Question

Divide 24 in three parts such that they are in AP and their product is 440.

Answer 1

Hint:
As given in the question the three parts are in AP then let the three parts of 24 is $a - d, a, a + d$. Then submission of all three parts is equal to 24 from here we will calculate a. then the product of three parts is 440 from this we will calculate d.

Complete step by step solution:
Given three parts of 24 is in AP then let us assume three numbers to be $a - d, a, a + d$
So, $a - d + a + a + d = 24$
Therefore $a = 8$
Now, product of all three parts are 440 then
$\left( {a - d} \right) \times \left( a \right) \times \left( {a + d} \right) = 440$
Substituting $a = 8$
$\left( {8 - d} \right) \times \left( 8 \right) \times \left( {8 + d} \right) = 440$
We know that ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
\[\left( {{8^2} - {d^2}} \right) = 55\]
${d^2} = 64 - 55 = 9$
Taking root on both side we get
$d = 3$

So, three parts of 24 are 5, 8, 11.

Note:
For consecutive three terms to be in AP $a - d,a,a + d$
For 4 consecutive terms in AP =$a - 2d,a - d,a + d,a + 2d$ and so one …
Where a is the 1st term of AP and d is the common difference