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Sia ? 6 years, 5 months ago
Let the required numbers be (a - d), a, (a + d)
Sum of these numbers = (a - d) + a + (a + d) = 3a
{tex}\therefore{/tex}sum of these squares = (a-d)2+a2+(a+d)2=3a2+2d2
According to the question,we are given that,
Sum of three numbers = 21
{tex}\therefore{/tex}3a = 21
a = 7
Also,sum of the squares of three numbers=165
{tex}\Rightarrow{/tex}3a2 + 2d2=165
{tex}\Rightarrow{/tex}3(7)2+2d2=165
{tex}\Rightarrow{/tex}2d2 =18
{tex}\Rightarrow{/tex}d2=9
{tex}\Rightarrow{/tex}d={tex} \pm {/tex}3
Thus, a=7 and d={tex} \pm {/tex}3
Hence, the required numbers are (4, 7, 10) or (10, 7, 4)
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Sia ? 6 years, 5 months ago
the additive inverse of a number is the number that, when added to a number , yields zero.
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Prince Bhatter 6 years, 5 months ago
0Thank You