Question

The sum of 4 consecutive numbers in an AP is 32 and ratio of the product of the first and last term to the product of middle 2 terms is 7:15.Find the numbers

Answer 1

The answer is given below :

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

Given that,

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

=> 4a = 32

=> a = 8

So, the numbers are

(8 - 3d), (8 - d), (8 + d) and (8 + 3d).

Given that :

(8 - 3d)(8 + 3d) : (8 - d)(8 + d) = 7 : 15

=> (64 - 9d²) : (64 - d²) = 7 : 15

=> (64 - 9d²)/(64 - d²) = 7/15

=> 960 - 135d² = 448 - 7d²

=> 128d² = 512

=> d² = 4

So, d = ± 2.

So, the numbers are :

2, 6, 10, 14

or,

14, 10, 6, 2.

Therefore, the four consecutive numbers are

2, 6, 10, 14.

Thank you for your question.

Answer 2

Answer:

→ 2, 6, 10, 14 .

Step-by-step explanation:

Note :- This question is come in CBSE class 10th board 2018 .

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

⇒ 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

Hence, it is solved .

THANKS

#BeBrainly.