Arithmetic Sum

$\begin{aligned} Q & = \frac {100+101+102+ \cdots +200}{1+2+3+ \cdots 99} & \small \color{#3D99F6} \text{Using } \frac {n(a+l)}2 \text{; see note.} \\ & = \frac {\frac {101(100+200)}2}{\frac {99(1+99)}2} \\ & = \frac {101(100+200)}{99(1+99)} \\ & = \frac {101(300)}{99(100)} \\ & = \frac {101}{33} \end{aligned}$

$\implies m+n = 101+33 = \boxed{134}$

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Note:

The numerator $100+101+102+ \cdots +200$ is an arithmetic progression of $n_1 = 101$ terms, with first term $a_1 = 100$ and last term $l_1 = 200$ . Similarly, the denominator $1+2+3+ \cdots 99$ has $n_2 = 99$ , $a_2 = 1$ and $l_2 = 99$ . The sum of an arithmetic progression is given by $S = \dfrac {n(a+l)}2$ .

Ratio of two simple sums of arithmetic sequences

Hence 1 + 2 + 3 +...+ 99 = 4950

And 100 + 101 +...+ 200 = 15150

We can then simplify the fraction of these two results :

Since 4950 = 10 × 5 × 3 × 33

And 15150 = 10 × 5 × 3 × 101

We get m = 101 and n = 33

Thus m + n = 134

First, we apply the formula regarding the sum of arithmetic sequence for each sum:

$S_1 = 100+101+...+209 = 0.5× (100+200) × 101 = 0.5× 300 × 101$

$S_2 = 1+2+3+...+99 = 0.5× (1+99) × 99 = 0.5× 100 × 99$

Their ratio:

$\frac {S_1}{S_2} = \frac {0.5× 300 × 101}{0.5× 100 × 99} = \frac {303}{99} = \frac {101}{33}$

Hence, m = 101 and n = 33, and our answer should be:

$m + n = 101 + 33 = \boxed {134}$

Ratio, m / n = [(100 + 200)(200 – 100 + 1) / 2] ÷ [(1 + 99)(99 – 1 + 1) / 2]
= [(100 + 200)(200 – 99)] / [(1 + 99)(99 – 0)]
= [(300)(101)] / [(100)(99)]
= 3 × 101 / 99
= 101 / 33

Answer
= m + n
= 134