Divide 32 into two parts such …
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Sia ? 6 years, 5 months ago
Let the four parts be (a - 3d), (a - d), (a + d) and (a + 3d). Then,
Sum of the numbers = 32
{tex}\Rightarrow{/tex}(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32 {tex}\Rightarrow{/tex} 4a = 32 {tex}\Rightarrow{/tex} a = 8
It is given that
{tex}\frac { ( a - 3 d ) ( a + 3 d ) } { ( a - d ) ( a + d ) } = \frac { 7 } { 15 }{/tex}
{tex}\Rightarrow \quad \frac { a ^ { 2 } - 9 d ^ { 2 } } { a ^ { 2 } - d ^ { 2 } } = \frac { 7 } { 15 }{/tex}
{tex}\Rightarrow \quad \frac { 64 - 9 d ^ { 2 } } { 64 - d ^ { 2 } } = \frac { 7 } { 15 } \Rightarrow{/tex}128d2 = 512 {tex}\Rightarrow{/tex} d2 = 4 {tex}\Rightarrow{/tex}d = ± 2
Thus, the four parts are a - 3d, a - d, a + d and a + 3d i.e., 2,6,10,14.
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