Question

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

Answer 1

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Your answer is given in the attachment

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Answer 2

$hi \: ..$
$here's \: \: your \: \: answer$
24 , 21 , 18 , .... is an given A.P

Here..

a = 24
d = 21 - 24 = -3

Sn = 78
n = ?

We know that ,
$sn = \frac{n}{2} (2a + (n - 1)d)$

$78 = \frac{n}{2} (2 \times 24 + (n - 1)-3)$

$78 = \frac{n}{2} (48 - 3n + 3)$

$78 = \frac{n}{2} (51 - 3n)$

$78 \times 2 = n(51 - 3n)$

$156 = 51n - 3 {n}^{2}$

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

$(dividing \: through \: by \: 3)$

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

${n}^{2} - 13n - 4n + 52 = 0$

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

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

$n - 13 = 0 \: \: or \: \: n - 4 = 0$

$n = 13 \: \: or \: \: n = 4$
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Thus , The number of terms that must be taken so that their sum is 78 are 13 or 4.
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$hope \: \: this \: \: helps...$