Question

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

Answer 1

Given AP : 24, 21, 18,....
a = 24
d = 21-24 = -3
Sn = 78
n = ?
Sn = n/2 [ 2a + (n-1)d]
⇒78 = n/2[ 2(24) + (n-1)(-3)
⇒78×2 = n[ 48 - 3n + 3]
⇒156 = n[51 - 3n]
⇒156 = 51n -3n²
⇒3n² - 51n + 156 = 0
⇒3[ n² - 17n + 52] = 0
⇒n² - 17n + 52 = 0
⇒n² - 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
Therefore 13 or 4 terms must be taken so that their sum is 78

Answer 2

Solution:

Given:

=> a = 24

=> d = 24 - 21 = -3

=> Sn = 78

To Find:

=> Number of terms taken so that sum is 78.

Formula used:

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

Now, put the values in the formula, we get

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

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

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

$\sf{\implies 156=n(51-3n)}$

=> 156 = 51n - 3n²

=> 3n² - 51n + 156 = 0

Take 3 common,

=> n² - 17n + 52 = 0

=> n² - 13n - 4n + 52 = 0

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

=> (n - 4) (n - 13)

=> n = 4, 13

So, the number of terms is either 4 or 13.