Question

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of
the first and the last terms to the product of the two middle terms is 7:15.
Find the numbers.
​

Answer 1

Answer:

2,6,10,14

Thanks

I hope this helps you

Please follow me and give me thanks....

Answer 2

Answer:

2,6,10,14 (or) 14,10,6,2

Step-by-step explanation:

Let the four consecutive numbers be (a - 3d), (a - d), (a + d), (a + 3d).

(i) Sum of four is 32:

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

⇒ 4a = 32

⇒ a = 8.

(ii) Ratio of product of 1st and last term to extremes is 7:15

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

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

⇒ 15a² - 135d² = 7a² - 7d²

⇒ 8a² = 128d²

⇒ 8(8)² = 128d²

⇒ 512 = 128d²

⇒ d² = 4

⇒ d = ±2

When a = 8, d = 2:

a - 3d = 2

a - d = 6

a + d = 10

a + 3d = 14

When a = 8, d = -2:

a - 3d = 14

a - d = 10

a + d = 6

a + 3d = 2

Therefore, the numbers are 2,6,10,14 (or) 14,10,6,2

Hope it helps!