Three numbers are in A.P. If the sum of these numbers be

In the given problem, the sum of three terms of an A.P is 27 and the product of the three terms is 648. We need to find the three terms.

Here,

Let the three terms be where a is the first term and d is the common difference of the A.P

So,

$(a-d)+a+(a+d)=27$

$3 a=27$

$a=9$.......(1)

Also,

$(a-d) a(a+d)=a+6$

$a\left(a^{2}-d^{2}\right)=648$ $\left[\right.$ Using $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$

$9\left(9^{2}-d^{2}\right)=648$

$81-d^{2}=72$

Further solving for d,

$81-d^{2}=72$

$81-72=d^{2}$

$d=\sqrt{9}$

$d=3 \ldots \ldots(2)$

Now, substituting (1) and (2) in three terms

First term $=a-d$

So,

$a-d=9-3$

$=6$

Also,

Second term = a

So,

$a=9$

Also,

Third term $=a+d$

So,

$a+d=9+3$

$=12$

Therefore, the three terms are 6,9 and 12 .