Question

solve this question for me

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 8
a² = 128d²

Put the value of a = 8 

8(8)² = 128d²128d² = 512d² = 512/128d² = 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

Answer 2

let the numbers be a-3d , a-d , a+d , a+3d
according to the ques
a-3d+a-d+a+d+a+3d = 32
4a = 32
a=8
as given in the ques.
(a-3d)(a+3d)/(a-d)(a+d)=7/15
a^2 - 9d^2/a^2 - d^2 = 7/15
8^2 - 9d^2/8^2 - d^2 =7/15
64-9d^2/64-d^2=7/15
960-135d^2=448-7d^2
512=128d^2
4=d^2
d = +2 , -2
taking d=2
numbers= 2,6,10,14
taking d= -2
numbers= 14,10,6,2


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