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Hint- Here, we will be assuming the two consecutive odd integers as $n$ and $\left( {n + 2} \right)$.

Let the two consecutive odd integers be n and $\left( {n + 2} \right)$.

Given, the sum of the two consecutive odd integers is 20

i.e., $n + \left( {n + 2} \right) = 20 \Rightarrow 2n + 2 = 20 \Rightarrow 2n = 18 \Rightarrow n = 9$

$ \Rightarrow \left( {n + 2} \right) = 9 + 2 = 11$

Therefore, the required two consecutive odd integers are 9 and 11.

Note- There always occur one even number in between two odd numbers and similarly, one odd number in between two even numbers that’s why if one odd number is $n$ then $\left( {n + 1} \right)$ will be an even number and $\left( {n + 2} \right)$ be will odd number.

Let the two consecutive odd integers be n and $\left( {n + 2} \right)$.

Given, the sum of the two consecutive odd integers is 20

i.e., $n + \left( {n + 2} \right) = 20 \Rightarrow 2n + 2 = 20 \Rightarrow 2n = 18 \Rightarrow n = 9$

$ \Rightarrow \left( {n + 2} \right) = 9 + 2 = 11$

Therefore, the required two consecutive odd integers are 9 and 11.

Note- There always occur one even number in between two odd numbers and similarly, one odd number in between two even numbers that’s why if one odd number is $n$ then $\left( {n + 1} \right)$ will be an even number and $\left( {n + 2} \right)$ be will odd number.