Question

pls answer this question very urgent ​

Answer 1

Step-by-step explanation:

Let the four parts be (a - 3d), (a - d), (a + d) and (a + 3d).

then, sum = 56

= (a - 3d) + (a - d) + ( a + d) + (a + 3d)

= 56

=4a = 56

=a=14

it is given that

$\frac{{a - 3d}}{{a - d}}\frac{{a + 3d}}{{a +d}} = \frac{5}{6}$

$= \frac{{a}^{2}}{{a}^{2}}\frac{{ - 9d}^{2}}{{ - d}^{2}} = \frac{5}{6}$

$= \frac{{{14}}^{2} }{ {{14}}^{2} } \frac{ { - 9d}^{2} }{ { - 9d}^{2} }$

$= \: 1176 - {54d}^{2} = 980 - {5d}^{2}$

$= 1176 \: - \: 980 = {54d}^{2} - {5d}^{2}$

$= {49}^{2} = 196$

$= {d}^{2} = 4$

$= d = 2$

Thus, the four parts are a - 3d, a - d, a + d, a + 3d i.e., 8, 12, 16, and 20.

Answer 2

Step-by-step explanation:

Let the four parts of 56 in AP be $a - 3d, a - d, a + d, a + 3d\\</p><p>\therefore a - 3d + a - d + a + d + a + 3d = 56\\</p><p>\therefore 4a = 56\\</p><p>\therefore a = \frac{56}{4}\\</p><p>\therefore a = {14}$

Now,

$Product\: of \:extremes \\</p><p>=(a - 3d)(a + 3d) \\</p><p>= a^2 - (3d)^2 \\ </p><p>= (14)^2 - 9d^2 \\ </p><p>= 196 - 9d^2 \\</p><p></p><p>Product\: of \:means \\</p><p>=(a - d)(a + d) \\</p><p>=a^2 - d^2 \\</p><p>=(14)^2 - d^2 \\</p><p>= 196 - d^2$

According to the given condition:

$\frac {Product\: of \:extremes}{Product\: of \:means}=\frac {5}{6}\\</p><p>\therefore \frac {196 - 9d^2}{196 - d^2}=\frac {5}{6}\\</p><p>\therefore 6({196 - 9d^2})=5({196 - d^2})\\</p><p>\therefore 6\times 196 - 54d^2=5\times196 - 5d^2\\</p><p>\therefore 6\times 196 - 5\times 196= 54d^2 - 5d^2</p><p>\\</p><p>\therefore 196 = 49d^2 \\</p><p>\therefore d^2= \frac {196}{49}\\</p><p>\therefore d^2= 4\\</p><p>\therefore d= \sqrt 4\\</p><p>\therefore d= 2\\</p><p>\implies\\ </p><p>a - 3d = 14 - 3\times 2= 14 - 6 = 8\\ a - d = 14 - 2= 12\\</p><p>a + d = 14 + 2= 16\\</p><p>a + 3d = 14 + 3\times 2= 14 + 6 = 20$

Thus, four parts of 56 are: 8, 12, 16 & 20.

Verification:

8 + 12 + 16 + 20 = 56