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प्रश्न
Solve:
1 + 4 + 7 + 10 + ... + x = 590.
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उत्तर
1 + 4 + 7 + 10 + ... + x = 590
Here, a = 1, d = 3,
\[\text { We know: } \]
\[ S_n = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]
\[ \Rightarrow 590 = \frac{n}{2}\left[ 2 \times 1 + (n - 1) \times \left( 3 \right) \right]\]
\[ \Rightarrow 590 \times 2 = n\left[ 2 + 3n - 3 \right]\]
\[ \Rightarrow 1180 = n\left( 3n - 1 \right)\]
\[ \Rightarrow 1180 = 3 n^2 - n\]
\[ \Rightarrow 3 n^2 - n - 1180 = 0\]
\[\text { By quadratic formula: } \]
\[ n = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[\text { Substituting a = 3, b = - 1 and c = - 1180, we get }: \]
\[ \Rightarrow n = \frac{1 \pm \sqrt{\left( 1 \right)^2 + 4 \times 3 \times 1180}}{2 \times 3} = \frac{- 118}{6}, 20\]
\[ \Rightarrow n = 20, \text { as } n \neq \frac{- 118}{6}\]
\[ \therefore a_n = x = a + (n - 1)d\]
\[ \Rightarrow x = 1 + (20 - 1)(3)\]
\[ \Rightarrow x = 1 + 60 - 3 = 58\]
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