Question

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

Answer 1

first term, a = 24
common difference, d = 21-24 = -3
let the number of terms to get sum 78 is n.

 $S_n=78\\ \\ \Rightarrow \frac{n}{2}(2a+(n-1)d)=78\\ \\ \Rightarrow \frac{n}{2}(2 \times 24+(n-1)(-3))=78\\ \\ \Rightarrow n(48-3n+3)=78 \times 2\\ \\ \Rightarrow -3n^2+51n-156=0\\ \\ \Rightarrow 3n^2-51n+156=0$

Solving the quadratic equation, we get n=4 and n=13.
So you can take either 4 terms or 13 terms to get the sum 78.

Answer 2

Answer:

Step-by-step explanation:

Hi!!!!☺☺☺

first term, a = 24

common difference, d = 21-24 = -3

let the number of terms to get sum 78 is n.

Sn = n/2 (2a +( n-1) d)

78 = n/2(48+ (-3)n + 3)

78 = n/2(51 + (-3)n)

78*2=n (51+ (-3)n)

156= 51n + -3n^2

3n^2-51n+156=0

n^2 - 17n +52 =0

n^2-13n-4n +52 =0

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

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

n = 4,13

 

Solving the quadratic equation, we get n=4 and n=13.

So you can take either 4 terms or 13 terms to get the sum 78.