Question

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

Answer 1

hey friend ur answer

Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.

So,

        a-3d+a-d+a+d+a+3d=32

        4a=32

        a=8 1

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

        15(a2-9d2) = 7(a2-d2)

         15a2-135d2 = 7a2-7d2

          8a2-128d2 = 0

          d2 = 4, ±2

Therefore d=4  or d=±2 1mark

So, when a=8 and d=2, the numbers are 2,6,10,14.

When a=8,d=-2 the numbers are 14,10,6,2




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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.