Answer:
2, 6, 10, 14
or
14, 10, 6, 2.
Step-by-step explanation:
Let the four consecutive numbers in A.P. be a - 3d, a - d, a + d, a + 3d
According to the first condition:
a - 3d + a - d + a + d + a + 3d =32
4a = 32
a = 32/4
a = 8
According to the second condition:
[tex] \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \\ \\ \frac{ {a}^{2} - (3d)^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15} \\ \\ \frac{ {a}^{2} - 9d^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15} \\ \\ \frac{ {8}^{2} - 9d^{2} }{ {8}^{2} - {d}^{2} } = \frac{7}{15} \\ \\ \frac{ 64 - 9d^{2} }{ 64 - {d}^{2} } = \frac{7}{15} \\ \\ 15(64 - 9 {d}^{2} ) = 7(64 - {d}^{2} ) \\ \\ 960 - 135 {d}^{2} = 448 - 7 {d}^{2} \\ \\ 960 - 448 = 135 {d}^{2} - 7 {d}^{2} \\ \\ 512 = 128 {d}^{2} \\ \\ {d}^{2} = \frac{512}{128} \\ \\ {d}^{2} = 4 \\ \\ d = \pm 2 \\ \\ when \: d = 2 \\ a - 3d = 8 - 3 \times 2 = 8 - 6 = 2 \\ a - d = 8 - 2 = 6 \\ a + d = 8 + 2 = 10 \\ a + 3d = 8 + 3 \times 2 = 8 + 6 = 14 \\\\ when \: d = - 2 \\ a - 3d = 8 - 3 \times ( -2 ) = 8 + 6 = 14 \\ a - d = 8 -( - 2 )= 8 + 2 = 10 \\ a + d = 8 + ( - 2 )= 8 - 2 = 6\\ a + 3d = 8 + 3 \times ( - 2 )= 8 - 6 = 2 \\[/tex]
Thus the four consecutive numbers of the A. P. are 2, 6, 10, 14 or 14, 10, 6, 2.
