In an A.P:
given a = 2 , d = 8 , `S_(n)= 90`, find n and `a_(n)`.

Answer

To solve the problem, we will use the formulas related to Arithmetic Progressions (A.P.).

### Given:
- First term \( a = 2 \)
- Common difference \( d = 8 \)
- Sum of the first \( n \) terms \( S_n = 90 \)

### Step 1: Use the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]

### Step 2: Substitute the known values into the formula.
Substituting \( S_n = 90 \), \( a = 2 \), and \( d = 8 \) into the formula, we get:
\[ 90 = \frac{n}{2} \times (2 \times 2 + (n - 1) \times 8) \]

### Step 3: Simplify the equation.
This simplifies to:
\[ 90 = \frac{n}{2} \times (4 + 8n - 8) \]
\[ 90 = \frac{n}{2} \times (8n - 4) \]
\[ 90 = \frac{n(8n - 4)}{2} \]
\[ 180 = n(8n - 4) \]

### Step 4: Rearrange the equation.
Rearranging gives us:
\[ 180 = 8n^2 - 4n \]
\[ 8n^2 - 4n - 180 = 0 \]

### Step 5: Simplify the quadratic equation.
Dividing the entire equation by 4:
\[ 2n^2 - n - 45 = 0 \]

### Step 6: Factor the quadratic equation.
Now we need to factor the quadratic equation. We look for two numbers that multiply to \( 2 \times -45 = -90 \) and add to \( -1 \). The numbers are \( -10 \) and \( 9 \).
\[ 2n^2 - 10n + 9n - 45 = 0 \]
\[ (2n^2 - 10n) + (9n - 45) = 0 \]
\[ 2n(n - 5) + 9(n - 5) = 0 \]
\[ (n - 5)(2n + 9) = 0 \]

### Step 7: Solve for \( n \).
Setting each factor to zero gives:
1. \( n - 5 = 0 \) → \( n = 5 \)
2. \( 2n + 9 = 0 \) → \( n = -\frac{9}{2} \) (not valid since \( n \) must be a positive integer)

Thus, \( n = 5 \).

### Step 8: Find \( a_n \) (the nth term).
The \( n \)-th term of an A.P. is given by:
\[ a_n = a + (n - 1)d \]
Substituting \( a = 2 \), \( n = 5 \), and \( d = 8 \):
\[ a_n = 2 + (5 - 1) \times 8 \]
\[ a_n = 2 + 4 \times 8 \]
\[ a_n = 2 + 32 \]
\[ a_n = 34 \]

### Final Answer:
- \( n = 5 \)
- \( a_n = 34 \)

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