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.
If (a + b), 2b, (b + c) are in HP, then which one of the following is correct?
What is the value of ab?
What is the value of xyz?
What is the value of pqr?
Which one of the following is correct?
x, y and z are