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 is7:15.Find the numbers.​

Answer 1

Answer:

2, 6, 10, 14

Step-by-step explanation:

let the AP be a, a + d, a + 2d, a + 3d

sum of the numbers is = 32

a + a + d + a + 2d + a + 3d = 32

4a + 6d = 32

2(2a + 3d) = 32

2a + 3d = 16 equ (1)

ratio of products of 1st & 4th to 2nd and 3rd = 7:15

$\frac{a(a + 3d)}{(a + d)(a + 2d)} = \frac{7}{15} \\ \frac{ {a}^{2} + 3ad }{ {a}^{2} + 3ad + 2 {d}^{2} } = \frac{7}{15} \\ 15( {a}^{2} + 3ad) = 7( {a}^{2} + 3ad + 2 {d}^{2} ) \\ 15 {a}^{2} + 45ad = 7 {a}^{2} + 21ad + 14 {d}^{2} \\ 15 {a}^{2} + 45ad - 7 {a}^{2} - 21ad - 14{d}^{2} = 0 \\ 8 {a}^{2} + 24ad - 14 {d}^{2} = 0 \\ 2(4 {a}^{2} + 12ad - 7 {d}^{2} ) = 0 \\ 4 {a}^{2} + 12ad - 7 {d}^{2} = 0 \\ 4 {a}^{2} - 2ad + 14ad - 7 {d}^{2} = 0 \\ 2a(2a - d) + 7d(2a - d) = 0 \\ 2a - d = 0 \: \: \: \: \: \: 2a + 7d = 0 \\ 2a = d \: \: \: \: \: \: \: \: 2a = - 7d$

substitute d = 2a in equ (1)

2a + 3d = 16

2a + 3(2a) = 16

2a + 6a = 16

8a = 16

a = 16/8 = 2

d = 2a

d = 2(2)

d = 4

the AP is

a = 2

a + d = 2 + 4 = 6

a + 2d = 2 + 2(4) = 2 + 8 = 10

a + 3d = 2 + 3(4) = 2 + 12 = 14

the numbers are

2, 6, 10, 14

hope you get your answer