Question

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 terms is 7:15. Find the numbers.....

Answer 1

Just look at the pic This helps u

Answer 2

Let the AP as :-

a - 3d , a - d , a + d , a + 3d

• Sum of AP is 32

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

4a = 32

a = 8

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$


[ Using identity ( a - b ) ( a + b ) = a² - b² ]

$\frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15}$

15a² - 135d² = 7a² - 7d²

15a² - 7a² = - 7d² + 135d²

8a² = 128d²

8 × 64 = 128d²

$\frac{8 \times 64}{128} = {d}^{2}$

4 = d²

√4 = d²

± 2 = d²

______________

To form the AP

If a = 8 and d = 2

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

2 , 6 , 10 , 14

______________

If a = 8 and d = - 2

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

14 , 10 , 6 , 2