For the following two (02) items : Let $(6+10+14 + ... \text{up to } m \text{ terms})$ $=(1+3+5+7+ ... \text{up to } n \text{ terms})$ where $m < 25$ and $n < 25$.
\(n^2 = 2m (m + 2)\)
To determine the relationship between \( m \) and \( n \), we need to analyze the given series for both sides of the equation:
The series \( 6 + 10 + 14 + \ldots \) to \( m \) terms is an arithmetic progression (AP) with the first term \( a = 6 \) and common difference \( d = 4 \). The sum of this AP, \( S_m \), is given by:
\(S_m = \frac{m}{2} \times (2a + (m-1)d)\)
Substituting the values:
\(S_m = \frac{m}{2} \times (2 \times 6 + (m - 1) \times 4)\) \(S_m = \frac{m}{2} \times (12 + 4m - 4) = \frac{m}{2} \times (4m + 8) = 2m(m + 2)\)
The series \( 1 + 3 + 5 + \ldots \) to \( n \) terms is another AP with the first term \( a' = 1 \) and common difference \( d' = 2 \). The sum of this AP, \( S_n \), is:
\(S_n = \frac{n}{2} \times (2a' + (n-1)d')\)
Substituting the values:
\(S_n = \frac{n}{2} \times (2 \times 1 + (n - 1) \times 2) = \frac{n}{2} \times (2 + 2n - 2) = \frac{n}{2} \times 2n = n^2\)
According to the problem statement, these two sums are equal:
\(2m(m + 2) = n^2\)
This is the desired relationship between \( m \) and \( n \), which matches the correct option.
Thus, the correct answer is:
\( n^2 = 2m(m + 2) \)
Let's verify the correctness by checking the boundary conditions (i.e., as both \( m \) and \( n \) are less than 25) to ensure that the derived formula holds true across valid values.
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