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

The Arithmetic Progression is: 24, 21, 18, ...

First term $a = 24$

Common difference $d = 21 - 24 = -3$

Required sum $S_n = 78$

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The formula for sum of n terms in an AP is:

$S_n = \dfrac{n}{2}[2a + (n-1)d]$

Substituting the values:

$78 = \dfrac{n}{2}[2(24) + (n-1)(-3)]$

$78 = \dfrac{n}{2}[48 - 3n + 3]$

$78 = \dfrac{n}{2}[51 - 3n]$

$156 = n[51 - 3n]$

$156 = 51n - 3n^2$

$3n^2 - 51n + 156 = 0$

$n^2 - 17n + 52 = 0$

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Factoring the quadratic equation:

$(n - 4)(n - 13) = 0$

Therefore $n = 4$ or $n = 13$

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For $n = 4$:

Terms: 24, 21, 18, 15

Sum $= 24 + 21 + 18 + 15 = 78$

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For $n = 13$:

$S_{13} = \dfrac{13}{2}[48 + 12(-3)]$

$S_{13} = \dfrac{13}{2}[48 - 36]$

$S_{13} = \dfrac{13}{2} \times 12$

$S_{13} = 78$

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Both answers are valid because after the 4th term (15), the terms continue: 12, 9, 6, 3, 0, -3, -6, -9, -12...

The sum of terms from 5th to 13th equals zero, so the sum of 13 terms equals the sum of the first 4 terms.

Therefore, both 4 and 13 terms can be taken to get a sum of 78.