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

Answer:

let the numbers be (a-3d),(a-d),(a+d),(a+3d)

according to question : (a-3d)+(a-d)+(a+d)+(a+3d)=32

⟹4a =32

⟹a = 32/4

⟹a = 8

also,

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

$⟹15 {a}^{2} - 135 {d}^{2} = 7 {a}^{2} - 7 {d}^{2}$

$⟹8 {a}^{2} = 128 {d}^{2}$

$⟹ {d}^{2} = \frac{8 {a}^{2} }{128}$

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

$⟹ {d}^{2} = 4$

$⟹d = ±2$

if d = +2,then the numbers are :2,6,10,14

if d = -2,then the numbers are :14,10,16,2

Answer 2

Answer:

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

Step-by-step explanation:

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

(i) Sum of four is 32:

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

⇒ 4a = 32

⇒ a = 8.

(ii) Ratio of product of 1st and last term to extremes is 7:15

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

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

⇒ 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 numbers are 2,6,10,14 (or) 14,10,6,2

Hope it helps!