
Split 207 into three parts such that they are in A.P. and the product of the two smaller parts is 4623.
Answer
616.5k+ views
Hint: Assume the three numbers to be equidistant from one another. Suppose, if the middle number is ‘a’, then assume the extreme numbers to be (a – d) and (a + d), where d is the common difference, since they are in arithmetic progression. Now, there are two conditions given in the question. Generate two simultaneous equations from them and solve for the values of ‘a’ and ‘d’, which are the unknown quantities. Thus, obtain the three parts.
Complete step-by-step solution -
Let us assume that among the three parts that 207 is split, the middlemost number is ‘a’.
Since, the three numbers are in A.P., we can easily assume the smaller number be (a – d) and the larger number be (a + d), where ‘d’ is the common difference of the AP.
Now, according to the first condition, the sum of the three numbers is equal to 207.
Thus,
$\left( \text{a }-\text{ d} \right)\text{ + a + }\left( \text{a + d} \right)\text{ = 207 }....\text{(i)}$
Again, according to the second condition, it is said that the product of two smaller numbers is 4623.
Therefore,
$\left( \text{a }-\text{ d} \right)\cdot \text{a = 4623 }....\left( \text{ii} \right)$
Now, simplifying the first equation (i), we get,
$\begin{align}
& \left( \text{a }-\text{ d} \right)\text{ + a + }\left( \text{a + d} \right)\text{ = 207} \\
& \Rightarrow \text{ 3a = 207} \\
& \therefore \text{ a = 69} \\
\end{align}$
Hence, we get the value of the middle value to be 69.
Putting the value of a = 69 in equation (ii), we get,
$\begin{align}
& \left( 69\text{ }-\text{ d} \right)\cdot 69\text{ = 4623} \\
& \Rightarrow \text{ }\left( 69-\text{d} \right)\text{ = }\dfrac{4623}{69} \\
& \Rightarrow \text{ }\left( 69-\text{d} \right)\text{ = 67} \\
& \therefore \text{ d = 69 }-\text{ 67 = 2} \\
\end{align}$
Therefore, we get the value of the common difference of the A.P. equal to 2.
Hence, the value of the smaller part = (69 – 2) = 67
The value of the larger part = (69 + 2) = 71
Thus, the three parts into which 207 is split are 67, 69 and 71.
Note: Observe that, if three numbers are in A.P., then the sum is always equal to thrice the value of the middle number. This also happens when an odd number of terms are in A.P, then their sum is always equal to the value of the (no. of terms x middlemost value).
Complete step-by-step solution -
Let us assume that among the three parts that 207 is split, the middlemost number is ‘a’.
Since, the three numbers are in A.P., we can easily assume the smaller number be (a – d) and the larger number be (a + d), where ‘d’ is the common difference of the AP.
Now, according to the first condition, the sum of the three numbers is equal to 207.
Thus,
$\left( \text{a }-\text{ d} \right)\text{ + a + }\left( \text{a + d} \right)\text{ = 207 }....\text{(i)}$
Again, according to the second condition, it is said that the product of two smaller numbers is 4623.
Therefore,
$\left( \text{a }-\text{ d} \right)\cdot \text{a = 4623 }....\left( \text{ii} \right)$
Now, simplifying the first equation (i), we get,
$\begin{align}
& \left( \text{a }-\text{ d} \right)\text{ + a + }\left( \text{a + d} \right)\text{ = 207} \\
& \Rightarrow \text{ 3a = 207} \\
& \therefore \text{ a = 69} \\
\end{align}$
Hence, we get the value of the middle value to be 69.
Putting the value of a = 69 in equation (ii), we get,
$\begin{align}
& \left( 69\text{ }-\text{ d} \right)\cdot 69\text{ = 4623} \\
& \Rightarrow \text{ }\left( 69-\text{d} \right)\text{ = }\dfrac{4623}{69} \\
& \Rightarrow \text{ }\left( 69-\text{d} \right)\text{ = 67} \\
& \therefore \text{ d = 69 }-\text{ 67 = 2} \\
\end{align}$
Therefore, we get the value of the common difference of the A.P. equal to 2.
Hence, the value of the smaller part = (69 – 2) = 67
The value of the larger part = (69 + 2) = 71
Thus, the three parts into which 207 is split are 67, 69 and 71.
Note: Observe that, if three numbers are in A.P., then the sum is always equal to thrice the value of the middle number. This also happens when an odd number of terms are in A.P, then their sum is always equal to the value of the (no. of terms x middlemost value).
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