Consider the real-valued function h: {0, 1, 2 | vedclass

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(A) Given the recurrence relation $h(p) = \frac{1}{2}\{h(p+1) + h(p-1)\}$,we can rewrite it as $2h(p) = h(p+1) + h(p-1)$,which implies $h(p+1) - h(p)

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$5.15$

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(A) Given the recurrence relation $h(p) = \frac{1}{2}\{h(p+1) + h(p-1)\}$,we can rewrite it as $2h(p) = h(p+1) + h(p-1)$,which implies $h(p+1) - h(p) = h(p) - h(p-1)$.This indicates that the sequence $h(0), h(1), \ldots, h(100)$ forms an Arithmetic Progression ($A$.$P$.).Let the common difference be $d$. Then $h(n) = h(0) + nd$.Using $h(100) = 20$ and $h(0) = 5$,we have $20 = 5 + 100d$.$100d = 15 \Rightarrow d = \frac{15}{100} = 0.15$.Thus,$h(1) = h(0) + d = 5 + 0.15 = 5.15$.

Consider the real-valued function $h: \{0, 1, 2, \ldots, 100\} \rightarrow \mathbb{R}$ 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:

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Statement-$I$: If the sum of $n$ terms of a sequence is $6n^2 + 3n + 1$,then it is an Arithmetic Progression $(AP)$.
Statement-$II$: The sum of $n$ terms of an Arithmetic Progression is always in the form $an^2 + bn$.

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