Question

The sum of three number in A.P is 9 and sum of their square is 59. Find the number

Answer 1

$\bold{Given,}$

$\mathsf{The \: sum \: of \: three \: number \: is \: 9}$

$\mathsf{Let \: the \: three \: numbers \: be}$

$\mathsf{(a+d),a \: and \: (a-d)}$

So,

$\mathsf{(a+d)+a +(a-d) =9}$

$\mathsf{a+d+a +a-d) =9}$

$\mathsf{3a =9}$

$\mathsf{a = \frac{9}{3}}$

$\huge \boxed{a = 3}$

$\bold{Given,}$

$\bold{The \: sum \: of \: square \: three \: number \: is \:59}$

$\mathsf{{(a + d)}^{2} + {a}^{2} + {(a - d)}^{2} = 59}$

${{(9+d)}^{2}+{9}^{2} +{(9-d)}^{2}=59}$

We know,

$\boxed{{(a+b)}^{2}= {a}^{2} + {b}^{2} + 2ab}$

$\boxed{{(a+b)}^{2}= {a}^{2} + {b}^{2} -2ab}$

So,

$\mathsf{{a}^{2} + {d}^{2} + 2ad + {a}^{2} + {a}^{2} + {d}^{2} -2ad= 59}$

Since,

$\huge \boxed{\mathsf{a = 3}}$

$\mathsf{{3}^{2} + {d}^{2} + 6d + {3}^{2} + {3}^{2} + {d}^{2} -6 d= 59}$

$\mathsf{9 + {d}^{2} + \cancel{6d} + 9+ 9+ {d}^{2} - \cancel{6d}= 59}$

$\mathsf{9 + 9+ 9+ 2 {d}^{2} =59}$

$\mathsf{27+ 2 {d}^{2} =59}$

$\mathsf{ 2 {d}^{2} =59-27}$

$\mathsf{ 2 {d}^{2} =32}$

$\mathsf{ {d}^{2} =\frac{32}{2}}$

$\mathsf{{d}^{2} = 16}$

$\mathsf{d = \sqrt{16}}$

$\boxed{ \mathsf{d= 4}}$

So, the numbers are,

$\boxed{ \mathsf{a+d = 3+4 =7}}$

$\boxed{ \mathsf{a-d = 3-4 =-1}}$

$\boxed{ \mathsf{a = 3}}$