How many terms of the AP: 24, 21, 18, ldots m | vedclass

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(A) Given the $AP$ is $24, 21, 18, \ldots$ where the first term $a = 24$ and the common difference $d = 21 - 24 = -3$.We are given the sum of $n$ term

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$4$ or $13$

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(A) Given the $AP$ is $24, 21, 18, \ldots$ where the first term $a = 24$ and the common difference $d = 21 - 24 = -3$.We are given the sum of $n$ terms $S_n = 78$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.Substituting the values,we get $78 = \frac{n}{2}[2(24) + (n - 1)(-3)]$.$78 = \frac{n}{2}[48 - 3n + 3] = \frac{n}{2}[51 - 3n]$.$156 = 51n - 3n^2$,which simplifies to $3n^2 - 51n + 156 = 0$.Dividing by $3$,we get $n^2 - 17n + 52 = 0$.Factoring the quadratic equation,we get $(n - 4)(n - 13) = 0$.Thus,$n = 4$ or $n = 13$.Both values are valid because the sum of the terms from $5$ to $13$ is zero.

How many terms of the $AP: 24, 21, 18, \ldots$ must be taken so that their sum is $78$?

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