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Question 55
If $$x_1=-1$$ and $$x_m=x_{m+1}+(m+1)$$ for every positive integer m, then $$X_{100}$$ equals
Solution
$$x_1=-1$$
$$x_1=x_2+2$$ => $$x_2=x_1-2$$ = -3
Similarly,
$$x_3=x_1-5=-6$$
$$x_4=-10$$
.
.
The series is -1, -3, -6, -10, -15......
When the differences are in AP, then the nth term is $$-\frac{n\left(n+1\right)}{2}$$
$$x_{100}=-\frac{100\left(100+1\right)}{2}=-5050$$
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