How many terms of the AP: $24, 21, 18, \dots$ must be taken so that their sum is $78$? Explain the double answer.
Step-by-Step Solution
Key Concept: Set $S_n = \dfrac{n}{2}[2(24) + (n-1)(-3)] = 78$, solve quadratic in $n$. Both roots are positive integers because terms from $5^{\text{th}}$ to $13^{\text{th}}$ sum to 0.
Here $a = 24, d = -3, S_n = 78$.
$\dfrac{n}{2}[48 + (n-1)(-3)] = 78 \Rightarrow \dfrac{n}{2}[51 - 3n] = 78 \Rightarrow 51n - 3n^2 = 156$. [1.0 Mark]
Divide by 3: $n^2 - 17n + 52 = 0 \Rightarrow (n - 4)(n - 13) = 0 \Rightarrow n = 4$ or $n = 13$. [1.0 Mark]
Explanation of double answer: The $5^{\text{th}}$ to $13^{\text{th}}$ terms are $12, 9, 6, 3, 0, -3, -6, -9, -12$, whose sum is $0$. Therefore, adding terms beyond $n=4$ up to $n=13$ does not change the total sum. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Forming quadratic equation $n^2 - 17n + 52 = 0$: 1.0 Mark
Solving for $n = 4$ and $n = 13$: 1.0 Mark
Explaining why sum of 5th to 13th terms is 0: 1.0 Mark
Correct Answer: