Question

In an arithmetic progression sum of the first 3 terms is 12 and their product is 28. Find the numbers​

Answer 1

Answer:

hi

Step-by-step explanation:

hi gheoejheiehehdidvehHbsuejidisbhwgsfatywuwopwkhrbdbdhshvevshbsbdhdhjekwkwnsheujebekshehejsox

Answer 2

Answer:

I HOPE U LIKE MY ANSWER

Step-by-step explanation:

let the three terms in A.P. be,

$(x-a),(x),(x+a)$

or,

$Given, \\(x-a)+x+(x+a)=12\\or, \ 3x=12\\or, \ x=4$

or,

$(4-a)(4+a)(4)=28\\or, \ 4(16-a^2)=28\\or, \ 16-a^2=7\\or, \ a^2=9\\or, \ a= \ +3,-3$

the numbers are,

$if, \ "a" \ is \ '3'\\4-a=4-3=1, \ 4, \ 4+a=4+3=7\\\\or, \ numbers \ are \ 1,4,7$

&

$if \ "a" \ is \ '(-3)'\\4-a=4-(-3)=7, \ 4, \ 4+a=4+(-3)=1\\\\or, \ numbers \ are \ 7,4,1$