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प्रश्न
Solve:
25 + 22 + 19 + 16 + ... + x = 115
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उत्तर
25 + 22 + 19 + 16 + ... + x = 115
Here, a = 25, d = \[-\] 3,
\[S_n = 115\]
\[\text { We know }: \]
\[ S_n = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]
\[ \Rightarrow 115 = \frac{n}{2}\left[ 2 \times 25 + (n - 1) \times \left( - 3 \right) \right]\]
\[ \Rightarrow 115 \times 2 = n\left[ 50 - 3n + 3 \right]\]
\[ \Rightarrow 230 = n\left( 53 - 3n \right)\]
\[ \Rightarrow 230 = 53n - 3 n^2 \]
\[ \Rightarrow 3 n^2 - 53n + 230 = 0\]
\[\text { By quadratic formula: } \]
\[ n = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[\text { Substituting a = 3, b = - 53 and c = 230, we get: } \]
\[n = \frac{53 \pm \sqrt{\left( 53 \right)^2 - 4 \times 3 \times 230}}{2 \times 3} = \frac{46}{6}, 10\]
\[ \Rightarrow n = 10, as n \neq \frac{46}{6}\]
\[ \therefore a_n = a + (n - 1)d\]
\[ \Rightarrow x = 25 + (10 - 1)( - 3)\]
\[ \Rightarrow x = 25 - 27 = - 2\]
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