Question

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

Answer 1

use formula Sn=n/2(2a+(n-1)d)
$78 = \frac{n}{2} (2 \times 24 + (n - 1)( - 3)) \\156 = n(48 - 3n + 3) \\ 156 = n(51 - 3n) \\ 52 = n(17 - n) \\ {n}^{2} - 17n + 52 = 0 \\ {n}^{2} - 13n - 4n + 52 = 0 \\ n(n - 13) - 4(n - 13) = 0 \\ (n - 4)(n - 13) = 0$
either n=13 or n=4
by putting and checking we get n=4
so we reject n=13.

hence 4 is the answer

Answer 2

AP Series = 24, 21, 18

a1 = 34

d = 21 - 24 = -3

Find the term an in term of n:

an = a1 + (n - 1)d

an = 24 + (-3)(n - 1)

an = 24 - 3n + 3

an = 27 - 3n

Formula for the sum of nth terms:

Sn = n/2 (a1 + an)

Find n:

78  = n/2 [24 + (27 - 3n) ]

78  = n/2 [24 + 27 - 3n ]

156 = n (51 - 3n)

3n² - 51n + 156 = 0

n² - 17n + 52 = 0

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

n = 4 or n = 13

Check validity:

When n = 4,

a4 = 27 - 3(4)

a4 = 15

S4 = 4/2(24 + 15)

S4 = 78 ------------------------ Verified

When n = 13

a13 = 27 - 3(13) = -12

s13 = 13/2(24 + (-12))

s13 = 78  ------------------------ Verified

Answer: We can take 4 terms or 13 terms to have a sum of 78