Question

The sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:5 find the numbers

Answer 1

Hlw mate!!

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

Hope it helpful

Answer 2

Error : There is minor error in your question .

→ It will be 7:15 instead of 7:5 .

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

= 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

Hence, it is solved .

THANKS

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