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CAT 2020 PYQsQuantitative Ability. Free, no login required.

Algebra

past year questions, CAT 2020

Progression & Series

past year questions, Algebra, CAT 2020

Medium

If x1=−1x_1 = -1 and xm=xm+1+(m+1)x_m = x_{m+1} + (m + 1) for every positive integer m,m, then x100x_{100} equals

Solution

✅ Correct Option: 3

We have:

x1=−1x_1 = -1 (our starting value)

xm=xm+1+(m+1)x_m = x_{m+1} + (m + 1) for every positive integer mm

This is a recurrence relation that tells us how each term relates to the next term.

From xm=xm+1+(m+1)x_m = x_{m+1} + (m + 1), we can rearrange to get:

xm+1=xm−(m+1)x_{m+1} = x_m - (m + 1)

This tells us: "To get the next term, subtract (m+1)(m + 1) from the current term"

Let's calculate the first few terms:

x2=x1−(1+1)=−1−2=−3x_2 = x_1 - (1 + 1) = -1 - 2 = -3

x3=x2−(2+1)=−3−3=−6x_3 = x_2 - (2 + 1) = -3 - 3 = -6

x4=x3−(3+1)=−6−4=−10x_4 = x_3 - (3 + 1) = -6 - 4 = -10

x5=x4−(4+1)=−10−5=−15x_5 = x_4 - (4 + 1) = -10 - 5 = -15

So our sequence is: −1,−3,−6,−10,−15,...-1, -3, -6, -10, -15, ...

Let's examine the pattern by looking at what we're subtracting:

x1=−1x_1 = -1

x2=−1−2x_2 = -1 - 2

x3=−1−2−3x_3 = -1 - 2 - 3

x4=−1−2−3−4x_4 = -1 - 2 - 3 - 4

For xnx_n, we have:

xn=−1−(2+3+4+...+n)x_n = -1 - (2 + 3 + 4 + ... + n)

The sum 2+3+4+...+n2 + 3 + 4 + ... + n can be written as:

(1+2+3+...+n)−1(1 + 2 + 3 + ... + n) - 1

Using the formula for sum of first nn natural numbers:

2+3+4+...+n=n(n+1)2−12 + 3 + 4 + ... + n = \dfrac{n(n+1)}{2} - 1

Therefore:

xn=−1−(n(n+1)2−1)x_n = -1 - \left(\dfrac{n(n+1)}{2} - 1\right)

xn=−1−n(n+1)2+1x_n = -1 - \dfrac{n(n+1)}{2} + 1

xn=−n(n+1)2x_n = -\dfrac{n(n+1)}{2}

Using our formula xn=−n(n+1)2x_n = -\dfrac{n(n+1)}{2}:

x100=−100(101)2=−101002=−5050x_{100} = -\dfrac{100(101)}{2} = -\dfrac{10100}{2} = -5050

Therefore, x100=−5050x_{100} = -5050

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