
Check if the given statement is correct or not “If $\sum{n}=55$, then $n=5$”.
(a) True
(b) False
Answer
615.3k+ views
Hint: Use the formula for calculating the sum of ‘n’ positive integers, which is $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$. Equate this formula to 55 and simplify to form a quadratic equation. Factorize the equation using splitting the middle term method. Solve to calculate the value of ‘n’ which satisfies the equation and check if the given statement is correct or not.
Complete step-by-step answer:
We have to check if the given statement is correct or not.
To do so, we will use the formula for calculating the sum of ‘n’ positive integers, which is $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$.
Thus, we have $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}=55$.
Cross multiplying the terms, we have $n\left( n+1 \right)=2\times 55=110$.
Simplifying the above equation, we have ${{n}^{2}}+n-110=0$. We observe that this is a quadratic equation with degree 2 as the degree of any polynomial is the highest exponent of any variable. We will now factorize this equation using splitting the middle term method.
We can rewrite the above equation as ${{n}^{2}}+11n-10n-110=0$.
Taking out the common terms, we have $n\left( n+11 \right)-10\left( n+11 \right)=0$.
Thus, we have $\left( n+11 \right)\left( n-10 \right)=0$. So, the possible values of ‘n’ which satisfy the given equation are $n=10,-11$.
Thus, we observe that $n=5$ doesn’t satisfy the given equation.
Hence, the given statement is incorrect, which is option (b).
Note: We can also solve this question by calculating the sum of the first 5 positive integers and check if they add up to 55 or not. We can prove the formula $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$ by applying induction on ‘n’. We can also find the roots of the quadratic equation by calculating its discriminant or completing the square method.
Complete step-by-step answer:
We have to check if the given statement is correct or not.
To do so, we will use the formula for calculating the sum of ‘n’ positive integers, which is $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$.
Thus, we have $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}=55$.
Cross multiplying the terms, we have $n\left( n+1 \right)=2\times 55=110$.
Simplifying the above equation, we have ${{n}^{2}}+n-110=0$. We observe that this is a quadratic equation with degree 2 as the degree of any polynomial is the highest exponent of any variable. We will now factorize this equation using splitting the middle term method.
We can rewrite the above equation as ${{n}^{2}}+11n-10n-110=0$.
Taking out the common terms, we have $n\left( n+11 \right)-10\left( n+11 \right)=0$.
Thus, we have $\left( n+11 \right)\left( n-10 \right)=0$. So, the possible values of ‘n’ which satisfy the given equation are $n=10,-11$.
Thus, we observe that $n=5$ doesn’t satisfy the given equation.
Hence, the given statement is incorrect, which is option (b).
Note: We can also solve this question by calculating the sum of the first 5 positive integers and check if they add up to 55 or not. We can prove the formula $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$ by applying induction on ‘n’. We can also find the roots of the quadratic equation by calculating its discriminant or completing the square method.
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