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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