Question

The sum of four cosechtive terms of an Ap are 32 and ratio of the product of their first and last terms to the product of the 2 middle terms is 7:15 find the numbers

Answer 1

Answer:

Four consecutive numbers are 2, 6, 10 and 14

Step-by-step explanation:

Solution:-

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 = 32/4

a = 8 ......(1)

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

15(a² - 9d²) = 7(a² - d²)

15a² - 135d² = 7a² - 7d²

15a² - 7a² = 135d² - 7d² 

8a² = 128d²

Putting the value of a = 8 in above we get.

8(8)² = 128d²

128d² = 512

d² = 512/128

d² = 4

d = 2

So, the four consecutive numbers are

8 - (3*2)

8 - 6 = 2

8 - 2 = 6

8 + 2 = 10

8 + (3*2)

8 + 6 = 14

Four consecutive numbers are 2, 6, 10 and 14