An `A.P.` consist of even number of terms `2n` having middle terms equal to `1` and `7` respectively. If `n` is the maximum value which satisfy `t_(1)t_(2n)+713 ge 0`, then the value of the first term of the series is

Options

A. (a) `17`
B. (b) `-15`
C. (c) `21`
D. (d) `-23`

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a clear mathematical approach.

### Step 1: Understand the Problem
We have an arithmetic progression (A.P.) with an even number of terms \(2n\). The middle terms are given as \(T_n = 1\) and \(T_{n+1} = 7\).

### Step 2: Find the Common Difference
In an A.P., the middle terms can be used to find the common difference \(d\). The common difference can be calculated as:
\[ d = T_{n+1} - T_n = 7 - 1 = 6 \]

### Step 3: Express the Last Term
The last term \(T_{2n}\) can be expressed in terms of the first term \(T_1\) and the common difference \(d\):
\[ T_{2n} = T_1 + (2n - 1) \cdot d = T_1 + (2n - 1) \cdot 6 \]

### Step 4: Set Up the Equation
We know the two middle terms, so we can set up the equation:
\[ T_n + T_{n+1} = 1 + 7 = 8 \]
This implies that:
\[ T_1 + T_{2n} = 8 \]
Substituting \(T_{2n}\) from the previous equation:
\[ T_1 + (T_1 + (2n - 1) \cdot 6) = 8 \]
Simplifying this gives:
\[ 2T_1 + (2n - 1) \cdot 6 = 8 \]

### Step 5: Solve for \(T_1\)
Rearranging the equation:
\[ 2T_1 + 12n - 6 = 8 \]
\[ 2T_1 = 14 - 12n \]
\[ T_1 = 7 - 6n \]

### Step 6: Use the Given Inequality
We are given the inequality:
\[ T_1 \cdot T_{2n} + 713 \geq 0 \]
Substituting \(T_1\) and \(T_{2n}\):
\[ (7 - 6n) \cdot (T_1 + (2n - 1) \cdot 6) + 713 \geq 0 \]
Substituting \(T_{2n}\):
\[ (7 - 6n) \cdot (7 - 6n + 12n - 6) + 713 \geq 0 \]
This simplifies to:
\[ (7 - 6n)(6n + 1) + 713 \geq 0 \]

### Step 7: Expand and Rearrange
Expanding the left side:
\[ 42n + 7 - 36n^2 - 6n + 713 \geq 0 \]
Combining like terms:
\[ -36n^2 + 36n + 720 \geq 0 \]
Dividing through by -1 (and flipping the inequality):
\[ 36n^2 - 36n - 720 \leq 0 \]
Dividing by 36:
\[ n^2 - n - 20 \leq 0 \]

### Step 8: Factor the Quadratic
Factoring:
\[ (n - 5)(n + 4) \leq 0 \]

### Step 9: Solve the Inequality
The critical points are \(n = 5\) and \(n = -4\). The solution to the inequality is:
\[ -4 \leq n \leq 5 \]
Thus, the maximum value of \(n\) is \(5\).

### Step 10: Find the First Term
Substituting \(n = 5\) back into the expression for \(T_1\):
\[ T_1 = 7 - 6 \cdot 5 = 7 - 30 = -23 \]

### Final Answer
The value of the first term of the series is:
\[ \boxed{-23} \]