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# ISI 2018 Objective Problem 8 | A Problem from Sequence

Try this beautiful Objective Sequence Problem appeared in ISI Entrance - 2018.

## Problem

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

(A) $$5.15$$
(B) $$5.5$$
(C) $$6$$
(D) $$6.15.$$

### Key Concepts

Sequence

Arithmetic Progression

## Suggested Book | Source | Answer

IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2018 , Objective problem number - 8

(B) $$5.15$$

## Try with Hints

Observe the following,

$$2 h(i)=h(i-1)+h(i+1)$$

$$2h(1) = h(0) + h(2)$$

$$2h(2) = h(1) + h(3)$$

and so on.

Therefore we have the following,

$$h(i+1)-h(i)=h(i)-h(i-1).$$

Means

$$h(0) , h(1) , h(2) , \ldots \ldots ,h(100)$$ are in Arithmetic Progression.

$$h(0)$$ and $$h(100)$$ are the first and last terms of the AP.

Common difference

$=\frac{h(100) - h(0)}{100}$

$=\frac{20 - 5}{100}= 0.15$

Therefore ,

$$h(1) = h(0) + 0.15 = 5.15$$

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