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

Last updated date: 13th Jun 2024
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Hint:To find the numbers in Arithmetic progression we divide the three terms of \[a-d,a,a+d\] and equate with the value of \[24\] and then we find the value of \[a\] and after finding the value of \[a\], we then find the value of \[d\] which is found by the product of the three numbers which is equivalent to \[440\].

Complete step by step solution:
Let us take the three numbers that are given in the question as \[a-d,a,a+d\] which is divided to form the total value of \[24\]. So let us find the equation by placing the values equivalent to \[24\] and then find the value of \[a\]. Placing the equation as:
\[\Rightarrow \left( a-d \right)+a+\left( a+d \right)=24\]
Removing the brackets, we get the value of the equation as:

\[\Rightarrow a-d+a+a+d=24\]
\[\Rightarrow a+a+a=24\]
\[\Rightarrow 3a=24\]
\[\Rightarrow a=8\]
Now that after placing the numbers together we have gotten the value of \[a\] after placing it with the value of \[24\], we now move to find the value of \[d\] by multiplying the three terms and equating it with \[440\], we get:
\[\Rightarrow a-d\times a\times a+d=440\]
\[\Rightarrow a\left( {{a}^{2}}-{{d}^{2}} \right)=440\]
Now placing the value of \[a=8\], we get the equation as:
\[\Rightarrow 8\left( {{8}^{2}}-{{d}^{2}} \right)=440\]
\[\Rightarrow d=\sqrt{9}\]
\[\Rightarrow d=\pm 3\]
Now that we know the value of \[a\] and \[d\], placing the values in the three terms as \[a-d,a,a+d\], we get the value of the three terms as:
\[\Rightarrow a-d,a,a+d\]
We have taken the value of \[d=3\] as either can be taken.
\[\Rightarrow 8-3,8,8+3\]
\[\Rightarrow 5,8,11\]
Therefore, the value of the three numbers are given as, \[5,8,11\].

The terms in arithmetic progression is written as: \[...(a-2),(a-1),a,(a+1),(a+2)...\] where as for geometric progression it is written as \[ar,a{{r}^{2}},a{{r}^{3}},....\] as in geometric progression, the terms move exponentially.