Question

Find four number in A.P such that the sum of the first and last number is 8 and the product of second And third number is 12 give solution ple answer

Answer 1

$\huge\textbf{ANSWER}$
$<b>$
Let the four terms be 

a-2d , a-d , a+d, a+2d

now, it's given that a + 2d + a - 2d = 8

2a=8

a=4

Also we are given that,

$(a-d)(a+d)= {a}^{2} - {b}^{2} =12$

${4}^{2} -d^2=12$

$16-12=d^2 \: \\ \\ 4=d^2 \: \\ \\ {2}^{2} = {d}^{2} \\ \\ 2 = d$

d=2

Now the four numbers are

0,2,6,8

THANKS