cat 2021 Question 16

Question

Consider a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots$ such that $x_{n+1}=x_{n}+n-1$ for all n≥1. If $x_{1}$=−1 then $x_{100}$ is equal to

Accepted Answer

Given $x_{n+1} = x_n + n - 1$ and x1 = -1

Considering

x1 = -1. (1)

x2 = x1+1-1 = x1 + 0 (2)

x3 = x2 + 2 - 1 =x2 + 1 (3)

x4 = x3 + 3 - 1 = x3 + 2 (4)

x100 = x99 + 98 (100)

Adding the LHS and RHS for the hundred equations we have:

(x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)

Subtracting this we have :

(x1+...........x100) - (x1+............. x 99) = $\frac{(98⋅99)}{2}$ - 1

x100 = 4851 - 1 = 4850

Alternatively

x1 = −1

x2 = x1 + 1 − 1 = x1 = −1

x3 = x2 + 2 − 1 = x2 + 1 = −1 + 1 = 0

x4 = x3 + 3 − 1 = x3 + 2 = 0 + 2 = 2

x5 = x4 + 4 − 1 = x4 + 3 = 2 + 3 = 5

......

If we observe the series, it is a series that has a difference between the consecutive terms in an AP.

Such series are represented as $t(n)=a+bn+cn^2$

We need to find t(100).

t(1) = -1

a + b + c = -1

t(2) = -1

a + 2b + 4c = -1

t(3) = 0

a + 3b + 9c = 0

Solving we get,

b + 3c = 0

b + 5c = 1

c = 0.5

b = -1.5

a = 0

Now,

$t(100)=(−1.5)100+(0.5)100^2 = −150+5000=4850$

Question 16.