Question

first answer will be selected as brainleist.....

Answer 1

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)
ATQ,

 a-3d + a - d + a + d + a + 3d = 32
⇒4a = 32
⇒a = 8 ....... (1)

Also, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15        (Note: (a+b)(a-b)= a² - b² )
15(a² - 9d²) = 7(a² - d²)            
15a² - 135d² = 7a² - 7d²
15a² - 7a² = 135d² - 7d² 
8a² = 128d²

Put the value of a = 8 

8(8)² = 128
d²128
d² = 512
d² = 512/128
d² = 4d = 2

So, a= 8 and d= 2

So the four consecutive numbers will be
(a - 3d) = 8 - (3*2)
⇒8 - 6 = 2

( a - d) = 8 - 2 = 6

(a + d) = 8 + 2 = 10

( a + 3d) = 8 + (3*2)
⇒8 + 6 = 14

Four consecutive numbers are 2, 6, 10 and 14