Question

The sum of four consecutive numbers it an AP is 32 and the ratio of the product of the
and the latter to the product of two middle terms is 7:15. Find the numbers.​

Answer 1

Answer:

S4 = 32

let the terms be a-d, a, a+d, a+2d

now sum will be,

a-d+a+a+d+a+2d = 32

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

4a + 2d = 32

2a + d = 16

Answer 2

Answer:

Hey there here is your answer,

Let four consecutive numbers in AP be as:

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

Then,

According to question,

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

4a = 32

a = 8

Also,

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \\ \frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15} \\ \frac{ {8}^{2} - 9 {d}^{2} }{ {8}^{2} - {d}^{2} } = \frac{7}{15} \\ 15(64 - 9 {d}^{2} ) = 7(64 - {d}^{2} ) \\ 960 - 135 {d}^{2} = 448 - 7 {d}^{2} \\ 135 {d}^{2} - 7 {d}^{2} = 960 - 448 \\ 128 {d}^{2} = 512 \\ {d}^{2} = \frac{512}{128} \\ {d}^{2} = 4 \\ d = 2$

Now, a = 8 and d = 2,

then,

Numbers are,

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

= 2, 6, 10, 14

Hence, AP: 2, 6, 10, 14

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