Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

Answer

To find the sum of the first 51 terms of an arithmetic progression (A.P.) where the second term (T2) is 14 and the third term (T3) is 18, we can follow these steps:

### Step 1: Identify the terms of the A.P.
In an A.P., the nth term can be expressed as:
- T2 = A + D (where A is the first term and D is the common difference)
- T3 = A + 2D

Given:
- T2 = 14
- T3 = 18

### Step 2: Set up the equations
From the information provided:
1. A + D = 14 (Equation 1)
2. A + 2D = 18 (Equation 2)

### Step 3: Solve for D
Subtract Equation 1 from Equation 2:
(A + 2D) - (A + D) = 18 - 14
This simplifies to:
D = 4

### Step 4: Solve for A
Now that we have D, substitute it back into Equation 1 to find A:
A + 4 = 14
A = 14 - 4
A = 10

### Step 5: Calculate the sum of the first 51 terms
The formula for the sum of the first n terms (S_n) of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]

For our case:
- n = 51
- A = 10
- D = 4

Substituting these values into the formula:
\[ S_{51} = \frac{51}{2} \times (2 \times 10 + (51 - 1) \times 4) \]
\[ S_{51} = \frac{51}{2} \times (20 + 50 \times 4) \]
\[ S_{51} = \frac{51}{2} \times (20 + 200) \]
\[ S_{51} = \frac{51}{2} \times 220 \]
\[ S_{51} = 51 \times 110 \]
\[ S_{51} = 5610 \]

### Final Answer:
The sum of the first 51 terms of the A.P. is **5610**.

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