If in A.P., 3rd term is 18 and 7th term is 30, the sum of its 17 terms is

Options

A. 600
B. 612
C. 624
D. none of these

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (A.P.).

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \).

### Step 2: Write the expressions for the 3rd and 7th terms.
The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
So, for the 3rd term (\( T_3 \)):
\[ T_3 = a + 2d = 18 \quad \text{(Equation 1)} \]
And for the 7th term (\( T_7 \)):
\[ T_7 = a + 6d = 30 \quad \text{(Equation 2)} \]

### Step 3: Set up the equations.
From Equation 1:
\[ a + 2d = 18 \]
From Equation 2:
\[ a + 6d = 30 \]

### Step 4: Subtract Equation 1 from Equation 2.
Subtracting the first equation from the second gives:
\[ (a + 6d) - (a + 2d) = 30 - 18 \]
This simplifies to:
\[ 4d = 12 \]

### Step 5: Solve for \( d \).
Dividing both sides by 4:
\[ d = 3 \]

### Step 6: Substitute \( d \) back into one of the equations to find \( a \).
Using Equation 1:
\[ a + 2(3) = 18 \]
This simplifies to:
\[ a + 6 = 18 \]
So,
\[ a = 12 \]

### Step 7: Calculate the sum of the first 17 terms.
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 17 \):
\[ S_{17} = \frac{17}{2} \times (2(12) + (17-1)(3)) \]
Calculating the terms inside the parentheses:
\[ S_{17} = \frac{17}{2} \times (24 + 48) = \frac{17}{2} \times 72 \]
Now, calculating:
\[ S_{17} = \frac{17 \times 72}{2} = 17 \times 36 = 612 \]

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

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