Arithmetic Progressions
RD Sharma
CBSE
Grade 10
Question:
How many terms of the A.P. $9, 17, 25, \dots$ must be taken to give a sum of $636$?
Step-by-Step Solution
Key Concept: $S_n = \dfrac{n}{2}[2(9) + (n-1)8] = 636 \Rightarrow \dfrac{n}{2}[18 + 8n - 8] = 636 \Rightarrow n(4n + 5) = 636 \Rightarrow 4n^2 + 5n - 636 = 0 \Rightarrow (4n + 53)(n - 12) = 0 \Rightarrow n = 12$.
$\dfrac{n}{2}[18 + 8n - 8] = 636 \Rightarrow n(4n + 5) = 636 \Rightarrow 4n^2 + 5n - 636 = 0$. [1.0 Mark]
$(4n + 53)(n - 12) = 0$. [1.0 Mark]
Since $n$ must be a positive integer, $n = 12$. $12$ terms must be taken. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Forming quadratic $4n^2 + 5n - 636 = 0$: 1.0 Mark
Factoring into $(4n + 53)(n - 12) = 0$: 1.0 Mark
Solving $n = 12$: 1.0 Mark
Correct Answer:
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