Question

Find the series of the first 10 consecutive odd numbers and also find the sum of them?
(a) $1+3+5+7+9+11+13+15+17+19=100$,
(b) $1+3+5+7+9+11+13+16+17+19=101$,
(c) $1+3+5+7+10+11+13+15+17+19=101$,
(d) $1+3+6+6+9+11+13+15+17+19=100$.

Answer 1

Hint: We start solving this by recalling the general equation to represent the odd natural numbers. We substitute the values $1,2,3,......10$ in place of n of the general equation to get the first 10 consecutive odd natural numbers. We then find the sum of the first n-consecutive odd numbers and then substitute 10 in place of n to get the required result.

Complete step-by-step answer:
According to the problem we need to find the series and sum of the series of the first 10 consecutive positive odd numbers.
We know that the general equation to the odd natural numbers is $2n-1$, $\left( n\ge 1 \right)$.
We need first 10 consecutive odd numbers. Let us substitute $n=1,2,3,4,.....,10$.
The odd number we get on substituting $n=1$ is $\left( 2\left( 1 \right)-1 \right)=2-1$.
The odd number we get on substituting $n=1$ is 1.
The odd number we get on substituting $n=2$ is $\left( 2\left( 2 \right)-1 \right)=4-1$.
The odd number we get on substituting $n=2$ is 3.
The odd number we get on substituting $n=3$ is $\left( 2\left( 3 \right)-1 \right)=6-1$.
The odd number we get on substituting $n=3$ is 5.
The odd number we get on substituting $n=4$ is $\left( 2\left( 4 \right)-1 \right)=8-1$.
The odd number we get on substituting $n=4$ is 7.
The odd number we get on substituting $n=5$ is $\left( 2\left( 5 \right)-1 \right)=10-1$.
The odd number we get on substituting $n=5$ is 9.
The odd number we get on substituting $n=6$ is $\left( 2\left( 6 \right)-1 \right)=12-1$.
The odd number we get on substituting $n=6$ is 11.
The odd number we get on substituting $n=7$ is $\left( 2\left( 7 \right)-1 \right)=14-1$.
The odd number we get on substituting $n=7$ is 13.
The odd number we get on substituting $n=8$ is $\left( 2\left( 8 \right)-1 \right)=16-1$.
The odd number we get on substituting $n=8$ is 15.
The odd number we get on substituting $n=9$ is $\left( 2\left( 9 \right)-1 \right)=18-1$.
The odd number we get on substituting $n=9$ is 17.
The odd number we get on substituting $n=10$ is $\left( 2\left( 10 \right)-1 \right)=20-1$.
The odd number we get on substituting $n=10$ is 19.
The sum of the series of the first 10 consecutive positive numbers is represented as $1+3+5+7+9+11+13+15+17+19$.
Let us find the sum of n consecutive odd positive numbers. We need to find $\sum\limits_{r=1}^{n}{\left( 2r-1 \right)}$.
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2r-1 \right)}=\sum\limits_{r=1}^{n}{2r-\sum\limits_{r=1}^{n}{1}}$.
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2r-1 \right)}=2\sum\limits_{r=1}^{n}{r-\sum\limits_{r=1}^{n}{1}}$ ---(1).
We know that sum of first n natural numbers is $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$ and sum of n times of a given number is $\sum\limits_{r=1}^{n}{a}=an$. We use this results in equation (1).
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2r-1 \right)}=2\times \left( \dfrac{n\left( n+1 \right)}{2} \right)-n$.
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2r-1 \right)}={{n}^{2}}+n-n$.
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2r-1 \right)}={{n}^{2}}$.
We have found the sum of first n consecutive odd positive numbers is ${{n}^{2}}$. We substitute 10 in place of n to get the sum of first 10 consecutive odd positive numbers.
$\Rightarrow 1+3+5+7+9+11+13+15+17+19={{10}^{2}}$.
$\Rightarrow 1+3+5+7+9+11+13+15+17+19=100$.
We have found the series and sum of the series of the first 10 consecutive positive odd numbers as $1+3+5+7+9+11+13+15+17+19=100$.
∴ The series and sum of the series of the first 10 consecutive positive odd numbers as $1+3+5+7+9+11+13+15+17+19=100$.
The correct option for the given problem is (a).

Note: We consider positive odd numbers if nothing particular is mentioned about the odd numbers in the problem. If we have taken the negative odd numbers, then the result will differ from the problem. We can also perform addition directly to the obtained numbers to get the required results. We can also expect problems to find the sum of consecutive even numbers and first n natural numbers.