sum of natural numbers from 1 to N is 36 find value of n
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sum of natural numbers from 1 to N is 36 find value of n
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Answer:
Question :
The sum of natural numbers from 1 to n is 36. Find value of n.
[tex]\begin{gathered}\end{gathered}[/tex]
Given :
[tex]\begin{gathered}\end{gathered}[/tex]
To Find
[tex]\begin{gathered}\end{gathered}[/tex]
Using Formula :
[tex]{\longrightarrow{\small{\underline{\boxed{\sf{\red{{S_n} = \dfrac{n}{2} \bigg[2a + \bigg(n - 1 \bigg)d\bigg]}}}}}}}[/tex]
[tex]\begin{gathered}\end{gathered}[/tex]
Solution :
Finding the value of n by substituting the values in the formula :
[tex]{:\implies{\small{\sf{{S_n} = \dfrac{n}{2} \bigg[2a + \bigg(n - 1 \bigg)d\bigg]}}}}[/tex]
Substituting the given values
[tex]{:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 \times 1 + \bigg(n - 1 \bigg)1\bigg]}}}}[/tex]
[tex]{:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 + \bigg(n - 1 \bigg)1\bigg]}}}}[/tex]
Multiplying (n - 1) by 1
[tex]{:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 + \bigg(n \times 1 - 1 \times 1\bigg)\bigg]}}}}[/tex]
[tex]{:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 + \bigg(n - 1\bigg)\bigg]}}}}[/tex]
Now, taking RHS (2) to LHS and opening round brackets
[tex]{:\implies{\small{\sf{36 \times 2 = n \bigg[2 + n - 1\bigg]}}}}[/tex]
[tex]{:\implies{\small{\sf{72 = n \bigg[2 + n - 1\bigg]}}}}[/tex]
Now, subtracting 1 from 2
[tex]{:\implies{\small{\sf{72 = n \bigg[1 + n \bigg]}}}}[/tex]
Now, multiplying n into 1 + n
[tex]{:\implies{\small{\sf{72 = \bigg[1 \times n + n \times n \bigg]}}}}[/tex]
[tex]{:\implies{\small{\sf{72 = \bigg[ n + {n}^{2} \bigg]}}}}[/tex]
Opening square brackets
[tex]{:\implies{\small{\sf{72 = n + {n}^{2}}}}}[/tex]
Now, making a equation
[tex]{:\implies{\small{\sf{{n}^{2} + n - 72 = 0}}}}[/tex]
[tex]{:\implies{\small{\sf{{n}^{2} + 9n - 8n - 72 = 0}}}}[/tex]
[tex]{:\implies{\small{\sf{ n(n + 9) - 8(n + 9) = 0}}}}[/tex]
[tex]{:\implies{\small{\sf{(n - 8)(n + 9) = 0}}}}[/tex]
Solving for (n - 8) = 0
[tex]{:\implies{\small{\sf{n - 8 = 0}}}}[/tex]
[tex]{:\implies{\small{\sf{n = 0 + 8}}}}[/tex]
[tex]{\therefore{\sf{\red{ \: n = 8}}}}[/tex]
Solving for (n + 9) = 0
[tex]{:\implies{\small{\sf{n + 9 = 0}}}}[/tex]
[tex]{:\implies{\small{\sf{n = 0 - 9}}}}[/tex]
[tex]{\therefore{\sf{\red{n = - 9}}}}[/tex]
Since, the number of terms n can’t be negative.
Hence, The value of n is 8.
[tex] \rule{220pt}{2.5pt}[/tex]
Learn More :
★ Formula to find the numbers of term of an AP:
[tex]\longrightarrow{\small{\underline{\boxed{\sf{\purple{n= \bigg[ \dfrac{(l - a)}{d} \bigg]}}}}}}[/tex]
★ Formula to find the tsum of first n terms of an AP:
[tex]\longrightarrow{\small{\underline{\boxed{\sf{\purple{S_n= \dfrac{n}{2} \big(a + l \big)}}}}}}[/tex]
★ Formula to find the sum of squares of first n natural numbers of an AP:
[tex]\longrightarrow{\small{\underline{\boxed{\sf{\purple{S = \dfrac{n(n + 1)(2n + 1)}{6} }}}}}}[/tex]
★ Formula to find the nth term of an AP is the square of the number of terms:
[tex]\longrightarrow{\small{\underline{\boxed{\sf{\purple{S = {(n)}^{2} }}}}}}[/tex]
★ Formula to find the sum of of an AP:
[tex]\longrightarrow{\small{\underline{\boxed{\sf{\purple{S = n(n+1)}}}}}}[/tex]
[tex]{\rule{220pt}{3pt}}[/tex]