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# Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.

Given:

Three numbers are in A.P.

The sum of these numbers is 27 and the product is 648.

To do:

We have to find the numbers.

Solution:

Let the first three terms of the AP be $a−d,\ a,\ a+d$.

According to the problem,

$a−d+a+a+d=27\ ......( i)$

$( a−d)(a)( a+d)=648\ .....(ii)$

From $(i)$, we get

$3a=27$

$\Rightarrow a=\frac{27}{3}=9$

From $(ii)$, we get

$a(a^2−d^2)=648\ .....(iii)$

On putting $a=9$ in equation $(iii)$, we get,

$9(9^2−d^2)=648$

$\Rightarrow 81−d^2=\frac{648}{9}$

$\Rightarrow 81-72=d^2$

$\Rightarrow d^2=9$

$\Rightarrow d=3$

This implies,

$a-d=9-3=6, a=9, a+d=9+3=12, a+2d=9+2(3)=9+6=15$

The required A.P is $6, 9, 12, 15......$

The first three terms of the A.P. are $6, 9$ and $12$.

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