Question

can someone solve this for me

Answer 1

hey dear

here is your answer

Solution

Let the four consecutive number in AP

Be ( a - 3d) , (a - d) , ( a+ d) and ( a + 3d)

According to question

a - 3d + a - d + a + d + a + 3d = 32

4a = 32
a = 32/4
a = 8

NOW

( a - 3 d) ( a + 3d) / ( a -d) ( a + d) = 7/15

15 ( a^2 - 9d^2) = 7 ( a^2 - d^2)

15a^2 - 135d^2 = 7a^2 - 7d ^2

15a^2 - 7a^2 = 135d^2 - 7d^2

8a^2 = 128 d ^2

Putting the value of a = 8 we get

8 ( 8)^2 = 128d^2

128d^2 = 512

d^2 = 512/ 128

d^2 = 4

d = √4

d = 2

Four consecutive number are

8 - (3*2)
8 - 6 =2

8 - 2 = 6

8 + 2 = 10

8 + (3*2)

8 +6 = 14


so four consecutive number are 2, 6, 10, 14

it is the answer


hope it helps

thank you