h(0)}{100}\]

\[=\frac{20 - 5}{100}= 0.15\]

Therefore ,

\(h(1) = h(0) + 0.15 = 5.15\)

ISI Entrance Program at Cheenta

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Try this beautiful Objective Sequence Problem appeared in ISI Entrance - 2018.

Problem

Consider the real valued function \(h:\{0,1,2, \ldots, 100\} \longrightarrow \mathbb{R}\) such that \(h(0)=5, h(100)=20\) and satisfying \(h(i)=\frac{1}{2}(h(i+1)+h(i-1))\), for every \(i=1,2, \ldots, 99\). Then the value of \(h(1)\) is:

(A) \(5.15\)
(B) \(5.5\)
(C) \(6\)
(D) \(6.15.\)


Key Concepts


Sequence

Arithmetic Progression

Suggested Book | Source | Answer


IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2018 , Objective problem number - 8

(B) \(5.15\)

Try with Hints


Hint 1

Observe the following,

\(2 h(i)=h(i-1)+h(i+1)\)

\(2h(1) = h(0) + h(2) \)

\(2h(2) = h(1) + h(3) \)

and so on.

Hint 2

Therefore we have the following,

\(h(i+1)-h(i)=h(i)-h(i-1).\)

Hint 3

Means

\(h(0) , h(1) , h(2) , \ldots \ldots ,h(100) \) are in Arithmetic Progression.

Hint 4

\(h(0) \) and \( h(100)\) are the first and last terms of the AP.

Conclusion

Common difference

\[=\frac{h(100) - h(0)}{100}\]

\[=\frac{20 - 5}{100}= 0.15\]

Therefore ,

\(h(1) = h(0) + 0.15 = 5.15\)

ISI Entrance Program at Cheenta

Subscribe to Cheenta at Youtube


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