Question

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

• A. 5.15
• B. 5.5
• C. 6
• D. 6.15

Answer 1

Answer: (A)

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