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Arithmetic Progressions
CH05 Question Bank
CBSE_CH05_QUESTION_BANK
Grade 10

Question:

How many terms of the AP $24, 21, 18, \ldots$ must be taken so that their sum is $78$? Explain why there are two possible answers.

Step-by-Step Solution

Key Concept: Set up the sum formula as a quadratic in n and solve; check both roots against the context.
$a=24,d=-3$. $S_n=\dfrac n2[2(24)+(n-1)(-3)]=78\Rightarrow \dfrac n2[48-3n+3]=78\Rightarrow n(51-3n)=156$. [1.0 Mark]

$51n-3n^2=156\Rightarrow3n^2-51n+156=0\Rightarrow n^2-17n+52=0\Rightarrow(n-4)(n-13)=0$. [1.0 Mark]

$n=4$ or $n=13$. Both are valid because the terms become negative after the $9$th term ($a_9=24-24=0$, $a_{10}=-3,\ldots$); the extra negative terms added between $n=4$ and $n=13$ exactly cancel out, bringing the sum back to $78$. [1.0 Mark]

Correct Answer:
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