How Cheenta works to ensure student success?
Explore the Back-Story

# 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

Hint 1

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.

Hint 2

Therefore we have the following,

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

Hint 3

Means

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

Hint 4

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

Conclusion

Common difference

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

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

Therefore ,

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

ISI Entrance Program at Cheenta

## Subscribe to Cheenta at Youtube

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

Hint 1

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.

Hint 2

Therefore we have the following,

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

Hint 3

Means

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

Hint 4

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

Conclusion

Common difference

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

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

Therefore ,

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

ISI Entrance Program at Cheenta

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.