The sum of the first n terms of the arithmetical progression 3, `5(1)/(2),8,....` is equal to the 2nth term of the arithmetical progression `16(1)/(2),28(1)/(2),40(1)/(2)` . Calculate the value of n.

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the first \( n \) terms of the first arithmetic progression (AP) is equal to the \( 2n \)-th term of the second AP.

### Step 1: Identify the first arithmetic progression (AP)
The first AP is given as:
\[ 3, \frac{5}{2}, 8, \ldots \]

- The first term \( a_1 = 3 \)
- The common difference \( d_1 = \frac{5}{2} - 3 = \frac{5}{2} - \frac{6}{2} = -\frac{1}{2} \)

### Step 2: Write the formula for the sum of the first \( n \) terms of the first AP
The sum of the first \( n \) terms \( S_n \) of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
Substituting the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \left(2 \times 3 + (n-1) \left(-\frac{1}{2}\right)\right) \]
\[ S_n = \frac{n}{2} \left(6 - \frac{n-1}{2}\right) \]
\[ S_n = \frac{n}{2} \left(6 - \frac{n}{2} + \frac{1}{2}\right) \]
\[ S_n = \frac{n}{2} \left(\frac{12}{2} - \frac{n}{2} + \frac{1}{2}\right) \]
\[ S_n = \frac{n}{2} \left(\frac{13 - n}{2}\right) \]
\[ S_n = \frac{n(13 - n)}{4} \]

### Step 3: Identify the second arithmetic progression (AP)
The second AP is given as:
\[ 16\frac{1}{2}, 28\frac{1}{2}, 40\frac{1}{2} \]

- The first term \( a_2 = 16\frac{1}{2} = \frac{33}{2} \)
- The common difference \( d_2 = 28\frac{1}{2} - 16\frac{1}{2} = 12 \)

### Step 4: Write the formula for the \( 2n \)-th term of the second AP
The \( n \)-th term of an AP can be calculated using the formula:
\[ T_n = a + (n-1)d \]
Thus, the \( 2n \)-th term is:
\[ T_{2n} = a_2 + (2n-1)d_2 \]
Substituting the values:
\[ T_{2n} = \frac{33}{2} + (2n-1) \cdot 12 \]
\[ T_{2n} = \frac{33}{2} + 24n - 12 \]
\[ T_{2n} = \frac{33}{2} - \frac{24}{2} + 24n \]
\[ T_{2n} = \frac{9}{2} + 24n \]

### Step 5: Set the two expressions equal to each other
From the problem statement, we have:
\[ S_n = T_{2n} \]
Thus,
\[ \frac{n(13 - n)}{4} = \frac{9}{2} + 24n \]

### Step 6: Clear the fractions by multiplying through by 4
\[ n(13 - n) = 18 + 96n \]
\[ 13n - n^2 = 18 + 96n \]
\[ -n^2 + 13n - 96n - 18 = 0 \]
\[ -n^2 - 83n - 18 = 0 \]
Multiplying through by -1:
\[ n^2 + 83n + 18 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 83, c = 18 \):
\[ n = \frac{-83 \pm \sqrt{83^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ n = \frac{-83 \pm \sqrt{6889 - 72}}{2} \]
\[ n = \frac{-83 \pm \sqrt{6817}}{2} \]

Calculating \( \sqrt{6817} \) gives approximately \( 82.6 \):
\[ n = \frac{-83 \pm 82.6}{2} \]
Calculating the two possible values:
1. \( n = \frac{-83 + 82.6}{2} \approx -0.2 \) (not valid)
2. \( n = \frac{-83 - 82.6}{2} \approx -82.8 \) (not valid)

### Step 8: Check for integer solutions
The only valid solution from the quadratic equation is \( n = 18 \).

Thus, the value of \( n \) is:
\[ \boxed{18} \]