Question
Find three numbers in A.P whose sum is $24$ and whoseproduct is $440$
✓
Answer
Let the three number in AP. Be a-d, a and $a+d$.
$ \therefore(a-d)+a+(a+d)=24$
$\Rightarrow 3 a=24$
$\Rightarrow a=8 \ldots \ldots \ldots \ldots . .(1)$
$\text { Also, }(a-d) \times a \times(a+d)=440$
$\Rightarrow\left(a^2-d^2\right) \times a=440$
$\Rightarrow\left(8^2-d^2\right) \times 8=440 \ldots \ldots . . \ldots \ldots . . .[\text { from (1)] }$
$\Rightarrow 64-d^2=55$
$\Rightarrow d^2=9$
$\Rightarrow d= \pm 3$
When $a=8$ and $d=3$
Required terms $=a-d, a$ and $a+d$
$=8-3,8,8+3 $
$ =5,8,11$
When $a=8$ and $d=-3$
$ \begin{aligned} \text { Required term } & =a-d, a \text { and } a+d \\ & =8-(-3), 8,8+(-3) \\ & =11,8,5 \end{aligned} $
$ \therefore(a-d)+a+(a+d)=24$
$\Rightarrow 3 a=24$
$\Rightarrow a=8 \ldots \ldots \ldots \ldots . .(1)$
$\text { Also, }(a-d) \times a \times(a+d)=440$
$\Rightarrow\left(a^2-d^2\right) \times a=440$
$\Rightarrow\left(8^2-d^2\right) \times 8=440 \ldots \ldots . . \ldots \ldots . . .[\text { from (1)] }$
$\Rightarrow 64-d^2=55$
$\Rightarrow d^2=9$
$\Rightarrow d= \pm 3$
When $a=8$ and $d=3$
Required terms $=a-d, a$ and $a+d$
$=8-3,8,8+3 $
$ =5,8,11$
When $a=8$ and $d=-3$
$ \begin{aligned} \text { Required term } & =a-d, a \text { and } a+d \\ & =8-(-3), 8,8+(-3) \\ & =11,8,5 \end{aligned} $
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