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Arithmetic Progressions
RD Sharma
CBSE
Grade 10

Question:

Solve for $x$:
$1 + 4 + 7 + 10 + \dots + x = 287$.

Step-by-Step Solution

Key Concept: $a = 1, d = 3$. $x = a_n = 1 + (n-1)3 = 3n - 2$. $S_n = \dfrac{n}{2}(1 + 3n - 2) = \dfrac{n(3n - 1)}{2} = 287 \Rightarrow 3n^2 - n - 574 = 0 \Rightarrow (3n + 41)(n - 14) = 0 \Rightarrow n = 14$. $x = 3(14) - 2 = 40$.
$x = 3n - 2$. $S_n = \dfrac{n(3n - 1)}{2} = 287 \Rightarrow 3n^2 - n - 574 = 0$. [1.0 Mark]
$(3n + 41)(n - 14) = 0 \Rightarrow n = 14$. [1.0 Mark]
$x = 3(14) - 2 = 40$. Value of $x$ is $40$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Forming quadratic $3n^2 - n - 574 = 0$: 1.0 Mark
Solving $n = 14$: 1.0 Mark
Finding $x = 40$: 1.0 Mark

Correct Answer:
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