consecutive numbeĺin APJs 3Q and

1.	consecutive numbeĺin APJs 3Q and
Answer» Let the four consecutive numbers in AP be; (a - 3d), (a - d), (a + d) and (a + 3d) So, according to the question a - 3d + a - d + a + d + a + 3d = 32 4a = 32 a = 8 Now, (a - 3d)(a + 3d) / (a - d)(a + d) = 7/15 15(a² - 9d²) = 7(a² - d²) 15a² - 135d² = 7a² - 7d² 8a² = 128d² Putting the value of a = 8, we get d² = 4 d = ±2 So, the four consecutive numbers are 2, 6, 10, 14 or 14, 10, 6, 2 yes it is right

Answer»

Let the four consecutive numbers in AP be;

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

So, according to the question

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

4a = 32

a = 8

Now,

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

15(a² - 9d²) = 7(a² - d²)

15a² - 135d² = 7a² - 7d²

8a² = 128d²

Putting the value of a = 8, we get

d² = 4

d = ±2

So, the four consecutive numbers are 2, 6, 10, 14 or 14, 10, 6, 2

yes it is right

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