Question

Consider the real valued function \(h:\{0,1,2 \ldots . .100\} \rightarrow R\) such that \(h(0)=5, h(100)=20\) and satisfying \(h(p)=\frac{1}{2}\{h(p+1)+h(p-1)\}\) for every \(p=1,2 \ldots . .99\). Then the value of \(h(1)\) is
(A) \(5.15\)
(B) \(5.5\)
(C) 6
(D) \(6.15\)

Answer 1

Ans: (A)
Hint \(: h(p)=\frac{1}{2}(h(p+1)+h(p-1)) \Rightarrow h(p-1), h(p), h(p+1)\) are in A.P.
$$
\begin{aligned}
&h(100)=h(0)+99 d \\
&\Rightarrow \frac{20-5}{99}=d \Rightarrow d=\frac{15}{99} \Rightarrow h(1)=h(0)+d=5+\frac{15}{99}=5.15
\end{aligned}
$$

Answer 2

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