(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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(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, $h(1) = h(0) + d = 5 + 0.15 = 5.15$.

Class 11 Mathematics Sequences and Series Arithmetic progression

Medium

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:

WBJEE 2021 All WBJEE

A
$5.15$
B
$5.5$
C
$6$
D
$6.15$

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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:

Similar Questions

Let the sequence $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ Then $a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n - 1}^2 - a_{2n}^2 = $

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