Question

find 3 numbers
numbers in A.P whose sum is 24 and where
product is 440​

Answer 1

Answer:

bruh

Step-by-step explanation:

moment

Answer 2

Let’s assume the terms in the A.P to be (a – d), a, (a + d) with common difference as d.

Given conditions,

Sn = 24

(a – d) + a + (a + d) = 3a = 24

a = 24/3 = 8

Therefore:

Product of terms = 440

(a – d) x a x (a + d) = 440

a(a2 – d2) = 440

8(64 – d2) = 440

(64 – d2) = 55

d2 = 9

d = ± 3

Hence,

When a = 8 and d = 3, we have

A.P. = 5, 8, 11

And, when a = 8 and d = -3 we have

A.P. = 11, 8, 5