Question

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CLASS-X

The sum of four 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 term is 7:15. Find the numbers. ​

Answer 1

Answer:

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

Step-by-step explanation:

Let four consecutive terms of an AP be a-3d, a - d, a + d and a + 3d.

(i) Sum of four numbers is 32:

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

⇒ 4a = 32

⇒ a = 8.

(ii) Product of 1st and last to middle is 7:15.

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

⇒ 15(a² - 9d²) = 7(a² - d²)

⇒ 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 four consecutive numbers are 2,6,10,14 (or) 14,10,6,2.

Hope it helps!

Answer 2

Answer:

The 4 numbers are 14 , 10 , 6 , 2 .

Step-by-step explanation:

Let the 4 terms be a - 3 d , a - d , a + d , a + 3 d

Given :

The sum of the terms = 32 .

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

= 4 a = 32

= a = 32/4

= a = 8

So a = 8

Given :

ratio of product of first and last term to middle terms = 7 : 15

So :

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

= >  ( a² - 9 d² ) : ( a² - d²  ) = 7 : 15

put a = 8 :

= > ( 64 - 9 d² )/( 64 - d² ) = 7/15

= > 15 ( 64 - 9 d² ) = 7 ( 64 - d² )

= > 960 - 135 d² = 448 - 7 d²

= > 128 d² = 960 - 448

= > 128 d² = 512

= > d² = 512/128

= > d² = 4

= > d = ± 2

The 2 numbers will be :

8 + 6 , 8 + 2 , 8 - 2 , 8 - 6

= 14 , 10 , 6 , 2

The 4 numbers are 14 , 10 , 6 , 2 .