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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 $12$.
Updated on: 10-Oct-2022
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