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Question
Sum of 1 to n natural numbers is 36, then find the value of n.
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Solution
It is given that,
a = 1
d = 1
Sn = 36
Now,
\[S_n = \frac{n}{2}\left( 2a + \left( n - 1 \right)d \right)\]
\[ \Rightarrow 36 = \frac{n}{2}\left( 2\left( 1 \right) + \left( n - 1 \right)\left( 1 \right) \right)\]
\[ \Rightarrow 36 \times 2 = n\left( 2 + n - 1 \right)\]
\[ \Rightarrow 72 = n\left( n + 1 \right)\]
\[ \Rightarrow n^2 + n - 72 = 0\]
\[ \Rightarrow n^2 + 9n - 8n - 72 = 0\]
\[ \Rightarrow n\left( n + 9 \right) - 8\left( n + 9 \right) = 0\]
\[ \Rightarrow \left( n + 9 \right)\left( n - 8 \right) = 0\]
\[ \Rightarrow n = - 9\text { or n} = 8\]
\[ \Rightarrow n = 8 \left( \because n \neq - 9 \right)\]
Hence, the value of n is 8.
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