Question

If \(x_m +1\) and \(x_m=x_{m+1}+(m+1)\) for every positive integer \(m\), then \(x_{100 }\) equals

• A. -5151
• B. -5150
• C. -5051
• D. -5050

Answer 1

The Correct Option is D

The correct answer is (D): \(-5050\)

\(x_m+1=x_m-(m+1)\)

\(x^2=x_1-2=-1-2=-3\)

\(x_3=x_2-3=-1-2-3=-6\)

Similarly,

\(x_n=-(1+2+3+...+n)=-\frac{n(n+1)}{2}\)

Hence \(x_{100}=-\frac{100(101)}{2}=-5050\)

Answer 2

Given,
\(x_1=-1\;and\;x_m=x_{m+1}+(m+1)\)
\(x_{m+1}=x_m-(m+1)\)
Put m value= 1, 2, 3, 4…..
m=1 and x₁=-1 (given)
\(x_{2}=x_1-(2)\)
\(x_{2}=-1-2\)

Now m=2
\(x_{3}=x_2-(2+1)\)
\(x_{3}=x_2-3\)   Now put x₂ value
\(x_{3}=-1-2-3\)
Same for m=3,4,5,6…
\(x_{4}=-1-2-3-4\)
\(x_{5}=-1-2-3-4-5\)
\(x_{6}=-1-2-3-4-5-6\)
And so on,
\(x_{100}=-1-2-3-4-5-6....-100\)

\(x_{100}=-(1+2+3+4+5+6....+100)\)

Sum of n positive numbers \(\frac{n(n+1)}{2}\)

\(x_{100}=-\frac{100(100+1)}{2}\)

\(x_{100}=-50(101)\)

\(x_{100}=-5050\)

So, the correct option is (D): -5050