Question

helo friend solve this question​

Answer 1

Answer:

14, 10, 6, and 2 or 2, 6, 10 and 14

Step-by-step explanation:

Let's assume the four consecutive numbers in Arithmetic Progression are:

x, x+a, x+2a and x+3a

According to the problem:

x + x+a + x+2a + x+3a = 32

=>4x + 6a = 32

=>2x + 3a = 16.

∴ x = $\frac{16 - 3a}{2}$

And  x(x+3a):(x+a)(x+2a) = 7:15

=>$\frac{x(x + 3a)}{(x+a)(x+2a)} = \frac{7}{15}$

=>$15x^{2}$ $+45ax$ $= 7x^{2}$ $+21ax$$+14a^{2}$

=>$8x^{2}$ $+24ax$ = $+14a^{2}$

=>$4x^{2}$ $+ 12ax$ $=7a^{2}$

=>$4(\frac{16-3a}{2} )^{2}$ $+12a.\frac{16-3a}{2}$ $=7a^{2}$

=>256 + $9a^{2}$ - 96a + 96a - $18a^{2}$ = $7a^{2}$

=>$16a^{2}$ = 256

=>$a^{2}$ = 16

∴ a = ±4

∴ x = $\frac{1}{2}$(16±12) = 14, 2

The numbers are:

14, 10, 6, and 2 or 2, 6, 10 and 14