Question

The sum of 3 numbers in an A.P. is 90 and their product is 24000 then find the numbers.

Answer 1

Given:-

Sum of 3 A.P = 90

Product = 24000

Now,

Numbers = $\alpha - \beta , \alpha , \alpha + \beta$

Sum = 90

$( \alpha - \beta ) + \alpha + ( \alpha + \beta ) = 90$

$3 \alpha = 90$

$\alpha = 30$

Again,

Product= 24000

$( \alpha - \beta ) \times \alpha \times ( \alpha + \beta ) = 24000$

$\alpha \times ( { \alpha }^{2} - { \beta }^{2} ) = 24000$

$30 \times ( {30}^{2} - { \beta }^{2} ) = 24000$

$900 - { \beta }^{2} = 800$

${ \beta }^{2} = 100$

$\beta = \sqrt{100}$

$\beta = \pm10$

Now,

A.P = 30-10,30,30+10 = 20,30,40

Or, If we take negative.

A.P = 30+10,30,30-10 = 40,30,20

Answer 2

Step-by-step explanation:

let,

the numbers are - (a-d), a, (a+d)

then,

the sum of the three nos

a-d+a+a+d = 90

=> 3a = 90

=> a = 30

and the product

(a-d). a . (a+d) = 24000

=> (a² - d²)a = 24000

=> (30² - d²). 30 = 24000

=> (900 - d²) = 24000/30 = 800

=> - d² = - 100

=> d² = 10²

=> d = 10

there for,

nos are

a-d = 30-10 = 20

a = 30

a+d = 30+10 = 40

please mark me as brainlist

and follow me and rate me