Try this beautiful Objective Sequence Problem appeared in ISI Entrance - 2018.
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) \(6.15.\)
Sequence
Arithmetic Progression
IIT mathematics by Asit Das Gupta
ISI UG Entrance - 2018 , Objective problem number - 8
(B) \(5.15\)
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.
Therefore we have the following,
\(h(i+1)-h(i)=h(i)-h(i-1).\)
Means
\(h(0) , h(1) , h(2) , \ldots \ldots ,h(100) \) are in Arithmetic Progression.
\(h(0) \) and \( h(100)\) are the first and last terms of the AP.
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
