If in A.P., 3rd term is 18 and 7 term is 30, then find sum of its first 17 terms

Answer

To solve the problem step by step, we will first identify the terms of the arithmetic progression (A.P.) using the information given about the 3rd and 7th terms, and then we will calculate the sum of the first 17 terms.

### Step 1: Write the formulas for the terms of A.P.
In an A.P., the nth term can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up equations based on the given terms.
From the problem, we know:
- The 3rd term \( a_3 = 18 \)
- The 7th term \( a_7 = 30 \)

Using the formula for the nth term:
1. For the 3rd term:
\[ a + 2d = 18 \] (Equation 1)

2. For the 7th term:
\[ a + 6d = 30 \] (Equation 2)

### Step 3: Solve the equations simultaneously.
We have two equations:
1. \( a + 2d = 18 \)
2. \( a + 6d = 30 \)

We can subtract Equation 1 from Equation 2 to eliminate \( a \):
\[ (a + 6d) - (a + 2d) = 30 - 18 \]
This simplifies to:
\[ 4d = 12 \]
Dividing both sides by 4 gives:
\[ d = 3 \]

### Step 4: Substitute \( d \) back to find \( a \).
Now that we have \( d \), we can substitute it back into Equation 1:
\[ a + 2(3) = 18 \]
This simplifies to:
\[ a + 6 = 18 \]
Subtracting 6 from both sides gives:
\[ a = 12 \]

### Step 5: Calculate the sum of the first 17 terms.
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 17 \):
\[ S_{17} = \frac{17}{2} \times (2(12) + (17-1)(3)) \]
Calculating inside the parentheses:
\[ S_{17} = \frac{17}{2} \times (24 + 48) \]
\[ S_{17} = \frac{17}{2} \times 72 \]
Now, calculating further:
\[ S_{17} = 17 \times 36 = 612 \]

### Final Answer:
The sum of the first 17 terms is \( S_{17} = 612 \).

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