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Best answer

Let the sum of the odd positive integers from $1$ to $100$ be $’x’,$ and the sum of the even positive integers from $150$ to $200$ be $’ y’.$

Now, $x = 1 + 3 + 7 + \ldots + 99$

Here, we have $a = 1,d = 2,l = 99$

$T_{n} = l = a+(n-1)d$

$\implies 99 = 1 + (n-1)2$

$\implies 2n – 2 = 98$

$\implies n = 50$

$\therefore x = \frac{50}{2} (1 + 99) = 2500 \quad [\because S_{n} = \frac{n}{2}(a + l)]$

And, $y = 150 + 152 + 154 + \ldots + 200$

Here, we have $a = 150,d = 2,l = 200$

$T_{n} = l = a+(n-1)d$

$200 = 150 + (n – 1)2$

$\implies 2n - 2 = 50$

$\implies n = 26$

$\therefore y = \frac{26}{2} (150 + 200) = 13 \cdot 350 = 4550$

Now, the required ratio $ = \dfrac{x}{y} = \dfrac{2500}{4550} = \dfrac{50}{91}.$

So, the correct answer is $(C).$

Now, $x = 1 + 3 + 7 + \ldots + 99$

Here, we have $a = 1,d = 2,l = 99$

$T_{n} = l = a+(n-1)d$

$\implies 99 = 1 + (n-1)2$

$\implies 2n – 2 = 98$

$\implies n = 50$

$\therefore x = \frac{50}{2} (1 + 99) = 2500 \quad [\because S_{n} = \frac{n}{2}(a + l)]$

And, $y = 150 + 152 + 154 + \ldots + 200$

Here, we have $a = 150,d = 2,l = 200$

$T_{n} = l = a+(n-1)d$

$200 = 150 + (n – 1)2$

$\implies 2n - 2 = 50$

$\implies n = 26$

$\therefore y = \frac{26}{2} (150 + 200) = 13 \cdot 350 = 4550$

Now, the required ratio $ = \dfrac{x}{y} = \dfrac{2500}{4550} = \dfrac{50}{91}.$

So, the correct answer is $(C).$