Find 4 numbers in AP such …

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Sia ? 6 years, 6 months ago
Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d)
Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)
According to the question, sum of the numbers=32
{tex}\therefore{/tex}4a = 32 {tex}\Rightarrow{/tex} a = 8
Sum of the squares of these numbers
=(a-3d)2+(a-d)2+(a+d)2+(a+3d)2=4(a2+5d{tex}^2{/tex})
Now, sum of the squares of numbers=336
{tex}\therefore{/tex}4(a2+5d2)=336
{tex}\Rightarrow{/tex}a2+5d2=84 [{tex}\because {/tex}a=8]
{tex}\Rightarrow{/tex}5d2= 84-64
{tex}\Rightarrow{/tex}5d2=20
{tex}\Rightarrow{/tex}d2=4
{tex}\Rightarrow{/tex}d={tex} \pm {/tex}2
Hence, the required numbers (2, 6, 10, 14).
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