Find the nth term of an A.P. whose 6th and 8th terms are 12 and 22 respectively :

Options

A. 7n-30
B. `n^2 - 24`
C. `5n-18`
D. none of these

Answer

To find the nth term of an arithmetic progression (A.P.) where the 6th term (T6) is 12 and the 8th term (T8) is 22, we can follow these steps:

### Step 1: Write the formulas for the 6th and 8th terms
The nth term of an A.P. can be expressed as:
\[ T_n = A + (n-1)D \]
where \( A \) is the first term and \( D \) is the common difference.

For the 6th term:
\[ T_6 = A + 5D = 12 \]
(Equation 1)

For the 8th term:
\[ T_8 = A + 7D = 22 \]
(Equation 2)

### Step 2: Set up the equations
From Equation 1:
\[ A + 5D = 12 \]

From Equation 2:
\[ A + 7D = 22 \]

### Step 3: Subtract Equation 1 from Equation 2
To eliminate \( A \), we can subtract Equation 1 from Equation 2:
\[ (A + 7D) - (A + 5D) = 22 - 12 \]
This simplifies to:
\[ 2D = 10 \]

### Step 4: Solve for D
Now, divide both sides by 2:
\[ D = 5 \]

### Step 5: Substitute D back into one of the equations to find A
We can substitute \( D = 5 \) back into Equation 1:
\[ A + 5(5) = 12 \]
This simplifies to:
\[ A + 25 = 12 \]
Now, subtract 25 from both sides:
\[ A = 12 - 25 \]
\[ A = -13 \]

### Step 6: Write the formula for the nth term
Now that we have both \( A \) and \( D \), we can write the formula for the nth term:
\[ T_n = A + (n-1)D \]
Substituting \( A = -13 \) and \( D = 5 \):
\[ T_n = -13 + (n-1) \cdot 5 \]
Expanding this gives:
\[ T_n = -13 + 5n - 5 \]
Combining like terms:
\[ T_n = 5n - 18 \]

### Final Answer
Thus, the nth term of the A.P. is:
\[ T_n = 5n - 18 \]